Applications of the Derivative

Tangent and Normal Lines

In calculus, the derivative tells us more than just a rate of change. It also helps us find the slope of a curve at a point. This lets us write the equation of the tangent line, which touches the curve at that point, and the normal line, which is perpendicular to the tangent line.

These ideas are important in graph sketching, motion, approximation, and many applications of derivatives. If you know the point on the curve and the derivative at that point, you can build the exact equation of both lines.

1. What is a tangent line?

A tangent line to a curve at a given point is the line that has the same direction as the curve at that point. In calculus, that direction is measured by the derivative.

If a curve is given by \(y=f(x)\), then the slope of the tangent line at \(x=a\) is

$$m_{\text{tangent}}=f'(a).$$

If the point of tangency is \((a,f(a))\), then the tangent line is found using the point-slope form of a line:

$$y-f(a)=f'(a)(x-a).$$

2. What is a normal line?

A normal line is a line that is perpendicular to the tangent line at the same point.

Recall that for two non-vertical perpendicular lines, their slopes are negative reciprocals. So if the tangent slope is \(m\), then the normal slope is

$$m_{\text{normal}}=-\frac{1}{m}.$$

That means if \(m_{\text{tangent}}=f'(a)\), then

$$m_{\text{normal}}=-\frac{1}{f'(a)},$$

as long as \(f'(a)\neq 0\).

The normal line through \((a,f(a))\) is then

$$y-f(a)=\left(-\frac{1}{f'(a)}\right)(x-a).$$

3. Step-by-step method

To find the tangent and normal lines to \(y=f(x)\) at \(x=a\), follow these steps:

1. Find the point on the curve by calculating \(f(a)\). So the point is \((a,f(a))\).

2. Differentiate the function to get \(f'(x)\).

3. Substitute \(x=a\) into \(f'(x)\) to get the tangent slope: \(f'(a)\).

4. Use point-slope form to write the tangent line.

5. Find the normal slope using the negative reciprocal of the tangent slope.

6. Use point-slope form again to write the normal line.

4. Important special cases

• If \(f'(a)=0\), then the tangent line is horizontal.

• A horizontal tangent has slope \(0\), so the normal line is vertical.

• A vertical line cannot be written in slope-intercept form. Instead, write it as \(x=a\).

So if \(f'(a)=0\), then:

• Tangent line: \(y=f(a)\)

• Normal line: \(x=a\)

5. Worked Example 1: Basic polynomial

Find the equations of the tangent line and normal line to \(y=x^2+3x\) at \(x=1\).

Step 1: Find the point on the curve.

$$f(1)=1^2+3(1)=4$$

So the point is \((1,4)\).

Step 2: Differentiate.

$$f'(x)=2x+3$$

Step 3: Find the tangent slope at \(x=1\).

$$f'(1)=2(1)+3=5$$

So the tangent slope is \(5\).

Step 4: Write the tangent line.

$$y-4=5(x-1)$$

Simplifying:

$$y-4=5x-5$$ $$y=5x-1$$

Step 5: Find the normal slope.

$$m_{\text{normal}}=-\frac{1}{5}$$

Step 6: Write the normal line.

$$y-4=-\frac{1}{5}(x-1)$$

This is the equation of the normal line. In slope-intercept form, it becomes

$$y=-\frac{1}{5}x+\frac{1}{5}+4$$ $$y=-\frac{1}{5}x+\frac{21}{5}$$

Answer:

• Tangent line: \(y=5x-1\)

• Normal line: \(y-4=-\frac{1}{5}(x-1)\)

6. Worked Example 2: A function with a fraction

Find the tangent line and normal line to \(f(x)=\frac{2}{x}\) at \(x=1\).

Step 1: Find the point.

$$f(1)=\frac{2}{1}=2$$

So the point is \((1,2)\).

Step 2: Differentiate.

Rewrite first:

$$f(x)=2x^{-1}$$

Now differentiate:

$$f'(x)=-2x^{-2}=-\frac{2}{x^2}$$

Step 3: Find the tangent slope.

$$f'(1)=-\frac{2}{1^2}=-2$$

Step 4: Tangent line.

$$y-2=-2(x-1)$$

Simplify:

$$y-2=-2x+2$$ $$y=-2x+4$$

Step 5: Normal slope.

The negative reciprocal of \(-2\) is \(\frac{1}{2}\).

$$m_{\text{normal}}=\frac{1}{2}$$

Step 6: Normal line.

$$y-2=\frac{1}{2}(x-1)$$

Answer:

• Tangent line: \(y=-2x+4\)

• Normal line: \(y-2=\frac{1}{2}(x-1)\)

7. Worked Example 3: Horizontal tangent and vertical normal

Find the tangent line and normal line to \(y=x^3-3x+2\) at \(x=1\).

Step 1: Find the point.

$$f(1)=1^3-3(1)+2=0$$

So the point is \((1,0)\).

Step 2: Differentiate.

$$f'(x)=3x^2-3$$

Step 3: Find the tangent slope.

$$f'(1)=3(1)^2-3=0$$

This means the tangent line is horizontal.

Step 4: Tangent line.

A horizontal line through \((1,0)\) is

$$y=0$$

Step 5: Normal line.

The normal line is perpendicular to a horizontal line, so it is vertical.

A vertical line through \((1,0)\) is

$$x=1$$

Answer:

• Tangent line: \(y=0\)

• Normal line: \(x=1\)

8. Worked Example 4: Point not directly given by x-value

Find the tangent line and normal line to \(y=x^2-4x+5\) at the point \((2,1)\).

Here, the point is already given, but we should still check that it lies on the curve.

$$2^2-4(2)+5=4-8+5=1$$

Yes, \((2,1)\) is on the curve.

Step 1: Differentiate.

$$f'(x)=2x-4$$

Step 2: Find the tangent slope at \(x=2\).

$$f'(2)=2(2)-4=0$$

So the tangent is horizontal.

Step 3: Tangent line.

A horizontal line through \((2,1)\) is

$$y=1$$

Step 4: Normal line.

The normal is vertical through \((2,1)\), so

$$x=2$$

Answer:

• Tangent line: \(y=1\)

• Normal line: \(x=2\)

9. Common mistakes to avoid

• Using the function instead of the derivative for the slope. The tangent slope comes from \(f'(a)\), not \(f(a)\).

• Forgetting to find the point first. You need both the slope and the point to write a line.

• Incorrect negative reciprocal. If the tangent slope is \(3\), the normal slope is \(-\frac{1}{3}\), not just \(\frac{1}{3}\).

• Trying to write a vertical normal line in slope-intercept form. Vertical lines must be written as \(x=\text{constant}\).

• Substituting the wrong x-value. Be careful to evaluate both \(f(a)\) and \(f'(a)\) at the same point.

10. Quick formula review

For \(y=f(x)\) at \(x=a\):

• Point on curve: \((a,f(a))\)

• Tangent slope: \(f'(a)\)

• Tangent line:

$$y-f(a)=f'(a)(x-a)$$

• Normal slope:

$$m_{\text{normal}}=-\frac{1}{f'(a)} \quad \text{if } f'(a)\neq 0$$

• Normal line:

$$y-f(a)=-\frac{1}{f'(a)}(x-a)$$

If \(f'(a)=0\):

• Tangent line: \(y=f(a)\)

• Normal line: \(x=a\)

11. Why tangent lines matter

Tangent lines are useful because they give a very close linear approximation to a curve near a point. This is why they appear in applications of derivatives, including estimating function values, analyzing motion, and understanding the shape of graphs.

Normal lines are also important, especially in geometry and physics, because they describe directions perpendicular to a curve at a point.

12. Brief summary

To find a tangent line, first find the point on the curve and then use the derivative to get the slope. To find a normal line, use the negative reciprocal of the tangent slope. Always use point-slope form, and remember that if the tangent is horizontal, the normal is vertical.

Rectilinear Motion Kinematics is the study of motion along a straight line. In calculus, this topic connects directly to derivatives because the position of a moving particle can be written as a function of time, and then differentiated to find velocity and acceleration.

This idea is important because it shows how calculus describes change in the real world. If you know where an object is at each moment, you can use derivatives to determine how fast it is moving and how its speed is changing.

In this lesson, you will learn how to move from position to velocity to acceleration, how to interpret positive and negative values, and how to analyze the motion of a particle on a line.

1. Position, Velocity, and Acceleration

Suppose a particle moves along a straight line, and its position at time \(t\) is given by a function \(s(t)\). The variable \(s\) tells us where the particle is relative to a chosen origin on the line.

If \(s(t)\) is increasing, the particle is moving in the positive direction. If \(s(t)\) is decreasing, the particle is moving in the negative direction.

The velocity of the particle is the derivative of position with respect to time:

$$v(t)=s'(t)$$

Velocity tells both how fast and in which direction the particle is moving.

The acceleration of the particle is the derivative of velocity, or the second derivative of position:

$$a(t)=v'(t)=s''(t)$$

Acceleration tells how velocity is changing over time.

So the chain of ideas is:

• Position: \(s(t)\)
• Velocity: \(v(t)=s'(t)\)
• Acceleration: \(a(t)=s''(t)\)

2. Average Velocity and Instantaneous Velocity

Before using derivatives, it helps to remember the meaning of average velocity. Over a time interval from \(t=a\) to \(t=b\), the average velocity is:

$$\frac{s(b)-s(a)}{b-a}$$

This measures the overall change in position divided by the change in time.

The instantaneous velocity at a particular time is found by taking the limit of average velocity over smaller and smaller intervals. This is exactly the derivative:

$$v(t)=\lim_{h\to 0}\frac{s(t+h)-s(t)}{h}$$

So velocity is the rate of change of position.

3. Interpreting the Sign of Velocity

The sign of velocity is very important in rectilinear motion.

• If \(v(t)>0\), the particle moves in the positive direction.
• If \(v(t)<0\), the particle moves in the negative direction.
• If \(v(t)=0\), the particle is temporarily at rest.

A common mistake is to confuse velocity with speed. Velocity can be positive or negative because it includes direction. Speed is always non-negative and is the magnitude of velocity:

$$\text{speed}=|v(t)|$$

For example, if \(v(t)=-6\), the particle is moving left or backward at a speed of \(6\) units per second.

4. Interpreting the Sign of Acceleration

The sign of acceleration tells whether velocity is increasing or decreasing, but you must be careful when interpreting whether the particle is speeding up or slowing down.

• If velocity and acceleration have the same sign, the particle is speeding up.
• If velocity and acceleration have opposite signs, the particle is slowing down.

Why? Because speed depends on the size of velocity, not just its sign.

For example:

• If \(v=4\) and \(a=2\), the particle moves right and speeds up.
• If \(v=4\) and \(a=-2\), the particle moves right and slows down.
• If \(v=-4\) and \(a=-2\), the particle moves left and speeds up.
• If \(v=-4\) and \(a=2\), the particle moves left and slows down.

5. Displacement vs Distance Traveled

These two ideas are related, but they are not the same.

Displacement is the change in position:

$$\text{displacement}=s(b)-s(a)$$

Displacement can be positive, negative, or zero.

Distance traveled is the total length of the path traveled. If the particle changes direction, distance traveled is larger than the absolute value of displacement.

To find distance traveled on an interval, you must look for times when velocity is zero and check whether direction changes. Then add the lengths of each part of the trip.

In calculus notation, distance traveled over \([a,b]\) can be written as:

$$\int_a^b |v(t)|\,dt$$

At this level, you will often find it by breaking the motion into intervals where velocity keeps the same sign.

