Function Theory and Transformations

Lesson: Relations and Mappings

In mathematics, we often want to describe how one quantity is connected to another. For example, the number of hours studied may be connected to test scores, or the side length of a square may be connected to its area. These connections are called relations and, in special cases, functions.

This lesson will help you understand what relations and mappings are, how to represent them using sets, tables, and graphs, and how to decide whether a relation is a function. A very important tool for graphs is the vertical line test.

1. What is a relation?

A relation is any set of ordered pairs. An ordered pair is written as \\( (x, y) \\), where the first value is the input and the second value is the output.

For example, the set

$$\{(1,2),(2,4),(3,6)\}$$

is a relation. It connects each first number to a second number.

You can think of a relation as a rule or connection between elements of two sets:

• the input values
• the output values

2. Domain and range

Every relation has a domain and a range.

• Domain: the set of all input values, or all first coordinates
• Range: the set of all output values, or all second coordinates

For the relation

$$\{(1,2),(2,4),(3,6)\}$$

the domain is

$$\{1,2,3\}$$

and the range is

$$\{2,4,6\}$$

If a value repeats, we only write it once in the set. Sets do not list duplicates.

3. What is a mapping?

A mapping shows how each input is paired with an output. It is often drawn using a mapping diagram, where values in one set have arrows pointing to values in another set.

For example, if the relation is

$$\{(1,3),(2,5),(4,5)\}$$

then the mapping is:

• \\(1 \to 3\\)
• \\(2 \to 5\\)
• \\(4 \to 5\\)

A mapping helps us see clearly which inputs are connected to which outputs.

4. Different ways to represent relations

Relations can be shown in several ways. You should be comfortable moving between these forms.

1. Ordered pairs: \\( (x,y) \\)
2. Tables: list inputs and outputs in rows or columns
3. Mapping diagrams: arrows from domain to range
4. Graphs: points drawn on the coordinate plane

All of these can describe the same relation.

5. What is a function?

A function is a special type of relation.

In a function, each input has exactly one output.

This means:

• an input can point to only one output
• different inputs can point to the same output
• but one input cannot point to two different outputs

So every function is a relation, but not every relation is a function.

For example,

$$\{(1,4),(2,5),(3,6)\}$$

is a function because each input appears only once.

But

$$\{(1,4),(1,5),(2,6)\}$$

is not a function because the input \\(1\\) has two outputs: \\(4\\) and \\(5\\).

6. How to tell if a relation is a function

There are different ways to check.

Using ordered pairs or a table:

Look at the input values. If any input is matched with more than one output, the relation is not a function.

Using a mapping diagram:

Check the arrows from each input. If one input has two arrows going to different outputs, it is not a function.

Using a graph:

Use the vertical line test.

7. The vertical line test

The vertical line test is used to decide if a graph represents a function.

Rule: If any vertical line crosses the graph at more than one point, then the graph is not a function.

This works because a vertical line represents one fixed input value \\(x\\). If that line hits the graph more than once, then the same input has more than one output.

For example:

• A straight line like \\(y = 2x + 1\\) passes the vertical line test, so it is a function.
• A circle like \\(x^2 + y^2 = 9\\) fails the vertical line test, so it is not a function.

For the circle, some \\(x\\)-values have two different \\(y\\)-values, one above the \\(x\\)-axis and one below it.

8. One-to-one, many-to-one, and one-to-many

Mappings can be described in different ways.

• One-to-one: each input has one unique output, and no outputs are shared
• Many-to-one: different inputs share the same output
• One-to-many: one input has more than one output

Only the first two can be functions. A one-to-many mapping is not a function.

Example of many-to-one:

$$\{(1,5),(2,5),(3,7)\}$$

This is a function because each input still has only one output.

Example of one-to-many:

$$\{(2,1),(2,4),(3,5)\}$$

This is not a function because input \\(2\\) has two outputs.

9. Worked Example 1: Find domain and range

Consider the relation

$$\{(-1,2),(0,3),(2,3),(4,7)\}$$

Step 1: Identify the domain.

The domain is the set of first coordinates:

$$\{-1,0,2,4\}$$

Step 2: Identify the range.

The range is the set of second coordinates:

$$\{2,3,7\}$$

Notice that \\(3\\) appears twice in the ordered pairs, but it is written only once in the range.

Answer:

• Domain: \\(\{-1,0,2,4\}\\)
• Range: \\(\{2,3,7\}\\)

10. Worked Example 2: Decide whether a relation is a function

Determine whether

$$\{(1,2),(2,4),(3,6),(2,5)\}$$

is a function.

Step 1: Look at the inputs.

The inputs are \\(1, 2, 3, 2\\).

Step 2: Check whether any input has more than one output.

The input \\(2\\) is paired with \\(4\\) and also with \\(5\\).

Conclusion: This relation is not a function.

Reason: One input has two different outputs.

11. Worked Example 3: Read a mapping diagram

Suppose a mapping diagram shows:

• \\( -2 \to 1 \\)
• \\( 0 \to 1 \\)
• \\( 3 \to 4 \\)

a) Write the relation as ordered pairs.

The relation is

$$\{(-2,1),(0,1),(3,4)\}$$

b) Is it a function?

Yes. Each input has exactly one output.

c) Is it many-to-one?

Yes. The inputs \\(-2\\) and \\(0\\) both map to \\(1\\).

12. Worked Example 4: Use the vertical line test

We compare two graphs.

Graph A: a parabola opening upward, such as

$$y = x^2$$

Any vertical line crosses this graph at only one point.

So Graph A is a function.

Graph B: a circle, such as

$$x^2 + y^2 = 4$$

A vertical line through \\(x=1\\) crosses the circle at two points.

So Graph B is not a function.

13. Important ideas to remember

• A relation is any set of ordered pairs.
• A function is a relation where each input has exactly one output.
• The domain is the set of inputs.
• The range is the set of outputs.
• Repeated outputs are allowed in a function.
• Repeated inputs are allowed only if they always give the same output.
• On a graph, use the vertical line test to decide whether the relation is a function.

14. Common mistakes

• Mistake 1: Thinking that different inputs cannot have the same output. They can. That can still be a function.
• Mistake 2: Looking at repeated outputs instead of repeated inputs when checking for a function.
• Mistake 3: Forgetting that the vertical line test applies to graphs, not tables or lists.
• Mistake 4: Writing repeated values more than once in the domain or range when using set notation.

15. Brief summary

Relations and mappings describe connections between inputs and outputs. A function is a special relation in which every input has exactly one output. You can represent relations using ordered pairs, tables, mappings, and graphs, and you can use the vertical line test to decide from a graph whether a relation is a function.

Domain and Range Restrictions are about figuring out which input values a function is allowed to use, and which output values it can produce.

In 11th Grade Maths, this is an important skill because many functions look fine at first, but actually have values that are not allowed. These restrictions usually come from three common sources:

• Denominators: you cannot divide by zero.
• Even roots: you cannot take the square root, fourth root, and so on of a negative number if you are working in the real numbers.
• Logarithms: the value inside a logarithm must be positive.

This lesson will show you how to find the maximal real domain of a function. That means the largest possible set of real input values for which the function makes sense.

We will also look at how domain restrictions can affect the range, which is the set of possible output values.

Key definitions:

• Domain: the set of all allowed input values, usually the possible values of \(x\).
• Range: the set of all output values, usually the possible values of \(y\) or \(f(x)\).
• Restriction: a rule that excludes some values because the expression would not be defined in the real numbers.

When finding the domain, always ask: What would make this function invalid?

The three main checks are:

1. Does any denominator become zero?
2. Is there an even root with a negative value inside?
3. Is there a logarithm with a non-positive value inside?

1. Restrictions from denominators

If a function has a denominator, that denominator cannot equal zero.

For example, in the function

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

the denominator is \(x-4\). So we must avoid

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

This means the domain is all real numbers except \(4\).

We can write this as:

$$\text{Domain}: x\in \mathbb{R},\; x\ne 4$$

2. Restrictions from even roots

If a function contains an even root such as a square root, the expression inside must be greater than or equal to zero.

For example, in

$$g(x)=\sqrt{2x+6}$$

the expression under the root is \(2x+6\). So we need

$$2x+6\ge 0$$ $$2x\ge -6$$ $$x\ge -3$$

So the domain is:

$$\text{Domain}: x\ge -3$$

Notice that zero is allowed inside a square root, because \(\sqrt{0}=0\).

3. Restrictions from logarithms

If a function contains a logarithm, the expression inside the log must be strictly greater than zero.

For example, in

$$h(x)=\log(x+1)$$

we need:

$$x+1>0$$ $$x>-1$$

So the domain is:

$$\text{Domain}: x>-1$$

This is different from square roots. For logs, the inside cannot be zero. It must be positive.

How restrictions affect range

Sometimes the domain restriction directly changes what output values are possible.

For example, if

$$f(x)=\sqrt{x-2}$$

then the domain is \(x\ge 2\), because:

$$x-2\ge 0$$ $$x\ge 2$$

Now think about the outputs. A square root always gives a value that is zero or positive. So the range is:

$$\text{Range}: y\ge 0$$

In other cases, a value missing from the domain can also create a value missing from the range.

For example, in

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

the domain excludes \(x=3\). But the outputs also can never equal zero, because a fraction with numerator 1 can never become 0.

So the range is:

$$\text{Range}: y\in \mathbb{R},\; y\ne 0$$

Strategy for finding the maximal real domain

Use this step-by-step method:

1. Write down the function clearly.
2. Look for any denominator and set it not equal to zero.
3. Look for any even root and require the inside to be at least zero.
4. Look for any logarithm and require the inside to be greater than zero.
5. If there is more than one restriction, combine them.

Now let us work through some examples.

Worked Example 1: Denominator restriction

Find the domain of

$$f(x)=\frac{5}{x+2}$$

Step 1: Identify the denominator.

The denominator is \(x+2\).

Step 2: Denominator cannot be zero.

$$x+2\ne 0$$ $$x\ne -2$$

Answer:

$$\text{Domain}: x\in \mathbb{R},\; x\ne -2$$

Range: Since the numerator is 5, the function can never equal 0.

$$\text{Range}: y\in \mathbb{R},\; y\ne 0$$

Worked Example 2: Even root restriction

Find the domain and range of

$$g(x)=\sqrt{7-x}$$

Step 1: Require the expression inside the square root to be non-negative.

$$7-x\ge 0$$ $$-x\ge -7$$ $$x\le 7$$

Domain:

$$\text{Domain}: x\le 7$$

Step 2: Find the range.

A square root output is always non-negative, so:

$$y\ge 0$$

Also, as \(x\) gets smaller, \(7-x\) gets larger, so the square root can become as large as we want. Therefore:

$$\text{Range}: y\ge 0$$

Worked Example 3: Logarithm restriction

Find the domain of

$$h(x)=\ln(3x-12)$$

Step 1: The inside of the logarithm must be positive.

$$3x-12>0$$ $$3x>12$$ $$x>4$$

Answer:

$$\text{Domain}: x>4$$

About the range: A logarithm can produce any real number, so the range is all real numbers.

$$\text{Range}: y\in \mathbb{R}$$

Worked Example 4: Combined restrictions

Find the maximal real domain of

$$p(x)=\frac{\sqrt{x-1}}{x-5}$$

This function has two different restrictions.

Step 1: Square root restriction.

$$x-1\ge 0$$ $$x\ge 1$$

Step 2: Denominator restriction.

$$x-5\ne 0$$ $$x\ne 5$$

Step 3: Combine both conditions.

The function needs \(x\ge 1\), but also cannot use \(x=5\).

Answer:

$$\text{Domain}: x\ge 1,\; x\ne 5$$

In interval form, this is:

$$[1,5)\cup(5,\infty)$$

Common mistakes to avoid

• Forgetting the denominator restriction. Always check if any value makes the denominator zero.
• Using the wrong inequality for square roots. For even roots, use \(\ge 0\), not \(>0\).
• Using the wrong inequality for logs. For logarithms, use \(>0\), not \(\ge 0\).
• Stopping after finding one restriction. A function can have more than one restriction.
• Mixing up domain and range. Domain is about inputs; range is about outputs.

Quick comparison table

• For \(\frac{1}{expression}\): require \(expression\ne 0\)
• For \(\sqrt{expression}\): require \(expression\ge 0\)
• For \(\log(expression)\): require \(expression>0\)

Practice thinking

If you see a function like

$$f(x)=\log(\sqrt{x-2})$$

you must apply both ideas carefully.

First, because of the square root:

$$x-2\ge 0$$ $$x\ge 2$$

But the square root is inside a logarithm, so it must also be strictly positive:

$$\sqrt{x-2}>0$$

This means:

$$x-2>0$$ $$x>2$$

So the domain is:

$$\text{Domain}: x>2$$

This example shows why it is important to think about every part of the function.

Summary

To find the maximal real domain of a function, look for values that would make the function undefined. Denominators cannot be zero, even roots need non-negative values inside, and logarithms need positive values inside.

Once the domain is known, you can often determine the range by thinking about what outputs the function can actually produce. With practice, checking domain restrictions becomes a quick and reliable habit whenever you work with functions.

Function Notation and Evaluation is a way of writing rules that connect inputs and outputs. In 11th Grade Maths, functions are used to model patterns, relationships, and change. Understanding the notation helps you read a function correctly, substitute values accurately, and work with expressions such as \(f(x+h)\).

A function takes an input and gives exactly one output. If a function is named \(f\), then \(f(x)\) means “the output of function \(f\) when the input is \(x\).” The letter inside the parentheses tells you what is being used as the input.