6. How Derivatives Model Motion

If you are given a position function, you can completely study the motion by following these steps:

1. Differentiate \(s(t)\) to get velocity \(v(t)\).
2. Differentiate \(v(t)\) to get acceleration \(a(t)\).
3. Find where \(v(t)=0\) to locate times when the particle may stop or change direction.
4. Use the sign of \(v(t)\) to determine direction of motion.
5. Use the signs of \(v(t)\) and \(a(t)\) together to decide whether the particle is speeding up or slowing down.

This process is a direct application of derivatives to physical motion.

Worked Example 1: Finding Velocity and Acceleration

A particle moves along a line with position

$$s(t)=3t^2-2t+5$$

Find the velocity and acceleration.

Step 1: Differentiate position.

$$v(t)=s'(t)=\frac{d}{dt}(3t^2-2t+5)=6t-2$$

Step 2: Differentiate velocity.

$$a(t)=v'(t)=\frac{d}{dt}(6t-2)=6$$

Answer:

• Velocity: \(v(t)=6t-2\)
• Acceleration: \(a(t)=6\)

This means the particle’s velocity changes at a constant rate of \(6\) units per second squared.

Worked Example 2: Direction of Motion and Rest

A particle has position

$$s(t)=t^3-6t^2+9t$$

Find:

• the velocity and acceleration,
• the times when the particle is at rest,
• the intervals when the particle moves right or left.

Step 1: Find velocity.

$$v(t)=s'(t)=3t^2-12t+9$$

Factor:

$$v(t)=3(t^2-4t+3)=3(t-1)(t-3)$$

Step 2: Find acceleration.

$$a(t)=v'(t)=6t-12$$

Step 3: Find when the particle is at rest.

The particle is at rest when \(v(t)=0\):

$$3(t-1)(t-3)=0$$

So the particle is at rest at:

$$t=1 \quad \text{and} \quad t=3$$

Step 4: Determine direction of motion using the sign of \(v(t)\).

Check intervals:

• For \(t<1\), choose \(t=0\): \(v(0)=9>0\), so motion is in the positive direction.
• For \(1<t<3\), choose \(t=2\): \(v(2)=3(1)(-1)=-3<0\), so motion is in the negative direction.
• For \(t>3\), choose \(t=4\): \(v(4)=3(3)(1)=9>0\), so motion is in the positive direction.

Answer:

• Velocity: \(v(t)=3(t-1)(t-3)\)
• Acceleration: \(a(t)=6t-12\)
• At rest at \(t=1\) and \(t=3\)
• Moving right for \(t<1\) and \(t>3\)
• Moving left for \(1<t<3\)

This example shows that when velocity changes sign, the particle changes direction.

Worked Example 3: Speeding Up or Slowing Down

A particle has position

$$s(t)=t^3-3t^2+2t$$

Determine where the particle is speeding up and where it is slowing down for \(t\ge 0\).

Step 1: Find velocity and acceleration.

$$v(t)=s'(t)=3t^2-6t+2$$ $$a(t)=v'(t)=6t-6$$

Step 2: Find critical times.

Velocity is zero when:

$$3t^2-6t+2=0$$

Use the quadratic formula:

$$t=\frac{6\pm\sqrt{36-24}}{6}=\frac{6\pm\sqrt{12}}{6}=1\pm\frac{\sqrt{3}}{3}$$

Acceleration is zero when:

$$6t-6=0 \Rightarrow t=1$$

Step 3: Make a sign chart.

The important points are:

$$t=1-\frac{\sqrt{3}}{3}, \quad 1, \quad 1+\frac{\sqrt{3}}{3}$$

Now compare the signs of \(v(t)\) and \(a(t)\):

• On \(0<t<1-\frac{\sqrt{3}}{3}\): \(v>0\), \(a<0\) so the particle is slowing down.
• On \(1-\frac{\sqrt{3}}{3}<t<1\): \(v<0\), \(a<0\) so the particle is speeding up.
• On \(1<t<1+\frac{\sqrt{3}}{3}\): \(v<0\), \(a>0\) so the particle is slowing down.
• On \(t>1+\frac{\sqrt{3}}{3}\): \(v>0\), \(a>0\) so the particle is speeding up.

Answer:

• Speeding up on \(\left(1-\frac{\sqrt{3}}{3},1\right)\) and \(\left(1+\frac{\sqrt{3}}{3},\infty\right)\)
• Slowing down on \(\left(0,1-\frac{\sqrt{3}}{3}\right)\) and \(\left(1,1+\frac{\sqrt{3}}{3}\right)\)

Worked Example 4: Displacement and Distance Traveled

A particle moves with position

$$s(t)=t^3-6t^2+9t, \quad 0\le t\le 4$$

Find the displacement and the total distance traveled from \(t=0\) to \(t=4\).

Step 1: Find displacement.

$$s(4)=4^3-6(4^2)+9(4)=64-96+36=4$$ $$s(0)=0$$

So displacement is:

$$s(4)-s(0)=4-0=4$$

Step 2: Find where direction can change.

From Example 2,

$$v(t)=3(t-1)(t-3)$$

So the particle may change direction at \(t=1\) and \(t=3\).

Step 3: Find position at these times.

$$s(1)=1-6+9=4$$ $$s(3)=27-54+27=0$$

Now track the motion:

• From \(t=0\) to \(t=1\): position goes from \(0\) to \(4\), distance \(=4\)
• From \(t=1\) to \(t=3\): position goes from \(4\) to \(0\), distance \(=4\)
• From \(t=3\) to \(t=4\): position goes from \(0\) to \(4\), distance \(=4\)

Total distance traveled:

$$4+4+4=12$$

Answer:

• Displacement: \(4\)
• Total distance traveled: \(12\)

This example shows clearly that displacement and distance traveled are not usually the same.

7. Graphical Meaning

Graphs also help you understand motion.

• The graph of \(s(t)\) shows position over time.
• The slope of the position graph at a point is velocity.
• If the position graph is increasing, velocity is positive.
• If the position graph is decreasing, velocity is negative.
• If the position graph has a horizontal tangent, velocity is zero.

The graph of \(v(t)\) shows how velocity changes over time.

• If \(v(t)\) is above the time axis, motion is positive.
• If \(v(t)\) is below the time axis, motion is negative.
• The slope of the velocity graph is acceleration.

This means derivatives connect the graphs as well as the formulas.

8. Common Mistakes to Avoid

• Mixing up position and velocity: Position tells where the particle is; velocity tells how position changes.
• Forgetting direction: A negative velocity does not mean “wrong”; it means motion in the negative direction.
• Thinking acceleration always means speeding up: Acceleration can also make a particle slow down.
• Confusing displacement with distance: Distance must account for every part of the trip.
• Not checking sign changes: When finding intervals of motion, test the sign of velocity on each interval.

9. Problem-Solving Checklist

When working on a rectilinear motion problem, ask yourself:

1. What is the position function \(s(t)\)?
2. What is the velocity \(v(t)=s'(t)\)?
3. What is the acceleration \(a(t)=s''(t)\)?
4. When is \(v(t)=0\)?
5. Where is \(v(t)>0\) or \(v(t)<0\)?
6. Where do \(v(t)\) and \(a(t)\) have the same sign or opposite signs?
7. Does the question ask for displacement or total distance traveled?

Brief Summary

Rectilinear motion describes movement along a straight line using a position function \(s(t)\). By differentiating, we find velocity \(v(t)=s'(t)\), and by differentiating again, we find acceleration \(a(t)=s''(t)\).

The sign of velocity tells direction, and the relationship between velocity and acceleration tells whether the particle is speeding up or slowing down. To analyze motion completely, find where velocity is zero, test signs on intervals, and distinguish carefully between displacement and total distance traveled.

Related Rates is a type of calculus problem where two or more quantities are changing over time, and those quantities are connected by an equation.

In these problems, you are usually given the rate of change of one quantity and asked to find the rate of change of another quantity at a specific moment.

For example, if the radius of a balloon is increasing, then its volume is also changing. Since radius and volume are related by a formula, their rates of change are related too.

This is why the topic is called related rates: the rates are connected because the variables are connected.

Big idea: If variables are related by an equation, then their derivatives with respect to time are also related.

In related rates, time is usually the hidden variable. Even if an equation only shows quantities like length, radius, or angle, those quantities are often changing as time passes.

So if a variable such as \(x\), \(y\), \(r\), or \(V\) changes over time, we treat it as a function of time. Then we differentiate both sides of the equation with respect to \(t\).

1. The Main Strategy

Most related rates problems can be solved using the same process.

1. Draw a diagram if the problem is geometric.
2. Assign variables to the changing quantities.
3. Write an equation that relates those variables.
4. Differentiate implicitly with respect to time, usually \(t\).
5. Substitute the known values from the specific instant asked about.
6. Solve for the unknown rate.
7. Check the sign and units of your answer.

Important warning: Do not substitute the numerical values too early if the equation contains more than one changing variable. First differentiate, then substitute.

2. Common Notation

Rates of change with respect to time are written using derivatives such as:

• \(\dfrac{dx}{dt}\): rate of change of \(x\)
• \(\dfrac{dy}{dt}\): rate of change of \(y\)
• \(\dfrac{dr}{dt}\): rate of change of radius
• \(\dfrac{dV}{dt}\): rate of change of volume

If a quantity is increasing, its rate is positive. If it is decreasing, its rate is negative.

For example:

• A ladder sliding down means the height is decreasing, so \(\dfrac{dy}{dt} < 0\).
• A circle getting bigger means the radius is increasing, so \(\dfrac{dr}{dt} > 0\).

3. Why Implicit Differentiation Is Used

Suppose two variables satisfy

$$x^2 + y^2 = 25$$

If both \(x\) and \(y\) change with time, then when we differentiate with respect to \(t\), we get

$$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$$

This step uses the chain rule, because \(x\) and \(y\) are functions of time.

This is the heart of related rates: differentiate the relationship with respect to time.

4. Example 1: Expanding Circle

Problem: The radius of a circle is increasing at a rate of \(3\) cm/s. How fast is the area increasing when the radius is \(5\) cm?

Step 1: Identify the variables.

The area of a circle is related to the radius by

$$A = \pi r^2$$

We are given

$$\frac{dr}{dt} = 3$$

We want to find

$$\frac{dA}{dt}$$

when \(r=5\).

Step 2: Differentiate with respect to time.

$$\frac{dA}{dt} = 2\pi r\frac{dr}{dt}$$

Step 3: Substitute the known values.

$$\frac{dA}{dt} = 2\pi(5)(3) = 30\pi$$

Answer: The area is increasing at

$$30\pi\text{ cm}^2/\text{s}$$

What to notice: Even though the formula for area only uses \(r\), the derivative introduces \(\dfrac{dr}{dt}\) because the radius changes over time.

5. Example 2: Ladder Sliding Down a Wall

Problem: A ladder 10 m long leans against a wall. The bottom of the ladder slides away from the wall at \(2\) m/s. How fast is the top sliding down the wall when the bottom is 6 m from the wall?

Step 1: Define variables.

Let \(x\) be the distance from the wall to the bottom of the ladder, and let \(y\) be the height of the top of the ladder above the ground.

The ladder length is constant, so by the Pythagorean theorem,

$$x^2 + y^2 = 10^2 = 100$$

We are given

$$\frac{dx}{dt} = 2$$

We want \(\dfrac{dy}{dt}\) when \(x=6\).

Step 2: Find \(y\) at that instant.

$$6^2 + y^2 = 100$$$$36 + y^2 = 100$$$$y^2 = 64$$$$y = 8$$

Step 3: Differentiate with respect to time.

$$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$$

Simplify:

$$x\frac{dx}{dt} + y\frac{dy}{dt} = 0$$

Step 4: Substitute values.

$$6(2) + 8\frac{dy}{dt} = 0$$$$12 + 8\frac{dy}{dt} = 0$$$$8\frac{dy}{dt} = -12$$$$\frac{dy}{dt} = -\frac{12}{8} = -\frac{3}{2}$$

Answer: The top of the ladder is sliding down at

$$\frac{3}{2}\text{ m/s}$$

Since the derivative is negative, the height is decreasing. So a more complete statement is:

$$\frac{dy}{dt} = -1.5\text{ m/s}$$

What to notice: The negative sign matters. It tells us the top is moving downward.

6. Example 3: Changing Volume of a Sphere

Problem: Air is being pumped into a spherical balloon so that its radius increases at \(0.4\) cm/s. How fast is the volume increasing when the radius is \(6\) cm?

Step 1: Write the formula.

$$V = \frac{4}{3}\pi r^3$$

Given:

$$\frac{dr}{dt} = 0.4$$

Find:

$$\frac{dV}{dt}$$

when \(r=6\).

Step 2: Differentiate with respect to time.

$$\frac{dV}{dt} = \frac{4}{3}\pi \cdot 3r^2\frac{dr}{dt}$$$$\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}$$

Step 3: Substitute values.

$$\frac{dV}{dt} = 4\pi(6^2)(0.4)$$$$\frac{dV}{dt} = 4\pi(36)(0.4)$$$$\frac{dV}{dt} = 57.6\pi$$

Answer: The volume is increasing at

$$57.6\pi\text{ cm}^3/\text{s}$$

Approximation:

$$57.6\pi \approx 181.0\text{ cm}^3/\text{s}$$

7. Example 4: Two Sides of a Right Triangle Changing

Problem: In a right triangle, one leg has length \(x\), the other leg has length \(y\), and the hypotenuse is 13 cm. If \(x\) is increasing at \(5\) cm/s, how fast is \(y\) changing when \(x=12\) cm?

Step 1: Write the relationship.

$$x^2 + y^2 = 13^2 = 169$$

Given:

$$\frac{dx}{dt} = 5$$

Find:

$$\frac{dy}{dt}$$

when \(x=12\).

Step 2: Find \(y\) at that moment.

$$12^2 + y^2 = 169$$$$144 + y^2 = 169$$$$y^2 = 25$$$$y = 5$$

Step 3: Differentiate implicitly.

$$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$$$$x\frac{dx}{dt} + y\frac{dy}{dt} = 0$$

Step 4: Substitute values.

$$12(5) + 5\frac{dy}{dt} = 0$$$$60 + 5\frac{dy}{dt} = 0$$$$5\frac{dy}{dt} = -60$$$$\frac{dy}{dt} = -12$$

Answer: The other leg is decreasing at

$$12\text{ cm/s}$$

or, using sign,

$$\frac{dy}{dt} = -12\text{ cm/s}$$

What to notice: Since the hypotenuse is fixed, if one leg gets longer, the other must get shorter.

8. Common Types of Related Rates Problems

You will often see related rates in geometric settings such as:

• circles: radius, area, circumference
• spheres: radius, volume, surface area
• cones: radius, height, volume
• right triangles: ladder, shadow, distance problems
• rectangles: changing length, width, or area

The key is always the same: find the equation connecting the variables.

9. Common Mistakes to Avoid

• Forgetting that variables depend on time. When differentiating \(r^2\), the derivative is not just \(2r\); it is \(2r\dfrac{dr}{dt}\).
• Substituting numbers too early. Differentiate the equation first, then plug in values.
• Using the wrong sign. Decreasing quantities should have negative rates.
• Using the wrong equation. Start by carefully choosing the correct geometric formula.
• Forgetting units. Rates should include units such as cm/s, m/s, or cm³/s.

10. A Quick Problem-Solving Checklist

• What quantities are changing?
• What rate is given?
• What rate is asked for?
• What equation relates the variables?
• Did I differentiate with respect to time?
• Did I substitute values only after differentiating?
• Does my final answer include the correct sign and units?

11. Final Summary

Related rates problems involve quantities that change together over time.

To solve them, write an equation relating the variables, differentiate both sides with respect to time, and then use the given information to find the unknown rate.

The most important skill is recognizing that variables such as length, radius, area, and volume can all depend on time, even if time does not appear in the original formula.

With careful setup, correct differentiation, and attention to signs and units, related rates problems become a clear step-by-step process.

Linear Approximation and Differentials help us estimate the value of a function near a point where the function is easier to work with.

The main idea is simple: if a function is smooth and you stay close to a known point, then the graph of the function looks very much like its tangent line. Because lines are easier than curves, we can use the tangent line to estimate nearby function values.

This idea is especially useful when a calculator gives an awkward expression, when mental estimation is needed, or when we want to understand how a small change in one quantity affects another.

1. The Tangent Line as a Local Approximation

Suppose we know a function value at some point \(x=a\). The tangent line to \(f(x)\) at \(x=a\) has slope \(f'(a)\), so its equation is

$$L(x)=f(a)+f'(a)(x-a).$$

This line is called the linearization or linear approximation of \(f(x)\) at \(x=a\).

For values of \(x\) close to \(a\), we use

$$f(x)\approx L(x)=f(a)+f'(a)(x-a).$$

This means the function value is approximately equal to the value of its tangent line.

Why does this work?

If a function is differentiable at \(x=a\), then near that point the curve does not change direction sharply. Its tangent line matches both the function's value and its immediate rate of change, so for small changes in \(x\), the line gives a very good estimate.

2. Formula for Linear Approximation

To approximate \(f(x)\) near \(x=a\), follow these steps:

1. Choose a value \(a\) close to the \(x\)-value you want.
2. Compute \(f(a)\).
3. Compute \(f'(a)\).
4. Use the formula \(f(x)\approx f(a)+f'(a)(x-a)\).

It is best to choose \(a\) so that \(f(a)\) is easy to evaluate exactly.

3. Differentials

Differentials give a way to describe small changes in variables.

If \(y=f(x)\), then the differential of \(y\) is

$$dy=f'(x)\,dx.$$

Here:

• \(dx\) represents a small change in \(x\),
• \(dy\) represents the change predicted by the tangent line.

If \(\Delta x\) is the actual change in \(x\), then we usually let

$$dx=\Delta x.$$

The actual change in the function is

$$\Delta y=f(x+\Delta x)-f(x).$$

The differential gives the approximation

$$\Delta y\approx dy=f'(x)\,dx.$$

So:

• Actual change: \(\Delta y\)
• Estimated change using calculus: \(dy\)

This is why differentials are useful in estimation: they quickly approximate how much the output changes when the input changes a little.

4. Connection Between Linear Approximation and Differentials

These two ideas are closely connected.

Starting from \(y=f(x)\), if \(x\) changes by a small amount \(dx\), then

$$dy=f'(x)dx.$$

Adding this small change to the original function value gives

$$f(x+dx)\approx f(x)+dy=f(x)+f'(x)dx.$$

This is the same idea as linearization. The tangent line predicts the nearby value.

5. Worked Example 1: Approximating a Square Root

Use linear approximation to estimate \(\sqrt{4.1}\).

Let

$$f(x)=\sqrt{x}.$$

We choose \(a=4\) because \(\sqrt{4}=2\), which is easy to work with.

First find the derivative:

$$f'(x)=\frac{1}{2\sqrt{x}}.$$

Now evaluate at \(x=4\):

$$f(4)=2, \qquad f'(4)=\frac{1}{4}.$$

The linear approximation at \(x=4\) is

$$L(x)=2+\frac{1}{4}(x-4).$$

Now substitute \(x=4.1\):

$$L(4.1)=2+\frac{1}{4}(0.1)=2+0.025=2.025.$$

So

$$\sqrt{4.1}\approx 2.025.$$

This is very close to the actual value.

6. Worked Example 2: Estimating with Differentials

Use differentials to estimate the change in \(y=x^3\) when \(x\) changes from \(2\) to \(2.02\).

Let

$$y=x^3.$$

Then

$$\frac{dy}{dx}=3x^2,$$

so the differential is

$$dy=3x^2\,dx.$$

At \(x=2\), and since \(x\) changes from \(2\) to \(2.02\), we have

$$dx=0.02.$$

Now compute:

$$dy=3(2)^2(0.02)=3(4)(0.02)=12(0.02)=0.24.$$

So the estimated change in \(y\) is

$$\Delta y\approx dy=0.24.$$

This means

$$2.02^3\approx 2^3+0.24=8.24.$$

The differential gives a quick estimate without needing to cube \(2.02\) directly.

7. Worked Example 3: Linear Approximation for a Reciprocal

Approximate \(\dfrac{1}{0.98}\).

Let

$$f(x)=\frac{1}{x}.$$

Choose \(a=1\), since \(f(1)=1\).

Find the derivative:

$$f'(x)=-\frac{1}{x^2}.$$

Then

$$f(1)=1, \qquad f'(1)=-1.$$

The linearization at \(x=1\) is

$$L(x)=1- (x-1).$$

Simplify:

$$L(x)=2-x.$$

Now estimate at \(x=0.98\):

$$L(0.98)=2-0.98=1.02.$$

So

$$\frac{1}{0.98}\approx 1.02.$$

This makes sense because dividing by a number slightly less than 1 should give a result slightly greater than 1.

8. Worked Example 4: Estimating Error in Measurement

A sphere has radius \(r=10\) cm, measured with a possible error of \(dr=0.1\) cm. Estimate the resulting error in its volume.

The volume of a sphere is

$$V=\frac{4}{3}\pi r^3.$$

Differentiate:

$$dV=4\pi r^2\,dr.$$

Substitute \(r=10\) and \(dr=0.1\):

$$dV=4\pi (10)^2(0.1)=4\pi(100)(0.1)=40\pi.$$

So the estimated error in volume is

$$dV\approx 40\pi\text{ cm}^3.$$

This means a small measurement error in radius can cause a larger error in volume.

9. Important Notes About Accuracy

• Linear approximation works best when \(x\) is close to the chosen point \(a\).
• If \(x\) is far from \(a\), the tangent line may no longer match the curve well.
• The differential \(dy\) is an estimate of the true change \(\Delta y\), not usually the exact value.

In symbols:

$$\Delta y=f(x+\Delta x)-f(x), \qquad dy=f'(x)dx, \qquad \Delta y\approx dy.$$

10. How to Recognize When to Use This Idea

You should think about linear approximation or differentials when:

• you need to estimate a function value near a nice number,
• you are asked to approximate a change in a quantity,
• you see words like estimate, approximate, near, or small change,
• you want to model how measurement error affects another quantity.

11. Common Mistakes to Avoid

• Choosing a bad base point: pick \(a\) close to the target value and easy to evaluate.
• Using the derivative at the wrong point: in \(f(a)+f'(a)(x-a)\), the derivative must be evaluated at \(a\).
• Confusing \(dy\) and \(\Delta y\): \(dy\) is the estimate, while \(\Delta y\) is the actual change.
• Forgetting that the method is local: the approximation is good only nearby.

12. Quick Review of the Main Formulas

For a function \(f(x)\) near \(x=a\):

$$L(x)=f(a)+f'(a)(x-a)$$

and

$$f(x)\approx L(x).$$

If \(y=f(x)\), then the differential is

$$dy=f'(x)dx.$$

For a small change in \(x\):

$$\Delta y\approx dy.$$

Summary

Linear approximation uses the tangent line to estimate a function near a known point. Differentials use the derivative to estimate how much the output changes when the input changes slightly. Both ideas depend on the fact that smooth curves behave like lines over very small intervals.