For example, if \(f(x)=2x+3\), then the rule is “multiply the input by 2, then add 3.” If the input is 4, then:

$$f(4)=2(4)+3=8+3=11$$

This shows that function notation is not multiplication. The expression \(f(4)\) does not mean \(f \times 4\). It means “evaluate the function \(f\) at 4.”

There are two main skills in this topic:

• Reading and interpreting function notation
• Evaluating a function by substituting a value or expression into the rule

1. Reading function notation

If you see \(f(x)=x^2-5x+6\), then:

• \(f\) is the name of the function
• \(x\) is the input variable
• \(x^2-5x+6\) is the rule that gives the output

If a different letter is used inside the parentheses, the rule stays the same. For example, if:

$$f(x)=x^2-5x+6$$

then:

$$f(a)=a^2-5a+6$$

You simply replace every \(x\) in the rule with \(a\).

2. Evaluating functions using numbers

To evaluate a function, substitute the given value into the function rule everywhere you see the input variable. Then simplify carefully using the order of operations.

Worked Example 1

Given \(g(x)=3x-7\), find \(g(5)\).

Step 1: Replace \(x\) with 5.

$$g(5)=3(5)-7$$

Step 2: Simplify.

$$g(5)=15-7=8$$

So, \(g(5)=8\).

Worked Example 2

Given \(h(x)=x^2+4x-1\), find \(h(-2)\).

Step 1: Substitute \(-2\) for \(x\).

$$h(-2)=(-2)^2+4(-2)-1$$

Step 2: Simplify carefully. Notice the square of \(-2\) is positive.

$$h(-2)=4-8-1=-5$$

So, \(h(-2)=-5\).

When substituting negative numbers, using parentheses is very important. Writing \((-2)^2\) is different from writing \(-2^2\). Parentheses help avoid sign mistakes.

3. Evaluating functions using expressions

Sometimes the input is not a number. It may be another variable or an expression such as \(x+1\), \(2a\), or \(x+h\). The process is the same: replace the original input variable everywhere in the rule.

This skill is important because it helps students prepare for ideas involving change, rates of change, and algebraic manipulation.

Worked Example 3

Given \(f(x)=2x^2-3x+1\), find \(f(x+h)\).

Step 1: Replace every \(x\) with \(x+h\).

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

Step 2: Expand \((x+h)^2\).

$$f(x+h)=2(x^2+2xh+h^2)-3x-3h+1$$

Step 3: Distribute the 2.

$$f(x+h)=2x^2+4xh+2h^2-3x-3h+1$$

So,

$$f(x+h)=2x^2+4xh+2h^2-3x-3h+1$$

This example shows why parentheses matter. The entire expression \(x+h\) must be substituted everywhere the variable appears.

4. Evaluating more than one function

Sometimes different functions are given, and you may need to evaluate each one separately. Each function can have a different rule.

For example, if:

$$f(x)=x+2 \qquad \text{and} \qquad g(x)=x^2$$

then:

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

and

$$g(3)=3^2=9$$

Even though the input is the same, the outputs are different because the function rules are different.

5. Understanding function notation in context

Function notation can also describe real situations. For example, suppose \(C(t)=50+12t\) gives the cost in dollars of renting a bike for \(t\) hours. Then:

• \(C(0)=50\) means the starting cost is 50 dollars
• \(C(3)=50+12(3)=86\) means renting for 3 hours costs 86 dollars

In this way, the notation helps describe a changing quantity clearly and efficiently.

6. Common mistakes to avoid

• Thinking \(f(x)\) means multiplication: It means the value of the function at \(x\), not \(f \times x\).
• Forgetting parentheses: If substituting a negative number or an expression, always use parentheses.
• Replacing only one variable: When evaluating \(f(x+h)\), every \(x\) in the rule must be replaced by \((x+h)\).
• Not simplifying fully: After substitution, combine like terms and use the correct order of operations.

Worked Example 4

Given \(p(x)=x^2-4x+7\), find both \(p(3)\) and \(p(a-1)\).

Part A: Find \(p(3)\)

$$p(3)=3^2-4(3)+7$$ $$p(3)=9-12+7=4$$

So, \(p(3)=4\).

Part B: Find \(p(a-1)\)

Replace every \(x\) with \((a-1)\):

$$p(a-1)=(a-1)^2-4(a-1)+7$$

Now expand:

$$p(a-1)=a^2-2a+1-4a+4+7$$

Combine like terms:

$$p(a-1)=a^2-6a+12$$

So, \(p(a-1)=a^2-6a+12\).

7. A step-by-step method for evaluation

1. Write the function rule clearly.
2. Identify the input you are substituting.
3. Replace every occurrence of the variable with that input.
4. Use parentheses when needed.
5. Simplify carefully.

This method works whether the input is a number, a variable, or an expression.

8. Why \(f(x+h)\) matters

The expression \(f(x+h)\) is especially important because it represents the function evaluated at a new input that is slightly different from \(x\). It helps describe how outputs change when inputs change. This idea is used later when studying rates of change and the behavior of graphs.

For now, the main goal is to substitute correctly and simplify the result accurately.

Summary

Function notation is a compact way to describe a rule that matches each input to one output. To evaluate a function, substitute the given value or expression for the input variable everywhere in the rule. Be especially careful with negative numbers and expressions like \(x+h\), because correct use of parentheses makes the algebra accurate.

Piecewise-Defined Functions are functions made up of different rules for different parts of the input values. Instead of using one formula for every value of \(x\), a piecewise function uses one formula on one interval, and a different formula on another interval.

These functions are useful because many real-world situations change behavior depending on the input. For example, a taxi fare may have one price for the first few miles and a different rate after that. A piecewise function can model that change clearly.

In this lesson, you will learn how to read, evaluate, and graph piecewise-defined functions. You will also see how absolute value functions and step functions connect to this idea.

1. What is a piecewise-defined function?

A piecewise-defined function gives different formulas on different intervals of the domain. It is often written like this:

$$ f(x)= \begin{cases} \text{first rule} & \text{if condition on } x \\ \text{second rule} & \text{if condition on } x \\ \text{third rule} & \text{if condition on } x \end{cases} $$

For example,

$$ f(x)= \begin{cases} 2x+1 & \text{if } x<0 \\ x^2 & \text{if } x\ge 0 \end{cases} $$

This means:

• Use \(2x+1\) when \(x\) is less than 0.
• Use \(x^2\) when \(x\) is greater than or equal to 0.

2. How to evaluate a piecewise function

To evaluate a piecewise-defined function, follow these steps:

1. Look at the input value.
2. Decide which condition the input satisfies.
3. Use only the formula that matches that condition.
4. Substitute the value and simplify.

The most important step is choosing the correct rule. If you use the wrong interval, the answer will be wrong even if your algebra is correct.

Worked Example 1: Evaluating a piecewise function

Evaluate the function

$$ g(x)= \begin{cases} 3x-2 & \text{if } x<1 \\ x+4 & \text{if } x\ge 1 \end{cases} $$

Find \(g(-2)\), \(g(0)\), and \(g(3)\).