When used correctly, these tools make hard calculations much easier and help us understand errors, changes, and nearby values in real situations.

L'Hôpital's Rule is a useful calculus method for evaluating certain limits that at first seem impossible to compute directly. It is especially helpful when direct substitution gives an indeterminate form, such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\).

In this lesson, you will learn what L'Hôpital's Rule says, when you are allowed to use it, and how to apply it correctly. You will also see worked examples that become gradually more challenging.

Why do we need this rule? Sometimes, when finding a limit, plugging in the value of \(x\) gives a fraction like

$$\frac{0}{0} \quad \text{or} \quad \frac{\infty}{\infty}.$$

These forms do not tell us the actual value of the limit. They are called indeterminate because the true limit could be many different things. L'Hôpital's Rule gives us a way to simplify the problem by differentiating the top and bottom separately.

The rule: Suppose

$$\lim_{x \to a} \frac{f(x)}{g(x)}$$

gives either \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). If \(f\) and \(g\) are differentiable near \(a\), and \(g'(x) \neq 0\) near \(a\), then under the right conditions,

$$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$$

provided the limit on the right exists or can be determined.

In simple words: if the original limit gives \(0/0\) or \(\infty/\infty\), differentiate the numerator and denominator separately, then try the limit again.

Important: L'Hôpital's Rule does not mean differentiating the whole fraction using the quotient rule. Instead, you take the derivative of the numerator and the derivative of the denominator separately.

For example,

$$\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) \neq \frac{f'(x)}{g'(x)}$$

in general. L'Hôpital's Rule is a special rule for limits, not a rule for ordinary derivatives.

When can you use L'Hôpital's Rule?

• The expression must be a quotient.
• Direct substitution must give \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\).
• The numerator and denominator must be differentiable near the point.
• After differentiating, the new limit should exist or be easier to evaluate.

When should you not use it right away?

• If direct substitution already gives a real number.
• If the form is not \(0/0\) or \(\infty/\infty\).
• If algebra, factoring, rationalizing, or trig identities would solve the limit more simply.

A step-by-step method:

1. Substitute the limiting value into the expression.
2. Check whether the result is \(0/0\) or \(\infty/\infty\).
3. If it is, differentiate the numerator and denominator separately.
4. Evaluate the new limit.
5. If the new limit is still indeterminate, you may apply L'Hôpital's Rule again.

Now let us work through some examples.

Example 1: A basic \(0/0\) form

Evaluate

$$\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$$

First, substitute \(x=2\):

$$\frac{2^2 - 4}{2 - 2} = \frac{0}{0}$$

This is an indeterminate form, so L'Hôpital's Rule applies.

Differentiate the numerator and denominator separately:

$$\frac{d}{dx}(x^2 - 4) = 2x, \qquad \frac{d}{dx}(x - 2) = 1$$

Now compute the new limit:

$$\lim_{x \to 2} \frac{2x}{1} = 2(2) = 4$$

So,

$$\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = 4$$

This example could also be done by factoring, but it shows clearly how the rule works.

Example 2: A trigonometric limit

Evaluate

$$\lim_{x \to 0} \frac{\sin x}{x}$$

Substitute \(x=0\):

$$\frac{\sin 0}{0} = \frac{0}{0}$$

This is indeterminate, so use L'Hôpital's Rule.

Differentiate top and bottom:

$$\frac{d}{dx}(\sin x) = \cos x, \qquad \frac{d}{dx}(x) = 1$$

Now evaluate:

$$\lim_{x \to 0} \frac{\cos x}{1} = \cos 0 = 1$$

Therefore,

$$\lim_{x \to 0} \frac{\sin x}{x} = 1$$

This is one of the most important limits in calculus.

Example 3: A case that needs L'Hôpital's Rule more than once

Evaluate

$$\lim_{x \to 0} \frac{1 - \cos x}{x^2}$$

Substitute \(x=0\):

$$\frac{1 - \cos 0}{0^2} = \frac{0}{0}$$

Apply L'Hôpital's Rule once:

$$\frac{d}{dx}(1 - \cos x) = \sin x, \qquad \frac{d}{dx}(x^2) = 2x$$

The limit becomes

$$\lim_{x \to 0} \frac{\sin x}{2x}$$

Substitute again:

$$\frac{\sin 0}{2 \cdot 0} = \frac{0}{0}$$

It is still indeterminate, so apply L'Hôpital's Rule a second time:

$$\frac{d}{dx}(\sin x) = \cos x, \qquad \frac{d}{dx}(2x) = 2$$

Now the limit is

$$\lim_{x \to 0} \frac{\cos x}{2} = \frac{\cos 0}{2} = \frac{1}{2}$$

So,

$$\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$$

This example shows that you may need to use the rule more than once.

Example 4: An \(\infty/\infty\) form

Evaluate

$$\lim_{x \to \infty} \frac{3x^2 + 1}{5x^2 - 7x}$$

As \(x \to \infty\), both numerator and denominator go to infinity, so this is an \(\infty/\infty\) form.

Apply L'Hôpital's Rule:

$$\frac{d}{dx}(3x^2 + 1) = 6x, \qquad \frac{d}{dx}(5x^2 - 7x) = 10x - 7$$

Now evaluate the new limit:

$$\lim_{x \to \infty} \frac{6x}{10x - 7}$$

This still has the form \(\infty/\infty\), so apply the rule again:

$$\frac{d}{dx}(6x) = 6, \qquad \frac{d}{dx}(10x - 7) = 10$$

Then

$$\lim_{x \to \infty} \frac{6}{10} = \frac{3}{5}$$

Therefore,

$$\lim_{x \to \infty} \frac{3x^2 + 1}{5x^2 - 7x} = \frac{3}{5}$$

Common mistakes to avoid

• Using L'Hôpital's Rule when the form is not indeterminate. Always substitute first.
• Differentiating only the numerator or only the denominator. You must differentiate both separately.
• Using the quotient rule by mistake. L'Hôpital's Rule is not the same as differentiating a fraction normally.
• Stopping too soon. If the new limit is still \(0/0\) or \(\infty/\infty\), you may need to apply the rule again.

Connection to applications of derivatives

L'Hôpital's Rule is part of the broader topic of applications of derivatives because it uses derivatives to study how functions behave near a point or as \(x\) becomes very large. It helps approximate limits efficiently, which is useful when analyzing motion, rates of change, and the shape of graphs.

For example, when sketching a graph or studying a function's behavior near a special value, limits often appear. If those limits produce indeterminate forms, L'Hôpital's Rule can help you find the exact result.

Quick checklist before using the rule

• Is it a fraction?
• Did direct substitution give \(0/0\) or \(\infty/\infty\)?
• Can I differentiate the top and bottom?
• After differentiating, is the limit easier to compute?

Brief summary

L'Hôpital's Rule helps evaluate limits that give the indeterminate forms \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). To use it, differentiate the numerator and denominator separately, then evaluate the new limit. If needed, repeat the process. Always check the form first, and remember that this rule is for limits only.

Lesson: The Mean Value Theorem (MVT)

The Mean Value Theorem is one of the most important ideas in calculus. It connects an average rate of change over an interval to an instantaneous rate of change at some point inside that interval.

In simpler words, if a function behaves nicely on an interval, then at some point its tangent slope must match the slope of the secant line across the whole interval.

This theorem helps us understand motion, rates of change, and why derivatives give powerful information about a function. It is also used in graph analysis, estimating function behavior, and proving other calculus results.

1. Review: Average rate of change vs instantaneous rate of change

Before learning the theorem, we need to understand the two rates it compares.

• Average rate of change of a function from \(x=a\) to \(x=b\):

$$\frac{f(b)-f(a)}{b-a}$$

This is the slope of the secant line joining the points \((a,f(a))\) and \((b,f(b))\).

• Instantaneous rate of change at \(x=c\):

$$f'(c)$$

This is the slope of the tangent line to the graph at \(x=c\).

The Mean Value Theorem says that, under the right conditions, there is at least one number \(c\) between \(a\) and \(b\) such that these two slopes are equal.

2. Statement of the Mean Value Theorem

If a function \(f\) satisfies both of these conditions:

• \(f\) is continuous on the closed interval \([a,b]\)
• \(f\) is differentiable on the open interval \((a,b)\)

then there exists at least one number \(c\) in \((a,b)\) such that

$$f'(c)=\frac{f(b)-f(a)}{b-a}$$

This is the Mean Value Theorem.

3. What the theorem means visually

Imagine drawing the secant line between the endpoints of the graph on \([a,b]\). The theorem guarantees that somewhere between those endpoints, the graph has a tangent line parallel to that secant line.

So the function’s instantaneous slope matches its overall average slope at least once.

4. Why the conditions matter

The theorem only works if the function is:

• Continuous on \([a,b]\): the graph has no breaks, jumps, or holes on the interval, including the endpoints.
• Differentiable on \((a,b)\): the graph has no sharp corners, cusps, or vertical tangents inside the interval.

If either condition fails, the theorem may not apply.

5. Geometric and physical interpretation

Geometric interpretation: There is at least one point where the tangent line is parallel to the secant line connecting the endpoints.

Motion interpretation: If an object moves along a line and its position is given by \(s(t)\), then over a time interval \([a,b]\), its average velocity is

$$\frac{s(b)-s(a)}{b-a}$$

The Mean Value Theorem says that at some time \(c\in(a,b)\), the instantaneous velocity equals the average velocity:

$$s'(c)=\frac{s(b)-s(a)}{b-a}$$

This gives a very useful connection between overall motion and exact speed at a moment.

6. How to use the Mean Value Theorem

1. Check that the function is continuous on \([a,b]\).
2. Check that the function is differentiable on \((a,b)\).
3. Compute the average rate of change:

$$\frac{f(b)-f(a)}{b-a}$$

4. Find \(f'(x)\).
5. Set \(f'(c)\) equal to the average rate of change.
6. Solve for \(c\) in the interval \((a,b)\).

7. Worked Example 1: A basic polynomial

Find the value(s) of \(c\) guaranteed by the Mean Value Theorem for

$$f(x)=x^2+1 \quad \text{on } [1,5]$$

Step 1: Check the conditions

\(f(x)=x^2+1\) is a polynomial, so it is continuous and differentiable everywhere. Therefore, it satisfies the conditions of the Mean Value Theorem on \([1,5]\).

Step 2: Find the average rate of change

$$\frac{f(5)-f(1)}{5-1}=\frac{(25+1)-(1+1)}{4}=\frac{26-2}{4}=\frac{24}{4}=6$$

Step 3: Find the derivative

$$f'(x)=2x$$

Step 4: Set the derivative equal to the average rate of change

$$2c=6$$ $$c=3$$

Answer: The value guaranteed by the Mean Value Theorem is

$$c=3$$

This means that at \(x=3\), the tangent slope is the same as the secant slope from \(x=1\) to \(x=5\).

8. Worked Example 2: A cubic function

Find all values of \(c\) that satisfy the Mean Value Theorem for

$$f(x)=x^3-3x \quad \text{on } [-1,2]$$

Step 1: Check the conditions

This is a polynomial, so it is continuous on \([-1,2]\) and differentiable on \((-1,2)\).

Step 2: Compute the average rate of change

$$f(2)=8-6=2$$ $$f(-1)=-1+3=2$$ $$\frac{f(2)-f(-1)}{2-(-1)}=\frac{2-2}{3}=0$$

Step 3: Find the derivative

$$f'(x)=3x^2-3$$

Step 4: Solve \(f'(c)=0\)

$$3c^2-3=0$$ $$3c^2=3$$ $$c^2=1$$ $$c=\pm 1$$

But the theorem requires \(c\) to be in the open interval \((-1,2)\). So \(c=-1\) is not allowed because it is an endpoint.

Answer:

$$c=1$$

Important note: Even if solving the equation gives several values, only those inside the open interval count.

9. Worked Example 3: Rectilinear motion

A particle moves along a line with position function

$$s(t)=t^2+2t$$

for \(0\le t\le 4\). Show that at some time, the instantaneous velocity equals the average velocity, and find that time.

Step 1: Check the conditions

Since \(s(t)=t^2+2t\) is a polynomial, it is continuous on \([0,4]\) and differentiable on \((0,4)\). So the Mean Value Theorem applies.

Step 2: Find the average velocity

$$\frac{s(4)-s(0)}{4-0}=\frac{(16+8)-0}{4}=\frac{24}{4}=6$$

Step 3: Find the instantaneous velocity

$$s'(t)=2t+2$$

Step 4: Set instantaneous velocity equal to average velocity

$$2c+2=6$$ $$2c=4$$ $$c=2$$

Answer: At time

$$t=2$$

the instantaneous velocity equals the average velocity over the interval \([0,4]\).

This is a direct application of the theorem to motion along a straight line.

10. Worked Example 4: When MVT does not apply

Consider

$$f(x)=|x| \quad \text{on } [-2,2]$$

Does the Mean Value Theorem apply?

Step 1: Check continuity

The function \(|x|\) is continuous on \([-2,2]\).

Step 2: Check differentiability

The function has a sharp corner at \(x=0\), so it is not differentiable at \(x=0\).

Since differentiability fails on the open interval \((-2,2)\), the Mean Value Theorem does not apply.

Extra check: The average rate of change is

$$\frac{f(2)-f(-2)}{2-(-2)}=\frac{2-2}{4}=0$$

But the derivative of \(|x|\) is \(-1\) for \(x<0\) and \(1\) for \(x>0\). There is no point where the derivative equals \(0\). This shows why the differentiability condition is necessary.

11. Relationship to Rolle’s Theorem

A special case of the Mean Value Theorem happens when the endpoints have the same function value:

$$f(a)=f(b)$$

Then

$$\frac{f(b)-f(a)}{b-a}=0$$

So the Mean Value Theorem becomes

$$f'(c)=0$$

This special case is called Rolle’s Theorem. It says that if a smooth curve starts and ends at the same height, then somewhere in between it has a horizontal tangent.

12. Why MVT is useful

• It guarantees the existence of a point with a particular derivative value.
• It explains why average and instantaneous rates are connected.
• It helps analyze motion problems.
• It supports graph sketching by connecting slopes and behavior.
• It is used in proving whether functions are increasing or decreasing.

13. Common mistakes to avoid

• Forgetting to check the conditions. You must verify continuity and differentiability first.
• Using the closed interval incorrectly. The function must be continuous on \([a,b]\), but differentiable only on \((a,b)\).
• Including endpoints as possible values of \(c\). The value \(c\) must be strictly between \(a\) and \(b\).
• Mixing up the derivative with the average rate of change. The derivative is instantaneous; the secant slope is average over the interval.
• Stopping too early. After finding the average rate, you still need to solve \(f'(c)=\frac{f(b)-f(a)}{b-a}\).

14. Quick checklist for solving MVT questions

1. Write the function and interval clearly.
2. State whether the function is continuous on \([a,b]\).
3. State whether the function is differentiable on \((a,b)\).
4. Find \(f(a)\) and \(f(b)\).
5. Calculate the average rate of change.
6. Differentiate the function.
7. Solve for \(c\).
8. Make sure \(c\) is inside the interval.

15. Brief summary

The Mean Value Theorem says that if a function is continuous on a closed interval and differentiable on the corresponding open interval, then there is at least one interior point where the derivative equals the average rate of change over the interval.

In symbols, if the conditions are satisfied, then there exists some \(c\in(a,b)\) such that

$$f'(c)=\frac{f(b)-f(a)}{b-a}$$

You can think of this as a guarantee that somewhere on the interval, the tangent slope matches the secant slope. This idea is especially useful in motion problems, graph analysis, and understanding how derivatives describe change.

First Derivative Test and Extrema

In calculus, the first derivative helps us understand how a function is changing. Since the derivative gives the slope of the tangent line, it tells us whether a function is increasing, decreasing, or momentarily flat.

This idea is especially useful when we want to find extrema, which are the high and low points of a function. The first derivative test allows us to decide whether a critical point is a local maximum, a local minimum, or neither.

This lesson will show you how to:

• find critical points,
• determine where a function is increasing or decreasing,
• use sign changes in the derivative to classify extrema,
• connect these ideas to graphs and applied situations.

1. Review: What does the derivative tell us?

If a function is written as \(f(x)\), then its derivative is \(f'(x)\). The sign of \(f'(x)\) tells us the behavior of the original function:

• If \(f'(x) > 0\), then \(f(x)\) is increasing.
• If \(f'(x) < 0\), then \(f(x)\) is decreasing.
• If \(f'(x) = 0\), the graph has a horizontal tangent, so the point may be a turning point.

A point where \(f'(x)=0\) or where \(f'(x)\) does not exist is called a critical point, as long as \(x\) is in the domain of the function.

2. Local maximum and local minimum

A local maximum occurs when the function changes from increasing to decreasing. On a graph, the curve rises, reaches a peak, and then falls.

A local minimum occurs when the function changes from decreasing to increasing. On a graph, the curve falls, reaches a valley, and then rises.

These are called local extrema because they are highest or lowest only near that point, not necessarily on the entire graph.

3. The First Derivative Test

The first derivative test uses the sign of \(f'(x)\) before and after a critical point.

1. Find the critical points by solving \(f'(x)=0\) and checking where \(f'(x)\) is undefined.
2. Split the number line into intervals around those critical points.
3. Test the sign of \(f'(x)\) in each interval.
4. Use the sign change to classify each critical point.

The conclusions are:

• If \(f'(x)\) changes from positive to negative, then \(f(x)\) has a local maximum.
• If \(f'(x)\) changes from negative to positive, then \(f(x)\) has a local minimum.
• If \(f'(x)\) does not change sign, then the critical point is not a local extremum.

In sign-chart form:

$$ \begin{array}{c|ccc} x & \text{left of } c & c & \text{right of } c \\ \hline f'(x) & + & 0 & - \end{array} \Rightarrow \text{local maximum at } x=c $$ $$ \begin{array}{c|ccc} x & \text{left of } c & c & \text{right of } c \\ \hline f'(x) & - & 0 & + \end{array} \Rightarrow \text{local minimum at } x=c $$ $$ \begin{array}{c|ccc} x & \text{left of } c & c & \text{right of } c \\ \hline f'(x) & + & 0 & + \end{array} \Rightarrow \text{no local extremum at } x=c $$

4. Increasing and decreasing intervals

Before classifying extrema, we often identify where the function is increasing and decreasing.

• Find \(f'(x)\).
• Locate the critical points.
• Use test values in each interval to decide whether \(f'(x)\) is positive or negative.

This gives the monotonic intervals of the function:

• Increasing intervals: where \(f'(x)>0\)
• Decreasing intervals: where \(f'(x)<0\)

5. Worked Example 1: A basic polynomial

Find the critical points and classify the extrema of

$$f(x)=x^2-4x+1$$

Step 1: Differentiate.

$$f'(x)=2x-4$$

Step 2: Find critical points.

$$2x-4=0$$ $$2x=4$$ $$x=2$$

So the only critical point is at \(x=2\).

Step 3: Test the sign of \(f'(x)\).

Choose a value less than 2, such as \(x=1\):

$$f'(1)=2(1)-4=-2<0$$

So the function is decreasing for \(x<2\).

Choose a value greater than 2, such as \(x=3\):

$$f'(3)=2(3)-4=2>0$$

So the function is increasing for \(x>2\).

Step 4: Classify the critical point.

The derivative changes from negative to positive, so \(f(x)\) has a local minimum at \(x=2\).

To find the point on the graph, calculate \(f(2)\):

$$f(2)=2^2-4(2)+1=4-8+1=-3$$

Therefore, the local minimum point is \((2,-3)\).

Conclusion:

• Decreasing on \((-,2)\)
• Increasing on \((2,)\)
• Local minimum at \((2,-3)\)

6. Worked Example 2: A cubic with both a maximum and a minimum

Analyze

$$f(x)=x^3-3x^2+2$$

Step 1: Differentiate.

$$f'(x)=3x^2-6x$$

Factor the derivative:

$$f'(x)=3x(x-2)$$

Step 2: Find critical points.

$$3x(x-2)=0$$

So the critical points are:

$$x=0 \quad \text{and} \quad x=2$$

Step 3: Make a sign chart.

The critical points divide the number line into three intervals:

• \((-,0)\)
• \((0,2)\)
• \((2,)\)

Test each interval.

For \(x=-1\):

$$f'(-1)=3(-1)(-3)=9>0$$

So \(f(x)\) is increasing on \((-,0)\).

For \(x=1\):

$$f'(1)=3(1)(-1)=-3<0$$

So \(f(x)\) is decreasing on \((0,2)\).

For \(x=3\):

$$f'(3)=3(3)(1)=9>0$$

So \(f(x)\) is increasing on \((2,)\).

Step 4: Classify each critical point.

• At \(x=0\), \(f'(x)\) changes from positive to negative, so there is a local maximum.
• At \(x=2\), \(f'(x)\) changes from negative to positive, so there is a local minimum.

Step 5: Find the corresponding y-values.

$$f(0)=0-0+2=2$$ $$f(2)=8-12+2=-2$$

So the extrema are:

• Local maximum at \((0,2)\)
• Local minimum at \((2,-2)\)

Conclusion:

• Increasing on \((-,0)\cup(2,)\)
• Decreasing on \((0,2)\)

7. Worked Example 3: A critical point that is not an extremum

Consider

$$f(x)=x^3$$

Step 1: Differentiate.

$$f'(x)=3x^2$$

Step 2: Find critical points.

$$3x^2=0 \Rightarrow x=0$$

Step 3: Test the sign of the derivative.

If \(x=-1\), then

$$f'(-1)=3(-1)^2=3>0$$

If \(x=1\), then

$$f'(1)=3(1)^2=3>0$$

The derivative is positive on both sides of \(x=0\).

Step 4: Classify.

Since \(f'(x)\) does not change sign, \(x=0\) is not a local maximum or minimum.

Even though the tangent is horizontal there, the graph keeps increasing through the point. So \((0,0)\) is a critical point, but not an extremum.

This example is important because it shows that having \(f'(x)=0\) is not enough. You must check whether the sign changes.

8. Worked Example 4: An application to optimization

Suppose the profit, in hundreds of dollars, from selling a product is modeled by

$$P(x)=-x^2+6x+1$$

where \(x\) is the number of units sold in some scaled form. Find the value of \(x\) that gives the maximum profit.

Step 1: Differentiate.

$$P'(x)=-2x+6$$

Step 2: Find critical points.

$$-2x+6=0$$ $$-2x=-6$$ $$x=3$$

Step 3: Test the sign of \(P'(x)\).

If \(x=2\), then

$$P'(2)=-4+6=2>0$$

So profit is increasing before \(x=3\).

If \(x=4\), then

$$P'(4)=-8+6=-2<0$$

So profit is decreasing after \(x=3\).

Step 4: Classify.

The derivative changes from positive to negative, so \(x=3\) gives a local maximum. In this model, it is also the maximum profit value.

Now find the profit:

$$P(3)=-(3)^2+6(3)+1=-9+18+1=10$$

So the maximum profit is \(10\) hundred dollars, or \(\$1000\).