Step 1: Find \(g(-2)\)

Since \(-2<1\), use the rule \(3x-2\).

$$ g(-2)=3(-2)-2=-6-2=-8 $$

Step 2: Find \(g(0)\)

Since \(0<1\), use the same rule \(3x-2\).

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

Step 3: Find \(g(3)\)

Since \(3\ge 1\), use the rule \(x+4\).

$$ g(3)=3+4=7 $$

Answers: \(g(-2)=-8\), \(g(0)=-2\), and \(g(3)=7\).

3. Understanding the interval symbols

Piecewise functions often use inequality symbols. You need to read them carefully:

• \(x<2\) means values less than 2
• \(x\le 2\) means values less than or equal to 2
• \(x>2\) means values greater than 2
• \(x\ge 2\) means values greater than or equal to 2

The difference between \(<\) and \(\le\) matters when graphing. If the endpoint is included, we use a closed dot. If the endpoint is not included, we use an open dot.

4. How to graph a piecewise-defined function

To graph a piecewise function:

1. Take one rule at a time.
2. Graph that rule only on the interval where it is allowed.
3. Use open or closed dots at boundary points.
4. Repeat for each piece.

Each piece is just part of a familiar graph, such as a line, parabola, or constant function. The conditions tell you where each part belongs.

Worked Example 2: Graphing a piecewise function

Graph

$$ h(x)= \begin{cases} x+2 & \text{if } x<1 \\ 3 & \text{if } x\ge 1 \end{cases} $$

First piece: \(y=x+2\) for \(x<1\)

• This is a line with slope 1 and y-intercept 2.
• Graph only the part of the line where \(x<1\).
• At \(x=1\), the line would give \(y=3\), but \(x=1\) is not included here.
• So place an open dot at \((1,3)\).

Second piece: \(y=3\) for \(x\ge 1\)

• This is a horizontal line at height 3.
• Graph only for \(x\ge 1\).
• Since \(x=1\) is included, place a closed dot at \((1,3)\).

Notice that both pieces meet at \((1,3)\), but one side has an open dot and the other has a closed dot. The function value at \(x=1\) comes from the second piece, so the point is included.

5. Domain and range of a piecewise function

The domain is the set of all input values that the function allows. In many piecewise functions, the intervals together describe the entire domain.

The range is the set of output values the function actually produces. To find the range, look at the outputs from each piece and combine them.

For example, if

$$ f(x)= \begin{cases} -x & \text{if } x<0 \\ 2 & \text{if } x\ge 0 \end{cases} $$

then the domain is all real numbers, because every \(x\) is either less than 0 or greater than or equal to 0.

For \(x<0\), the outputs of \(-x\) are positive numbers. For \(x\ge 0\), the output is always 2. Since 2 is already positive, the range is all positive numbers:

$$ \text{Range: } y>0 $$

6. Piecewise functions and continuity on a graph

Some piecewise functions connect smoothly, and some have jumps or breaks. When the pieces meet at a point and the graph does not break, the function looks continuous there. When the graph suddenly jumps, it is not continuous at that point.

At the 11th Grade level, the main thing to notice is what happens at the boundary values between pieces. Check:

• Do the two pieces meet at the same point?
• Is that point included?
• Is there a gap or jump?

Worked Example 3: Evaluating and graphing with three pieces

Consider

$$ p(x)= \begin{cases} -1 & \text{if } x<-2 \\ x+1 & \text{if } -2\le x<2 \\ x^2 & \text{if } x\ge 2 \end{cases} $$

Part A: Evaluate \(p(-3)\), \(p(-2)\), and \(p(2)\)

For \(p(-3)\): since \(-3<-2\), use the first rule.

$$ p(-3)=-1 $$

For \(p(-2)\): since \(-2\le x<2\), use the second rule.

$$ p(-2)=-2+1=-1 $$

For \(p(2)\): since \(x\ge 2\), use the third rule.

$$ p(2)=2^2=4 $$

Part B: Describe the graph

• For \(x<-2\), graph the horizontal line \(y=-1\), with an open dot at \((-2,-1)\).
• For \(-2\le x<2\), graph the line \(y=x+1\). Put a closed dot at \((-2,-1)\) and an open dot at \((2,3)\).
• For \(x\ge 2\), graph the parabola \(y=x^2\) starting with a closed dot at \((2,4)\).

Notice what happens at \(x=2\): the middle piece approaches \(3\), but the function value is actually \(4\). This creates a jump in the graph.

7. Absolute value as a piecewise function

The absolute value function is one of the most important examples of a piecewise-defined function. Recall that absolute value gives the distance from 0, so it is never negative.

It can be written as

$$ |x|= \begin{cases} x & \text{if } x\ge 0 \\ -x & \text{if } x<0 \end{cases} $$

This means:

• If \(x\) is positive or zero, leave it alone.
• If \(x\) is negative, change its sign to make it positive.

So the graph of \(|x|\) is actually made from two lines. This is why it has a sharp V-shape.

Worked Example 4: Absolute value as a piecewise function

Write and evaluate the function \(f(x)=|x-2|\) as a piecewise-defined function.

First, think about when \(x-2\) is nonnegative or negative:

• If \(x-2\ge 0\), then \(x\ge 2\).
• If \(x-2<0\), then \(x<2\).

So

$$ f(x)=|x-2|= \begin{cases} x-2 & \text{if } x\ge 2 \\ -(x-2) & \text{if } x<2 \end{cases} $$

Simplify the second rule:

$$ f(x)= \begin{cases} x-2 & \text{if } x\ge 2 \\ -x+2 & \text{if } x<2 \end{cases} $$

Now evaluate a few values:

• \(f(5)=5-2=3\)
• \(f(2)=2-2=0\)
• \(f(0)=-0+2=2\)

The graph forms a V with vertex at \((2,0)\).

8. Step functions

A step function is a piecewise-defined function made of constant pieces. The graph looks like steps on a staircase.

For example, consider

$$ s(x)= \begin{cases} 1 & \text{if } 0\le x<2 \\ 2 & \text{if } 2\le x<4 \\ 3 & \text{if } 4\le x<6 \end{cases} $$

This function stays at 1 from \(x=0\) up to but not including \(x=2\). Then it jumps to 2, then jumps to 3.

When graphing a step function, pay very close attention to the endpoints:

• Use a closed dot where the interval begins if that endpoint is included.
• Use an open dot where the interval ends if that endpoint is not included.

9. Common mistakes to avoid

• Using the wrong rule: Always check which interval the input belongs to first.
• Ignoring equality: A value like \(x=2\) belongs in the piece with \(\le\) or \(\ge\), not the one with only \(<\) or \(>\).
• Graphing the whole line or curve: Only graph each formula on its given interval.
• Missing open or closed dots: Endpoints show whether a value is included.
• Mixing formulas together: A piecewise function uses one rule at a time, not all rules at once.

10. Strategy for solving problems

Whenever you see a piecewise-defined function, ask yourself these questions:

1. What are the intervals?
2. Which interval contains the input value?
3. What formula goes with that interval?
4. If I am graphing, what happens at the endpoint?