9. Using a sign chart efficiently

A sign chart helps organize your work clearly. For example, if

$$f'(x)=x(x-4)(x+1)$$

then the critical points are \(x=-1\), \(x=0\), and \(x=4\). These points split the number line into intervals:

$$(-,-1),\quad (-1,0),\quad (0,4),\quad (4,)$$

You do not need many values. One test value from each interval is enough to determine the sign of \(f'(x)\) on that entire interval.

This works because the sign of a polynomial factor only changes when you cross one of its zeros.

10. Common mistakes to avoid

• Forgetting undefined derivatives: A critical point can happen when \(f'(x)\) does not exist, not only when \(f'(x)=0\).
• Not checking sign changes: A derivative equal to zero does not automatically mean a maximum or minimum.
• Giving only x-values: Extrema are points on the graph, so include both \(x\) and \(y\) values when asked.
• Mixing up increasing and decreasing: Positive derivative means increasing, negative derivative means decreasing.
• Skipping intervals: If there are several critical points, test every interval created by them.

11. How this connects to graph sketching and motion

The first derivative test is very useful when sketching graphs. Once you know where a function increases and decreases, you can place peaks and valleys correctly and draw a more accurate graph.

It also helps in rectilinear motion. If a position function is \(s(t)\), then \(s'(t)\) is velocity. When velocity changes from positive to negative, the object changes direction and the position reaches a local maximum. When velocity changes from negative to positive, the position reaches a local minimum.

So the same derivative ideas work in pure graphing and in applied problems.

12. Step-by-step strategy for any problem

1. Find \(f'(x)\).
2. Solve \(f'(x)=0\) and identify where \(f'(x)\) is undefined.
3. List all critical points in order on a number line.
4. Test the sign of \(f'(x)\) in each interval.
5. State where the function is increasing and decreasing.
6. Use sign changes to classify each critical point as a local maximum, local minimum, or neither.
7. Substitute back into \(f(x)\) to get the full coordinates of each extremum.

Brief Summary

The first derivative test uses the sign of \(f'(x)\) around a critical point to classify extrema. If the derivative changes from positive to negative, the function has a local maximum. If it changes from negative to positive, the function has a local minimum. If there is no sign change, the critical point is not an extremum.

By finding critical points and testing intervals, you can also determine where a function is increasing or decreasing. This makes the first derivative test an important tool for graph sketching, optimization, and motion problems.

Second Derivative Test and Concavity

When we study a function, we do not only want to know whether it is increasing or decreasing. We also want to know how the graph bends. This bending is called concavity.

The second derivative helps us describe that bending. It also helps us decide whether a critical point is a local maximum or a local minimum. In this lesson, you will learn how to use the second derivative to analyze a graph carefully and find possible points of inflection.

1. Review: First and Second Derivatives

If a function is written as \(f(x)\), then:

• \(f'(x)\) tells us the slope of the graph.
• \(f''(x)\) tells us how the slope is changing.

If the slope is increasing, the graph bends upward. If the slope is decreasing, the graph bends downward.

2. What Does Concavity Mean?

A graph is concave up when it bends like a cup that could hold water. A graph is concave down when it bends like an upside-down cup.

• If \(f''(x) > 0\), the function is concave up.
• If \(f''(x) < 0\), the function is concave down.

This works because:

• \(f''(x) > 0\) means the slopes are getting larger.
• \(f''(x) < 0\) means the slopes are getting smaller.

3. Points of Inflection

A point of inflection is a point where the graph changes concavity.

That means the graph must switch:

• from concave up to concave down, or
• from concave down to concave up.

To find possible inflection points:

1. Find \(f''(x)\).
2. Solve \(f''(x) = 0\) or find where \(f''(x)\) is undefined.
3. Test intervals around those values to see whether the sign of \(f''(x)\) changes.

Important: If \(f''(x)=0\), that does not always mean there is an inflection point. The concavity must actually change.

4. The Second Derivative Test

The Second Derivative Test helps classify a critical point.

Suppose \(c\) is a critical number, so \(f'(c)=0\). Then:

• If \(f''(c) > 0\), then \(f(c)\) is a local minimum.
• If \(f''(c) < 0\), then \(f(c)\) is a local maximum.
• If \(f''(c) = 0\), the test is inconclusive.

Why does this make sense?

• Concave up means the graph bends upward, so a critical point tends to be a valley: a local minimum.
• Concave down means the graph bends downward, so a critical point tends to be a hill: a local maximum.

5. A Step-by-Step Method for Concavity and Inflection Points

1. Differentiate the function to get \(f'(x)\).
2. Differentiate again to get \(f''(x)\).
3. Find values where \(f''(x)=0\) or does not exist.
4. Split the number line into intervals using those values.
5. Choose a test point in each interval and determine the sign of \(f''(x)\).
6. State where the function is concave up and concave down.
7. If the sign changes, find the corresponding point on the graph. That point is an inflection point.

6. Worked Example 1: Basic Concavity

Find where \(f(x)=x^3-3x^2+2\) is concave up, concave down, and find any inflection points.

Step 1: Find the first and second derivatives.

$$f'(x)=3x^2-6x$$ $$f''(x)=6x-6$$

Step 2: Find possible inflection values.

Set the second derivative equal to zero:

$$6x-6=0$$ $$x=1$$

Step 3: Test intervals around \(x=1\).

For \(x<1\), choose \(x=0\):

$$f''(0)=6(0)-6=-6<0$$

So the function is concave down on \((-infty,1)\).

For \(x>1\), choose \(x=2\):

$$f''(2)=6(2)-6=6>0$$

So the function is concave up on \((1,infty)\).

Step 4: Find the inflection point.

Since the sign of \(f''(x)\) changes at \(x=1\), there is an inflection point there.

Now find the \(y\)-value:

$$f(1)=1^3-3(1)^2+2=1-3+2=0$$

Answer:

• Concave down on \((-infty,1)\)
• Concave up on \((1,infty)\)
• Inflection point at \((1,0)\)

7. Worked Example 2: Using the Second Derivative Test

Use the Second Derivative Test to classify the critical points of

$$f(x)=x^3-6x^2+9x+1$$

Step 1: Find critical points.

$$f'(x)=3x^2-12x+9$$

Set \(f'(x)=0\):

$$3x^2-12x+9=0$$ $$x^2-4x+3=0$$ $$(x-1)(x-3)=0$$

So the critical points are \(x=1\) and \(x=3\).

Step 2: Find the second derivative.

$$f''(x)=6x-12$$

Step 3: Test each critical point.

At \(x=1\):

$$f''(1)=6(1)-12=-6<0$$

So \(x=1\) gives a local maximum.

At \(x=3\):

$$f''(3)=6(3)-12=6>0$$

So \(x=3\) gives a local minimum.

Step 4: Find the function values.

$$f(1)=1-6+9+1=5$$ $$f(3)=27-54+27+1=1$$

Answer:

• Local maximum at \((1,5)\)
• Local minimum at \((3,1)\)

8. Worked Example 3: A Case Where \(f''(x)=0\) Does Not Guarantee an Inflection Point

Study the concavity of

$$f(x)=x^4$$

Step 1: Find the second derivative.

$$f'(x)=4x^3$$ $$f''(x)=12x^2$$

Step 2: Solve \(f''(x)=0\).

$$12x^2=0$$ $$x=0$$

Step 3: Test the sign of \(f''(x)\).

Choose \(x=-1\):

$$f''(-1)=12(-1)^2=12>0$$

Choose \(x=1\):

$$f''(1)=12(1)^2=12>0$$

The second derivative is positive on both sides of \(x=0\).

So the function is concave up everywhere, except that \(f''(0)=0\).

Because the concavity does not change, \((0,0)\) is not an inflection point.

Important lesson: A value where \(f''(x)=0\) is only a possible inflection point.

9. Worked Example 4: Concavity for a Rational Function

Determine the concavity and inflection points of

$$f(x)=\frac{1}{x}=x^{-1}, \quad x\ne 0$$

Step 1: Find the derivatives.

$$f'(x)=-x^{-2}=-\frac{1}{x^2}$$ $$f''(x)=2x^{-3}=\frac{2}{x^3}$$

Step 2: Find where \(f''(x)=0\) or undefined.

\(f''(x)=\frac{2}{x^3}\) is never zero, but it is undefined at \(x=0\).

Step 3: Test intervals.

For \(x<0\), choose \(x=-1\):

$$f''(-1)=\frac{2}{(-1)^3}=-2<0$$

So the function is concave down on \((-infty,0)\).

For \(x>0\), choose \(x=1\):

$$f''(1)=\frac{2}{1^3}=2>0$$

So the function is concave up on \((0,infty)\).

Is there an inflection point at \(x=0\)?

No. Even though concavity changes, \(x=0\) is not in the domain of the function, so there is no point on the graph there. Therefore, there is no inflection point.

Answer:

• Concave down on \((-infty,0)\)
• Concave up on \((0,infty)\)
• No inflection point

10. How Concavity Helps in Graph Sketching

Concavity is very useful when sketching graphs accurately. A good graph sketch often uses all of the following:

• intercepts,
• intervals where the function increases or decreases,
• local maximum and minimum points,
• concave up and concave down intervals,
• inflection points.

If you know both the first and second derivatives, your sketch can show not only where the graph rises or falls, but also exactly how it bends.

11. Common Mistakes to Avoid

• Mistake 1: Thinking \(f''(x)=0\) always gives an inflection point.
  Always check whether the sign of \(f''(x)\) changes.
• Mistake 2: Forgetting to check where \(f''(x)\) is undefined.
  These values can split the number line into intervals for testing.
• Mistake 3: Using the Second Derivative Test when the point is not critical.
  You must first have \(f'(c)=0\) for the test.
• Mistake 4: Calling a value an inflection point when it is not on the graph.
  The function must be defined there.

12. Quick Checklist

When solving a concavity problem, ask yourself:

• Did I find \(f''(x)\) correctly?
• Did I solve \(f''(x)=0\) and check where \(f''(x)\) is undefined?
• Did I test the sign of \(f''(x)\) on each interval?
• Did I confirm that concavity actually changes for an inflection point?
• If using the Second Derivative Test, did I first find the critical points from \(f'(x)=0\)?

13. Brief Summary

The second derivative tells us how a graph bends.

• If \(f''(x)>0\), the graph is concave up.
• If \(f''(x)<0\), the graph is concave down.
• A point of inflection occurs where concavity changes.
• The Second Derivative Test classifies critical points as local maxima or minima.

These ideas are essential for optimization, motion problems, and precise graph sketching. Once you can use the second derivative confidently, you can understand the shape of a function much more completely.

Curve Sketching Using Calculus is the process of using algebra and derivatives to draw an accurate graph of a function without relying on graphing technology.

Instead of guessing the shape, we use a sequence of mathematical facts: the domain, intercepts, symmetry, limits, asymptotes, increasing and decreasing intervals, local maximum and minimum points, concavity, and points of inflection.

This topic combines several important ideas from calculus. When all these ideas are used together, we can produce a graph that is precise and well-supported by reasoning.

Why this matters: Curve sketching helps you understand how a function behaves globally, not just at a few points. It is also closely connected to optimization, motion, and limit behavior.

1. The main goal of curve sketching

When sketching a curve, we want to answer questions like these:

• Where is the function defined?
• Where does the graph cross the axes?
• What happens near values where the function is undefined or very large?
• Where is the function increasing or decreasing?
• Where are the turning points?
• Where is the graph concave up or concave down?
• Are there any points of inflection?

Once we know these features, we can combine them into a reliable sketch.

2. A step-by-step method for sketching a curve

For a function \(f(x)\), the following order is very useful.

1. Find the domain.
2. Find intercepts.
   • For the \(y\)-intercept, compute \(f(0)\) if it exists.
   • For the \(x\)-intercepts, solve \(f(x)=0\).
3. Check for symmetry if helpful.
   • If \(f(-x)=f(x)\), the function is even and symmetric about the \(y\)-axis.
   • If \(f(-x)=-f(x)\), the function is odd and symmetric about the origin.
4. Study end behavior and limits.
   • Look at \(x \to \infty\), \(x \to -\infty\), and any points where the function is undefined.
   • This helps identify vertical, horizontal, or slant asymptotes.
5. Find the first derivative \(f'(x)\).
6. Find critical numbers.
   • These are values of \(x\) where \(f'(x)=0\) or where \(f'(x)\) does not exist, provided \(x\) is in the domain.
7. Use the sign of \(f'(x)\) to find increasing and decreasing intervals.
8. Identify local maximum and minimum points.
9. Find the second derivative \(f''(x)\).
10. Use the sign of \(f''(x)\) to determine concavity.
    • If \(f''(x)>0\), the graph is concave up.
    • If \(f''(x)<0\), the graph is concave down.
11. Find possible inflection points.
    • These occur where concavity changes.
    • Usually test values around where \(f''(x)=0\) or where \(f''(x)\) does not exist.
12. Combine all information into one sketch.

3. Important ideas in detail

Domain

The domain tells us which \(x\)-values are allowed. For example:

• A denominator cannot be zero.
• An even root, such as \(\sqrt{x}\), requires the inside to be nonnegative.
• Logarithms require a positive input.

If a value is not in the domain, the graph cannot exist there.

Intercepts

Intercepts are often the easiest points to plot.

• To find an \(x\)-intercept, set \(y=0\), so solve \(f(x)=0\).
• To find a \(y\)-intercept, set \(x=0\), so compute \(f(0)\).

Limits and asymptotes

Limits describe what the function approaches.

• If \(f(x)\to \infty\) or \(-\infty\) as \(x\to a\), then \(x=a\) is a vertical asymptote.
• If \(f(x)\to L\) as \(x\to \pm\infty\), then \(y=L\) is a horizontal asymptote.

These features strongly affect the shape of the graph.

Increasing and decreasing

The first derivative measures slope.

• If \(f'(x)>0\), the function is increasing.
• If \(f'(x)<0\), the function is decreasing.

A change from positive to negative usually means a local maximum. A change from negative to positive usually means a local minimum.

Concavity

The second derivative tells us how the slope is changing.

• If \(f''(x)>0\), slopes are increasing, so the graph bends upward: concave up.
• If \(f''(x)<0\), slopes are decreasing, so the graph bends downward: concave down.

Points of inflection

A point of inflection is a point where concavity changes. It is not enough to find where \(f''(x)=0\); you must also check that the sign of \(f''(x)\) actually changes.

4. Organizing information with a sign chart

A sign chart is a very useful tool. First, place critical numbers on a number line. Then test the sign of \(f'(x)\) in each interval.

For concavity, place values where \(f''(x)=0\) or undefined on a number line, and test the sign of \(f''(x)\) in each interval.

This makes it much easier to avoid mistakes.

Worked Example 1: A polynomial

Sketch the curve of

$$f(x)=x^3-3x^2+2$$

Step 1: Domain

This is a polynomial, so the domain is all real numbers:

$$(-\infty,\infty)$$

Step 2: Intercepts

The \(y\)-intercept is

$$f(0)=2$$

So the graph crosses the \(y\)-axis at \((0,2)\).

For the \(x\)-intercepts, solve

$$x^3-3x^2+2=0$$

Test simple roots. Since \(x=1\) works, factor:

$$x^3-3x^2+2=(x-1)(x^2-2x-2)$$

Then solve

$$x^2-2x-2=0$$

Using the quadratic formula:

$$x=\frac{2\pm\sqrt{4+8}}{2}=1\pm\sqrt{3}$$

So the \(x\)-intercepts are

$$x=1-\sqrt{3},\quad x=1,\quad x=1+\sqrt{3}$$

Step 3: First derivative

$$f'(x)=3x^2-6x=3x(x-2)$$

Critical numbers occur when \(f'(x)=0\):

$$3x(x-2)=0 \Rightarrow x=0,2$$

Step 4: Increasing/decreasing

Test the sign of \(f'(x)\):

• For \(x<0\), choose \(-1\): \(f'(-1)=3(-1)(-3)>0\), so increasing.
• For \(0<x<2\), choose \(1\): \(f'(1)=3(1)(-1)<0\), so decreasing.
• For \(x>2\), choose \(3\): \(f'(3)=3(3)(1)>0\), so increasing.

Therefore:

• Increasing on \(( -\infty,0)\) and \((2,\infty)\)
• Decreasing on \((0,2)\)

Step 5: Local extrema

At \(x=0\), the sign changes from positive to negative, so there is a local maximum.

$$f(0)=2$$

Local maximum at \((0,2)\).

At \(x=2\), the sign changes from negative to positive, so there is a local minimum.

$$f(2)=8-12+2=-2$$

Local minimum at \((2,-2)\).

Step 6: Second derivative

$$f''(x)=6x-6$$

Set it equal to zero:

$$6x-6=0 \Rightarrow x=1$$

Step 7: Concavity

• For \(x<1\), \(f''(x)<0\): concave down.
• For \(x>1\), \(f''(x)>0\): concave up.

Since concavity changes at \(x=1\), there is an inflection point.

$$f(1)=1-3+2=0$$

Inflection point at \((1,0)\).

Step 8: End behavior

Because the leading term is \(x^3\):

• As \(x\to -\infty\), \(f(x)\to -\infty\)
• As \(x\to \infty\), \(f(x)\to \infty\)

Conclusion for the sketch

The graph rises from the lower left, reaches a local maximum at \((0,2)\), falls through the inflection point \((1,0)\), reaches a local minimum at \((2,-2)\), and then rises to the upper right.

Worked Example 2: A rational function with asymptotes

Sketch the curve of

$$f(x)=\frac{x}{x^2-1}$$

Step 1: Domain

The denominator cannot be zero:

$$x^2-1\neq 0 \Rightarrow x\neq \pm 1$$

So the domain is

$$(-\infty,-1)\cup(-1,1)\cup(1,\infty)$$

Step 2: Intercepts

For the \(y\)-intercept:

$$f(0)=0$$

So the graph passes through \((0,0)\).

For the \(x\)-intercept, solve

$$\frac{x}{x^2-1}=0$$

This happens when \(x=0\), so the only \(x\)-intercept is \((0,0)\).

Step 3: Symmetry

Compute

$$f(-x)=\frac{-x}{x^2-1}=-f(x)$$

So the function is odd and symmetric about the origin.

Step 4: Vertical and horizontal asymptotes

Since the denominator is zero at \(x=\pm1\), there are vertical asymptotes:

$$x=-1,\quad x=1$$

Since the degree of the denominator is greater than the degree of the numerator,

$$\lim_{x\to\pm\infty} \frac{x}{x^2-1}=0$$

So the horizontal asymptote is

$$y=0$$

Step 5: First derivative

Using the quotient rule:

$$f'(x)=\frac{(x^2-1)(1)-x(2x)}{(x^2-1)^2}$$

$$f'(x)=\frac{x^2-1-2x^2}{(x^2-1)^2}=\frac{-(x^2+1)}{(x^2-1)^2}$$

Notice that \(x^2+1>0\) for all real \(x\), and the denominator squared is positive wherever defined. Therefore:

$$f'(x)<0$$

for all \(x\) in the domain.

So the function is decreasing on each interval of its domain:

• \(( -\infty,-1)\)
• \((-1,1)\)
• \((1,\infty)\)

There are no local maxima or minima because \(f'(x)\) is never zero.

Step 6: Second derivative

Differentiate

$$f'(x)=-(x^2+1)(x^2-1)^{-2}$$

This gives

$$f''(x)=\frac{2x(x^2+3)}{(x^2-1)^3}$$

Step 7: Concavity

Since \(x^2+3>0\), the sign depends on

$$\frac{x}{(x^2-1)^3}$$

Test intervals:

• On \(( -\infty,-1)\), \(f''(x)<0\): concave down.
• On \((-1,0)\), \(f''(x)>0\): concave up.
• On \((0,1)\), \(f''(x)<0\): concave down.
• On \((1,\infty)\), \(f''(x)>0\): concave up.

At \(x=0\), the function is defined and concavity changes, so \((0,0)\) is a point of inflection.

Conclusion for the sketch

The graph has three branches, vertical asymptotes at \(x=-1\) and \(x=1\), horizontal asymptote \(y=0\), passes through the origin, is decreasing on each domain interval, and changes concavity at the origin.

Worked Example 3: A function with one-sided limit behavior

Sketch the curve of

$$f(x)=x+\frac{1}{x}$$

Step 1: Domain

Since division by zero is not allowed, \(x\neq 0\). So the domain is

$$(-\infty,0)\cup(0,\infty)$$

Step 2: Intercepts

There is no \(y\)-intercept because \(f(0)\) is undefined.

For \(x\)-intercepts, solve

$$x+\frac{1}{x}=0$$

Multiply by \(x\):

$$x^2+1=0$$

This has no real solution, so there are no \(x\)-intercepts.

Step 3: Symmetry

$$f(-x)=-x-\frac{1}{x}=-f(x)$$

So the function is odd and symmetric about the origin.

Step 4: Asymptotes and limits

Since \(x=0\) is not in the domain and

$$\lim_{x\to 0^+}\left(x+\frac{1}{x}\right)=\infty$$

$$\lim_{x\to 0^-}\left(x+\frac{1}{x}\right)=-\infty$$

there is a vertical asymptote at

$$x=0$$

As \(x\to\infty\), \(1/x\to 0\), so \(f(x)\approx x\). As \(x\to -\infty\), the same idea holds. The graph behaves like the line \(y=x\) for large \(|x|\).

Step 5: First derivative

$$f'(x)=1-\frac{1}{x^2}=\frac{x^2-1}{x^2}$$

Critical numbers occur when

$$x^2-1=0 \Rightarrow x=\pm 1$$

Step 6: Increasing/decreasing

Since \(x^2>0\) for \(x\neq0\), the sign depends on \(x^2-1\).

• If \(|x|>1\), then \(f'(x)>0\), so increasing.
• If \(0<|x|<1\), then \(f'(x)<0\), so decreasing.

Therefore:

• Increasing on \(( -\infty,-1)\) and \((1,\infty)\)
• Decreasing on \((-1,0)\) and \((0,1)\)

Step 7: Local extrema

At \(x=-1\), the sign changes from positive to negative, so there is a local maximum.

$$f(-1)=-1-1=-2$$

Local maximum at \((-1,-2)\).

At \(x=1\), the sign changes from negative to positive, so there is a local minimum.

$$f(1)=1+1=2$$

Local minimum at \((1,2)\).

Step 8: Second derivative

$$f''(x)=\frac{2}{x^3}$$

Step 9: Concavity

• If \(x<0\), then \(f''(x)<0\): concave down.
• If \(x>0\), then \(f''(x)>0\): concave up.

Although concavity changes across \(x=0\), \(x=0\) is not in the domain, so there is no inflection point there.

Conclusion for the sketch

The left branch rises, reaches a local maximum at \((-1,-2)\), then falls toward \(-\infty\) as \(x\to0^-\). The right branch falls from \(+\infty\) as \(x\to0^+\), reaches a local minimum at \((1,2)\), and then rises again.