This simple strategy helps with both evaluation and graphing.

Brief Summary

A piecewise-defined function uses different formulas on different parts of its domain. To evaluate one, choose the rule that matches the input value. To graph one, graph each rule only on its interval and use open or closed dots at endpoints.

Absolute value functions and step functions are important examples of piecewise-defined functions. Once you learn to read the intervals carefully, these functions become much easier to understand and graph.

Algebraic Operations on Functions means combining two functions using the same operations you already know from algebra: addition, subtraction, multiplication, and division.

When we combine functions, we do more than just find a new formula. We also have to find the domain of the new function. The domain tells us which input values are allowed.

This is very important because a value might work in one function but not in the other. Also, division creates an extra restriction: we can never divide by zero.

In this lesson, you will learn how to:

• add, subtract, multiply, and divide functions,
• write the resulting function clearly,
• determine the domain of the new function, and
• check for values that must be excluded.

1. Reviewing what a function is

A function is a rule that assigns exactly one output to each allowed input. For example, if

\(f(x)=2x+3\), then plugging in \(x=4\) gives:

$$f(4)=2(4)+3=11$$

The domain of a function is the set of all input values for which the function is defined.

For many polynomial functions, such as \(f(x)=x^2-1\), the domain is all real numbers.

But some functions have restrictions. For example:

• In a rational function, the denominator cannot be zero.
• If square roots are involved, the expression inside the root must usually be non-negative.

2. Definitions of algebraic operations on functions

If \(f\) and \(g\) are functions, then we define:

$$ (f+g)(x)=f(x)+g(x) $$

$$ (f-g)(x)=f(x)-g(x) $$

$$ (fg)(x)=f(x)g(x) $$

$$ \left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)} \quad \text{as long as } g(x)\ne 0 $$

These operations combine the outputs of the functions, not the inputs.

For example, if \(f(x)=x+1\) and \(g(x)=x^2\), then:

$$ (f+g)(x)=(x+1)+x^2=x^2+x+1 $$

3. The key idea about domain

For any algebraic operation on functions, the input \(x\) must be allowed in both original functions.

So the domain of \((f+g)(x)\), \((f-g)(x)\), and \((fg)(x)\) is the intersection of the domains of \(f\) and \(g\).

That means we only keep values that work for both functions.

For division, we need one more condition:

• \(x\) must be in the domain of both functions, and
• \(g(x)\ne 0\), because division by zero is undefined.

4. How to find the domain of a combined function

Use this step-by-step method:

1. Find the domain of the first function.
2. Find the domain of the second function.
3. Keep only the values common to both domains.
4. If dividing, also remove any values that make the denominator function equal to zero.

5. Worked Example 1: Adding functions

Let

$$f(x)=2x-5 \qquad g(x)=x^2+1$$

Find \((f+g)(x)\) and its domain.

Step 1: Add the functions.

$$ (f+g)(x)=f(x)+g(x) $$

$$ (f+g)(x)=(2x-5)+(x^2+1) $$

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

Step 2: Find the domain.

Both \(f(x)=2x-5\) and \(g(x)=x^2+1\) are polynomials, so each has domain all real numbers.

The intersection of their domains is all real numbers.

Answer:

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

Domain: all real numbers

6. Worked Example 2: Subtracting functions with restricted domains

Let

$$f(x)=\frac{1}{x-3} \qquad g(x)=\sqrt{x+1}$$

Find \((f-g)(x)\) and its domain.

Step 1: Subtract the functions.

$$ (f-g)(x)=f(x)-g(x) $$

$$ (f-g)(x)=\frac{1}{x-3}-\sqrt{x+1} $$

Step 2: Find the domain of each function.

For \(f(x)=\frac{1}{x-3}\), the denominator cannot be zero.

So:

$$x-3\ne 0 \Rightarrow x\ne 3$$

For \(g(x)=\sqrt{x+1}\), the expression inside the square root must be non-negative.

So:

$$x+1\ge 0 \Rightarrow x\ge -1$$

Step 3: Take the intersection.

We need values with \(x\ge -1\) and \(x\ne 3\).

Answer:

$$ (f-g)(x)=\frac{1}{x-3}-\sqrt{x+1} $$

Domain: all \(x\) such that \(x\ge -1\) and \(x\ne 3\)

In interval form, this is:

$$ [-1,3)\cup(3,\infty) $$

7. Worked Example 3: Multiplying functions

Let

$$f(x)=x-2 \qquad g(x)=\frac{x+1}{x^2-4}$$

Find \((fg)(x)\) and its domain.

Step 1: Multiply the functions.

$$ (fg)(x)=f(x)g(x) $$

$$ (fg)(x)=(x-2)\left(\frac{x+1}{x^2-4}\right) $$

Factor the denominator:

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

So:

$$ (fg)(x)=(x-2)\left(\frac{x+1}{(x-2)(x+2)}\right) $$

$$ (fg)(x)=\frac{x+1}{x+2} $$

Important: Even though \((x-2)\) cancels in the formula, the original function \(g(x)\) was undefined at \(x=2\). It was also undefined at \(x=-2\).

So we must keep the original restrictions.

Step 2: Find the domain.

\(f(x)=x-2\) has domain all real numbers.

For \(g(x)=\frac{x+1}{x^2-4}\), the denominator cannot be zero:

$$x^2-4\ne 0$$

$$ (x-2)(x+2)\ne 0 $$

$$ x\ne 2, \; x\ne -2 $$

The intersection of the domains is all real numbers except \(-2\) and \(2\).

Answer:

$$ (fg)(x)=\frac{x+1}{x+2} $$

Domain: all real numbers except \(x=-2\) and \(x=2\)

8. Worked Example 4: Dividing functions

Let

$$f(x)=x^2-9 \qquad g(x)=x+3$$

Find \(\left(\frac{f}{g}\right)(x)\) and its domain.

Step 1: Write the quotient.

$$ \left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}=\frac{x^2-9}{x+3} $$

Factor the numerator:

$$ x^2-9=(x-3)(x+3) $$

So:

$$ \left(\frac{f}{g}\right)(x)=\frac{(x-3)(x+3)}{x+3} $$

$$ \left(\frac{f}{g}\right)(x)=x-3 $$

Step 2: Find the domain.

At first glance, the simplified expression \(x-3\) seems to allow all real numbers. But this is not correct for the original quotient.

Because we are dividing by \(g(x)=x+3\), we must have:

$$ x+3\ne 0 \Rightarrow x\ne -3 $$

Answer:

$$ \left(\frac{f}{g}\right)(x)=x-3 $$

Domain: all real numbers except \(x=-3\)

9. A very common mistake

Students often simplify an expression and then forget that some values were already excluded.

For example, in Example 4, the quotient simplified to \(x-3\), but \(x=-3\) is still not allowed.

This happens because simplification changes the appearance of the expression, but it does not bring back values that made the original function undefined.

10. Quick comparison of the four operations

• Addition: add the outputs, and use the common domain of both functions.
• Subtraction: subtract the outputs, and use the common domain of both functions.
• Multiplication: multiply the outputs, and use the common domain of both functions.
• Division: divide the outputs, use the common domain, and also exclude values where the denominator function equals zero.

11. Tips for success

• Always write the operation first, such as \((f+g)(x)=f(x)+g(x)\).
• Substitute the full expressions carefully with brackets.
• Simplify only after writing the full combined expression.
• Check the domain of each original function before giving your final answer.
• For division, test where the denominator function is zero.
• If factors cancel, keep any restrictions from the original expression.

12. Practice-style checks

If \(f(x)=\sqrt{x}\) and \(g(x)=\frac{2}{x-1}\), then:

• \((f+g)(x)=\sqrt{x}+\frac{2}{x-1}\)
• To find the domain, we need \(x\ge 0\) and \(x\ne 1\)

So the domain is:

$$ [0,1)\cup(1,\infty) $$

This example shows again that the domain comes from combining the restrictions of both functions.

Summary

Algebraic operations on functions work just like algebraic operations on expressions: we can add, subtract, multiply, and divide function outputs.

The most important idea is that the domain of the new function must include only values that are allowed in both original functions. For division, we must also exclude any value that makes the denominator zero.

If you remember to check domain restrictions before and after simplifying, you will avoid the most common errors.

Function Composition means combining two functions so that the output of one function becomes the input of another. It is like doing one machine first, then sending its result into a second machine.

If we have two functions, \(f(x)\) and \(g(x)\), then the composition of \(f\) with \(g\) is written as \((f \circ g)(x)\). This is read as “\(f\) of \(g\) of \(x\).”

By definition,

$$ (f \circ g)(x) = f(g(x)) $$

This means:

• First apply \(g\) to \(x\)
• Then apply \(f\) to the result

The order is very important. In general,

$$ (f \circ g)(x) \ne (g \circ f)(x) $$

So when composing functions, always work from the inside out.

Why composition matters: Many real situations happen in steps. For example, a quantity may be changed by one rule and then changed again by another rule. Composition lets us write these multi-step processes as one combined function.

Main Idea 1: How to compose functions

To find \((f \circ g)(x)\):

1. Start with the inside function, \(g(x)\).
2. Substitute that entire expression into \(f\).
3. Simplify carefully.

Similarly,

$$ (g \circ f)(x) = g(f(x)) $$

Main Idea 2: Composition is substitution

If \(f(x) = 2x+3\) and \(g(x)=x^2\), then

$$ (f \circ g)(x) = f(g(x)) = f(x^2) $$

Since \(f\) takes an input and changes it by the rule \(2(\text{input})+3\), replace the input with \(x^2\):

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

Main Idea 3: Order matters

Using the same functions,

$$ (g \circ f)(x)=g(f(x))=g(2x+3) $$

Now replace the input of \(g\) with \(2x+3\):

$$ g(2x+3)=(2x+3)^2 $$

This is different from \(2x^2+3\). That shows why the order of composition matters.

Main Idea 4: Domain in composition

The domain of a composition is the set of all inputs that make the composition defined. For \((f \circ g)(x)=f(g(x))\), two things must be true:

• \(x\) must be in the domain of \(g\)
• The value \(g(x)\) must be in the domain of \(f\)

This means a composition may have more restrictions than either function alone. You must check the inside function first, then see whether its output is allowed in the outside function.

For example, if

$$ f(x)=\frac{1}{x}, \qquad g(x)=x-2 $$

then

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

This function is undefined when \(x-2=0\), so \(x \ne 2\). Even though \(g(x)=x-2\) is defined for all real numbers, the composition is not.

Worked Example 1: Basic composition

Let

$$ f(x)=3x-1, \qquad g(x)=x+4 $$

Find \((f \circ g)(x)\).

Step 1: Write the composition formula.

$$ (f \circ g)(x)=f(g(x)) $$

Step 2: Substitute \(g(x)=x+4\) into \(f\).

$$ f(g(x))=f(x+4) $$

Step 3: Apply the rule for \(f\), which is \(3(\text{input})-1\).

$$ f(x+4)=3(x+4)-1 $$

Step 4: Simplify.

$$ 3(x+4)-1=3x+12-1=3x+11 $$

Answer:

$$ (f \circ g)(x)=3x+11 $$

Worked Example 2: Compare both orders

Let

$$ f(x)=x^2, \qquad g(x)=x-5 $$

Find both \((f \circ g)(x)\) and \((g \circ f)(x)\).

First: \((f \circ g)(x)=f(g(x))\)

$$ (f \circ g)(x)=f(x-5) $$

Since \(f\) squares its input,

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

So,

$$ (f \circ g)(x)=(x-5)^2 $$

Second: \((g \circ f)(x)=g(f(x))\)

$$ (g \circ f)(x)=g(x^2) $$

Since \(g\) subtracts 5 from its input,

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

So,

$$ (g \circ f)(x)=x^2-5 $$

Conclusion:

$$ (f \circ g)(x)=(x-5)^2 \ne x^2-5=(g \circ f)(x) $$

This is a clear example that composition usually depends on order.

Worked Example 3: Composition with domain restrictions

Let

$$ f(x)=\sqrt{x}, \qquad g(x)=2x-3 $$

Find \((f \circ g)(x)\) and state its domain.

Step 1: Compose the functions.

$$ (f \circ g)(x)=f(g(x))=f(2x-3) $$

Since \(f(x)=\sqrt{x}\), replace \(x\) with \(2x-3\):

$$ (f \circ g)(x)=\sqrt{2x-3} $$

Step 2: Find the domain.

A square root is only defined when the expression inside is greater than or equal to 0.

$$ 2x-3 \ge 0 $$ $$ 2x \ge 3 $$ $$ x \ge \frac{3}{2} $$

Answer:

$$ (f \circ g)(x)=\sqrt{2x-3}, \qquad \text{domain: } x \ge \frac{3}{2} $$

This restriction comes from the outside function \(f(x)=\sqrt{x}\), because the output of \(g(x)\) must be allowed inside the square root.

Worked Example 4: Rational composition and domain tracking

Let

$$ f(x)=\frac{1}{x+1}, \qquad g(x)=\frac{2}{x} $$

Find \((f \circ g)(x)\) and its domain.

Step 1: Start with the composition.

$$ (f \circ g)(x)=f(g(x))=f\left(\frac{2}{x}\right) $$

Step 2: Substitute into \(f(x)=\frac{1}{x+1}\).

$$ (f \circ g)(x)=\frac{1}{\frac{2}{x}+1} $$

Step 3: Simplify.

$$ \frac{1}{\frac{2}{x}+1} = \frac{1}{\frac{2+x}{x}} = \frac{x}{x+2} $$

So,

$$ (f \circ g)(x)=\frac{x}{x+2} $$

Step 4: Check the domain carefully.