5. How limits improve the accuracy of a sketch

Sometimes a graph can look similar in the middle but behave very differently near infinity or near undefined points. Limits tell us exactly what the function does in these regions.

For example, a rational function may flatten toward a horizontal asymptote, while another may grow without bound. A function with a denominator may shoot up or down near a vertical asymptote. These features must appear in a careful sketch.

6. Common mistakes to avoid

• Forgetting the domain. A graph cannot pass through points where the function is undefined.
• Assuming \(f''(x)=0\) always gives an inflection point. Concavity must actually change.
• Using only one test point when the domain is split. If asymptotes divide the domain, test each interval separately.
• Missing intercepts because of factoring mistakes. Solve carefully.
• Drawing turning points at critical numbers without checking signs. A critical number is not always a maximum or minimum.
• Forgetting one-sided limits near vertical asymptotes. Left and right behavior can be different.

7. A quick checklist for any curve sketch

• Domain
• Intercepts
• Symmetry
• Limits and asymptotes
• Critical numbers from \(f'(x)\)
• Increasing/decreasing intervals
• Local maxima/minima
• Concavity from \(f''(x)\)
• Inflection points
• Final sketch combining all features

8. Final summary

Curve sketching using calculus means building a graph from mathematical evidence. We begin with the domain, intercepts, and limit behavior, then use the first derivative to study monotonicity and extrema, and the second derivative to study concavity and inflection points.

When these ideas are combined carefully, you can draw highly accurate graphs without technology. The most important habit is to work step by step and justify each feature of the curve with algebra or calculus.

Applied Optimization is about using calculus to make the “best” choice in a real situation. In many problems, you are asked to find a maximum value, such as the largest volume, or a minimum value, such as the smallest cost or shortest distance.

These questions appear in geometry, business, physics, and engineering. The main idea is to turn a real-world situation into a mathematical model, then use derivatives to locate the best value.

In this lesson, you will learn how to:

• identify the quantity being optimized,
• write the important constraint equation,
• rewrite the quantity as a function of one variable,
• use derivatives to find critical points,
• decide whether the result is a maximum or minimum,
• interpret the answer in the context of the problem.

1. The general strategy for optimization

Most optimization problems follow the same pattern. If you stay organized, they become much easier.

1. Read carefully. Decide what quantity must be made as large or as small as possible.
2. Name variables. Draw a diagram if the problem involves geometry.
3. Write a constraint equation. This is the condition the variables must satisfy.
4. Write the objective function. This is the quantity you want to maximize or minimize.
5. Use the constraint to rewrite the objective in one variable.
6. Differentiate. Find critical points by solving \(f'(x)=0\) or checking where \(f'(x)\) is undefined.
7. Test the critical points. Use the first derivative test, second derivative test, or compare values at endpoints.
8. State the answer clearly. Include units and explain what the value means.

2. Important ideas to remember

Objective function: the quantity you want to optimize. Examples: area, volume, cost, time, distance.

Constraint: a relationship that limits the variables. Examples: fixed perimeter, fixed surface area, fixed amount of material.

Critical point: a value of the variable where the derivative is zero or undefined.

Feasible domain: the set of values that actually make sense in the problem. For example, a length cannot be negative.

In real-world optimization, the derivative gives candidates for the best value, but you must always check whether the answer fits the situation. A negative length or a value outside the allowed interval is not acceptable.

3. Why optimization uses derivatives

If a function has a local maximum or minimum at an interior point, its tangent line is often horizontal there. That means the derivative is zero.

So, if \(f(x)\) models the quantity we care about, then possible best values usually happen at:

• critical points where \(f'(x)=0\),
• points where \(f'(x)\) does not exist,
• endpoints of the allowed interval.

4. Worked Example 1: Maximum area with fixed perimeter

Problem: A rectangle has perimeter \(40\) m. What dimensions give the maximum area?

Step 1: Define variables.

Let the length be \(l\) and the width be \(w\).

Step 2: Write the constraint.

The perimeter is fixed:

$$2l+2w=40$$

Divide by 2:

$$l+w=20$$

So:

$$w=20-l$$

Step 3: Write the objective function.

The area is:

$$A=lw$$

Substitute \(w=20-l\):

$$A(l)=l(20-l)=20l-l^2$$

Step 4: Differentiate.

$$A'(l)=20-2l$$

Set the derivative equal to zero:

$$20-2l=0$$

$$l=10$$

Step 5: Find the other dimension.

$$w=20-10=10$$

Step 6: Confirm maximum.

$$A''(l)=-2$$

Since \(A''(l)<0\), the function is concave down, so this critical point gives a maximum.

Answer: The rectangle with maximum area is a square with dimensions \(10\text{ m} \times 10\text{ m}\).

Key idea: With a fixed perimeter, the rectangle of greatest area is a square.

5. Worked Example 2: Minimum cost of fencing

Problem: A farmer wants to build a rectangular pen beside a river. No fence is needed along the river. The farmer has \(200\) m of fencing for the other three sides. What dimensions minimize nothing? Actually, with fixed fencing, the natural question is to maximize area. Let us solve that useful optimization problem.

Step 1: Define variables.

Let \(x\) be the width perpendicular to the river, and \(y\) be the length parallel to the river.

Step 2: Write the constraint.

Only three sides are fenced, so:

$$2x+y=200$$

Thus:

$$y=200-2x$$

Step 3: Write the objective function.

The area is:

$$A=xy$$

Substitute:

$$A(x)=x(200-2x)=200x-2x^2$$

Step 4: Differentiate and find critical points.

$$A'(x)=200-4x$$

Set equal to zero:

$$200-4x=0$$

$$x=50$$

Step 5: Find \(y\).

$$y=200-2(50)=100$$

Step 6: Check maximum.

$$A''(x)=-4$$

Since \(A''(x)<0\), the area is maximized.

Answer: The pen should be \(50\) m wide and \(100\) m long.

6. Worked Example 3: Maximum volume of an open-top box

Problem: A square piece of cardboard has side length \(24\) cm. Squares of side length \(x\) are cut from each corner, and the sides are folded up to make an open-top box. What value of \(x\) gives the maximum volume?

Step 1: Understand the geometry.

Cutting out squares of side \(x\) leaves a box with:

• height \(x\),
• length \(24-2x\),
• width \(24-2x\).

Step 2: Write the objective function.

Volume:

$$V(x)=x(24-2x)^2$$

Step 3: Determine the domain.

We need positive dimensions:

$$x>0 \quad \text{and} \quad 24-2x>0$$

So:

$$0<x<12$$

Step 4: Expand before differentiating.

$$V(x)=x(576-96x+4x^2)$$

$$V(x)=576x-96x^2+4x^3$$

Step 5: Differentiate.

$$V'(x)=576-192x+12x^2$$

Set equal to zero:

$$576-192x+12x^2=0$$

Divide by 12:

$$48-16x+x^2=0$$

$$x^2-16x+48=0$$

Factor:

$$ (x-4)(x-12)=0 $$

So the critical points are \(x=4\) and \(x=12\).

Step 6: Check the feasible domain.

The domain is \(0<x<12\), so \(x=12\) is not allowed because it would make the base length zero. Only \(x=4\) works.

Step 7: State the result.

When \(x=4\):

• height \(=4\) cm,
• length \(=24-8=16\) cm,
• width \(=16\) cm.

The maximum volume is:

$$V(4)=4(16)^2=1024\text{ cm}^3$$

Answer: Cut out squares of side length \(4\) cm.

7. Worked Example 4: Minimum distance to a point

Problem: Find the point on the graph \(y=x^2\) that is closest to the point \((0,3)\).

This is a classic minimum-distance optimization problem. The distance formula includes a square root, but it is usually easier to minimize the square of the distance. Since square root is increasing, the value that minimizes distance also minimizes distance squared.

Step 1: Use a general point on the curve.

A point on \(y=x^2\) is \((x,x^2)\).

Step 2: Write the distance squared.

The distance from \((x,x^2)\) to \((0,3)\) is:

$$d=\sqrt{(x-0)^2+(x^2-3)^2}$$

So minimize:

$$D(x)=d^2=x^2+(x^2-3)^2$$

Step 3: Simplify.

$$D(x)=x^2+x^4-6x^2+9$$

$$D(x)=x^4-5x^2+9$$

Step 4: Differentiate.

$$D'(x)=4x^3-10x$$

Factor:

$$D'(x)=2x(2x^2-5)$$

Set equal to zero:

$$2x(2x^2-5)=0$$

So:

$$x=0 \quad \text{or} \quad x=\pm\sqrt{\frac{5}{2}}$$

Step 5: Compare values.

At \(x=0\):

$$D(0)=9$$

At \(x=\pm\sqrt{\frac{5}{2}}\):

$$x^2=\frac{5}{2}$$

$$D=\left(\frac{5}{2}\right)^2-5\left(\frac{5}{2}\right)+9$$

$$D=\frac{25}{4}-\frac{25}{2}+9$$

$$D=\frac{25}{4}-\frac{50}{4}+\frac{36}{4}=\frac{11}{4}$$

Since \(\frac{11}{4} < 9\), the minimum occurs at:

$$x=\pm\sqrt{\frac{5}{2}}$$

Step 6: Find the points.

Because \(y=x^2\):

$$y=\frac{5}{2}$$

Answer: The points on the parabola closest to \((0,3)\) are

$$\left(\sqrt{\frac{5}{2}},\frac{5}{2}\right) \quad \text{and} \quad \left(-\sqrt{\frac{5}{2}},\frac{5}{2}\right)$$

8. Optimization with cost models

Some applied optimization problems involve money instead of geometry. The process is still the same: define variables, write a model, differentiate, and test.

For example, suppose the cost of producing \(x\) items is \(C(x)\). If you want the lowest possible cost per item, you might minimize the average cost:

$$A(x)=\frac{C(x)}{x}$$

If the cost of materials, labor, or shipping depends on dimensions, then the objective function might include several terms. Your job is to combine them into one formula and optimize it.

Example idea: If a box has different material costs for the base and sides, then total cost depends on area of each part. You would write the cost function using those areas, then use the given volume or size constraint to reduce it to one variable.

9. Common mistakes in optimization problems

• Optimizing the wrong quantity. Always ask: What exactly must be maximized or minimized?
• Forgetting the constraint. Most problems cannot be solved until one variable is written in terms of another.
• Using too many variables. The objective function should usually end up with just one variable.
• Ignoring the domain. A critical point may be impossible in the real problem.
• Not checking whether the result is a maximum or minimum. A critical point alone is not enough.
• Forgetting units. Final answers should include meters, square centimeters, dollars, and so on.

10. How optimization connects to other derivative applications

Applied optimization is one part of the larger topic Applications of the Derivative. It connects closely with:

• rectilinear motion, where derivatives describe velocity and acceleration,
• approximate limits and local linear ideas, where derivatives describe change near a point,
• curve sketching, where first and second derivatives help identify maxima, minima, and concavity.

In fact, optimization often uses the same tools as curve sketching:

• first derivative to find increasing or decreasing behavior,
• second derivative to test concavity and classify critical points.

11. A quick checklist for solving any optimization problem

• What quantity am I optimizing?
• What variables should I define?
• What equation gives the constraint?
• Can I rewrite the objective as a function of one variable?
• What is the realistic domain?
• What are the critical points?
• Which point gives the maximum or minimum?
• Does my final answer make sense in the context?

12. Brief summary

Applied optimization uses derivatives to find the best possible value of a real-world quantity. The key steps are to identify the objective, write a constraint, reduce the model to one variable, and use derivatives to locate maxima or minima.

Whether the problem asks for maximum area, maximum volume, minimum distance, or lowest cost, the same method works. Careful setup is usually the most important part of the problem.