• First, \(g(x)=\frac{2}{x}\) requires \(x \ne 0\).
• Second, the output of \(g(x)\) goes into \(f\), and \(f\) is undefined when its input is \(-1\).

So we also need

$$ g(x) \ne -1 $$ $$ \frac{2}{x} \ne -1 $$ $$ 2 \ne -x $$ $$ x \ne -2 $$

Domain:

$$ x \ne 0, \quad x \ne -2 $$

Even though the simplified result is \(\frac{x}{x+2}\), the original composition also excludes \(x=0\) because the inside function \(g(x)\) is undefined there. This is why domain tracking must be done from the original functions, not just the final simplified expression.

Tips for success

• Work inside out: Evaluate the function in the parentheses first.
• Use parentheses carefully: If you substitute a binomial into a function, keep it grouped.
• Check both orders: \((f \circ g)(x)\) and \((g \circ f)(x)\) are usually different.
• Always check the domain: Make sure the inside function is defined, and its output is allowed in the outside function.

Common mistakes

• Mixing up the order of the functions.
• Forgetting to substitute the entire expression.
• Dropping parentheses too early, especially when squaring.
• Ignoring domain restrictions from the inside function.

Brief Summary

Function composition combines two functions into one by using the output of one as the input of another. The notation \((f \circ g)(x)=f(g(x))\) means apply \(g\) first, then \(f\). The order matters, and the domain of a composition must be checked carefully so that both the inside function and the outside function are defined.

Inverse Functions are functions that “undo” each other.

If a function takes an input and turns it into an output, then its inverse takes that output and gives you back the original input.

For example, if a function adds 5, then its inverse subtracts 5. If a function doubles a number, then its inverse halves it.

This idea is important because it helps us solve equations, understand graphs, and connect algebra with geometry.

1. What is an inverse function?

If a function is called \(f\), then its inverse is written \(f^{-1}\).

The inverse function reverses the effect of \(f\). This means:

$$f(f^{-1}(x)) = x \quad \text{and} \quad f^{-1}(f(x)) = x$$

These equations mean that applying a function and then its inverse brings you back to where you started.

Example idea: Let \(f(x)=3x-4\).

If you start with \(x=2\), then:

$$f(2)=3(2)-4=2$$

If \(f^{-1}\) is the inverse, then \(f^{-1}(2)=2\), because the inverse must undo the original function.

2. When does a function have an inverse?

Not every function has an inverse that is also a function.

For a function to have an inverse function, it must be one-to-one.

A function is one-to-one if different inputs always give different outputs.

In symbols, if \(f(a)=f(b)\), then a one-to-one function must have \(a=b\).

Horizontal Line Test

You can check whether a graph is one-to-one using the horizontal line test:

• If every horizontal line crosses the graph at most once, the function is one-to-one.
• If some horizontal line crosses the graph more than once, the function is not one-to-one.

This works because if a horizontal line hits the graph more than once, then the same output \(y\)-value comes from different inputs. That means the function cannot have an inverse that is also a function.

Examples:

• \(f(x)=2x+1\) is one-to-one, so it has an inverse.
• \(f(x)=x^2\) is not one-to-one over all real numbers, because \(f(2)=4\) and \(f(-2)=4\).

On the graph of \(y=x^2\), a horizontal line such as \(y=4\) crosses the parabola twice, so the function fails the horizontal line test.

3. How to find an inverse algebraically

To find the inverse of a function algebraically, follow these steps:

1. Write the function as \(y=f(x)\).
2. Swap \(x\) and \(y\).
3. Solve for \(y\).
4. Rewrite \(y\) as \(f^{-1}(x)\).

Swapping \(x\) and \(y\) makes sense because the inverse reverses inputs and outputs.

Worked Example 1: Linear function

Find the inverse of:

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

Step 1: Write \(y=2x+3\).

Step 2: Swap \(x\) and \(y\).

$$x=2y+3$$

Step 3: Solve for \(y\).

$$x-3=2y$$ $$y=\frac{x-3}{2}$$

Step 4: Write the inverse.

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

Check:

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

So the inverse is correct.

Worked Example 2: A fraction function

Find the inverse of:

$$f(x)=\frac{x+1}{4}$$

Step 1: Write \(y=\frac{x+1}{4}\).

Step 2: Swap \(x\) and \(y\).

$$x=\frac{y+1}{4}$$

Step 3: Solve for \(y\).

$$4x=y+1$$ $$y=4x-1$$

Step 4: Write the inverse.

$$f^{-1}(x)=4x-1$$

Check with composition:

$$f(f^{-1}(x))=\frac{(4x-1)+1}{4}=\frac{4x}{4}=x$$

4. Graphing inverse functions

The graph of a function and the graph of its inverse are reflections of each other across the line \(y=x\).

This happens because inverse functions switch inputs and outputs. So every point \((a,b)\) on the graph of \(f\) becomes the point \((b,a)\) on the graph of \(f^{-1}\).

Example: If \((2,7)\) is on the graph of \(f\), then \((7,2)\) is on the graph of \(f^{-1}\).

To sketch an inverse from a graph:

1. Draw the line \(y=x\).
2. Take key points from the original function.
3. Swap each point’s coordinates.
4. Plot the new points and sketch the reflected graph.

Worked Example 3: Inverse from points

Suppose the graph of \(f\) passes through the points:

$$(-1,0), \quad (0,2), \quad (3,5)$$

Then the inverse graph passes through:

$$ (0,-1), \quad (2,0), \quad (5,3) $$

Each point is reflected across \(y=x\) by switching the coordinates.

5. Functions that are not one-to-one

Some functions do not have inverses unless we restrict the domain.

A common example is:

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

Over all real numbers, this is not one-to-one because:

$$f(2)=4 \quad \text{and} \quad f(-2)=4$$

So its inverse would not be a function, since one output would come from two different inputs.

But if we restrict the domain to \(x\ge 0\), then \(f(x)=x^2\) becomes one-to-one.

Now we can find an inverse.

Worked Example 4: Restricting the domain

Find the inverse of \(f(x)=x^2\) with domain \(x\ge 0\).

Step 1: Write

$$y=x^2$$

Step 2: Swap \(x\) and \(y\).

$$x=y^2$$

Step 3: Solve for \(y\).

$$y=\pm \sqrt{x}$$

Because the original domain is \(x\ge 0\), the outputs of the inverse must also be nonnegative. So we choose the positive square root.

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

This inverse works only because the domain was restricted.

6. Domain and range of inverses

For a function and its inverse, the domain and range switch places.

• The domain of \(f\) becomes the range of \(f^{-1}\).
• The range of \(f\) becomes the domain of \(f^{-1}\).

Example: If \(f\) has domain \([0,\infty)\) and range \([2,\infty)\), then \(f^{-1}\) has domain \([2,\infty)\) and range \([0,\infty)\).

This is another result of inputs and outputs being reversed.

7. How to verify an inverse

After finding an inverse, it is a good idea to check it using composition.

If \(g(x)=f^{-1}(x)\), then both of these should be true:

$$f(g(x))=x$$ $$g(f(x))=x$$

If both simplify to \(x\), then the functions are inverses of each other.

Quick Example: Let

$$f(x)=5x-1 \quad \text{and} \quad f^{-1}(x)=\frac{x+1}{5}$$

Check:

$$f\left(f^{-1}(x)\right)=5\left(\frac{x+1}{5}\right)-1=x+1-1=x$$ $$f^{-1}(f(x))=\frac{(5x-1)+1}{5}=\frac{5x}{5}=x$$

So they are inverses.

8. Common mistakes to avoid

• Forgetting to check one-to-one behavior: A function must pass the horizontal line test to have an inverse function.
• Not swapping \(x\) and \(y\): This is a key step when finding an inverse algebraically.
• Ignoring domain restrictions: Some functions only have inverses after the domain is limited.
• Mixing up \(f^{-1}(x)\) and \((f(x))^{-1}\): \(f^{-1}(x)\) means inverse function, not reciprocal.

For example, if \(f(x)=2x\), then:

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

But the reciprocal of \(f(x)\) would be:

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

These are not the same.

Brief Summary

An inverse function reverses the action of the original function.

A function must be one-to-one to have an inverse function, and this can be checked with the horizontal line test.

To find an inverse algebraically, write \(y=f(x)\), swap \(x\) and \(y\), solve for \(y\), and then write the result as \(f^{-1}(x)\).

Graphically, a function and its inverse are reflections across the line \(y=x\).

Always check your answer by composition, and remember that the domain and range switch places.

Function Symmetry: Even and Odd Functions

In function theory, symmetry helps us understand how a graph behaves when we reflect or rotate it. Two important types of symmetry are even symmetry and odd symmetry.

These ideas connect algebra and graphs. Algebraically, we test symmetry by comparing \(f(-x)\) with \(f(x)\). Graphically, we look for mirror symmetry or rotational symmetry.

Learning to recognize even and odd functions is useful because it helps you analyze graphs more quickly, simplify calculations, and better understand how functions are structured.

1. What is an even function?

A function is even if replacing \(x\) with \(-x\) does not change the value of the function. In symbols, a function is even if

$$f(-x)=f(x)$$

for every \(x\) in the domain.

This means the graph looks exactly the same on the left and right sides of the \(y\)-axis. So an even function has symmetry about the \(y\)-axis.

Examples of common even functions include:

• \(f(x)=x^2\)
• \(f(x)=x^4+3\)
• \(f(x)=|x|\)

For example, with \(f(x)=x^2\):

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

So \(x^2\) is even.

2. What is an odd function?

A function is odd if replacing \(x\) with \(-x\) changes the sign of the output. In symbols, a function is odd if

$$f(-x)=-f(x)$$

for every \(x\) in the domain.

This means the graph has rotational symmetry about the origin. If you rotate the graph \(180^\circ\) around the origin, it matches itself.

Examples of common odd functions include:

• \(f(x)=x\)
• \(f(x)=x^3\)
• \(f(x)=x^5-2x\)

For example, with \(f(x)=x^3\):

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

So \(x^3\) is odd.

3. How to test whether a function is even, odd, or neither

To test a function, follow these steps:

1. Write the function as \(f(x)\).
2. Substitute \(-x\) for every \(x\) to find \(f(-x)\).
3. Simplify carefully.
4. Compare the result with \(f(x)\) and with \(-f(x)\).

Then decide:

• If \(f(-x)=f(x)\), the function is even.
• If \(f(-x)=-f(x)\), the function is odd.
• If neither is true, the function is neither.

Important: A function must satisfy the rule for all values in its domain. One or two matching values are not enough.

4. Domain matters

When talking about even or odd symmetry, the domain must also be symmetric about 0. That means if \(x\) is in the domain, then \(-x\) must also be in the domain.

For example, suppose a function is only defined for \(x\ge 0\). Then it usually cannot be classified as even or odd, because negative inputs are missing.

So before testing symmetry, make sure the domain includes both positive and negative values in matching pairs.

5. Graphical meaning of even and odd functions

Algebra and graphs tell the same story in different ways.

• Even function: symmetry about the \(y\)-axis. If \((a,b)\) is on the graph, then \((-a,b)\) is also on the graph.
• Odd function: rotational symmetry about the origin. If \((a,b)\) is on the graph, then \((-a,-b)\) is also on the graph.

These point patterns come directly from the equations:

$$f(-x)=f(x) \quad \text{for even functions}$$

$$f(-x)=-f(x) \quad \text{for odd functions}$$

If a graph does not show either of these symmetries, then the function is neither even nor odd.

6. Worked Example 1: An even function

Determine whether \(f(x)=3x^2-5\) is even, odd, or neither.

Step 1: Find \(f(-x)\).

$$f(-x)=3(-x)^2-5$$

$$f(-x)=3x^2-5$$

Step 2: Compare with \(f(x)\).

Since

$$f(-x)=3x^2-5=f(x)$$

the function is even.

Graph interpretation: The graph is symmetric about the \(y\)-axis.

7. Worked Example 2: An odd function

Determine whether \(f(x)=2x^3-x\) is even, odd, or neither.

Step 1: Find \(f(-x)\).

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

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

Step 2: Find \(-f(x)\).

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

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

Since

$$f(-x)=-f(x)$$

the function is odd.

Graph interpretation: The graph has rotational symmetry about the origin.

8. Worked Example 3: A function that is neither

Determine whether \(f(x)=x^2+x\) is even, odd, or neither.

Step 1: Find \(f(-x)\).

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

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

Step 2: Compare with \(f(x)\) and \(-f(x)\).

We have

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

and

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

Now compare:

• \(f(-x)=x^2-x \ne x^2+x=f(x)\)
• \(f(-x)=x^2-x \ne -x^2-x=-f(x)\)

So the function is neither even nor odd.

Graph interpretation: The graph has neither \(y\)-axis symmetry nor origin symmetry.

9. Worked Example 4: Using the graph idea

Suppose a graph contains the point \((4,7)\).

• If the function is even, then \((-4,7)\) must also be on the graph.
• If the function is odd, then \((-4,-7)\) must also be on the graph.

This is a quick way to recognize symmetry from a graph without doing algebra.

10. Patterns to notice in polynomial functions

For many polynomial functions, the exponents can give a clue:

• A polynomial with only even powers of \(x\) and constants is often even.
• A polynomial with only odd powers of \(x\) and no constant term is often odd.
• A mix of even and odd powers usually gives neither.

Examples:

• \(x^4+2x^2-1\) is even.
• \(x^5-3x\) is odd.
• \(x^3+x^2\) is neither.

This pattern is helpful, but you should still check by computing \(f(-x)\).

11. Common mistakes to avoid

• Forgetting parentheses: When finding \(f(-x)\), replace every \(x\) with \((-x)\). Write parentheses first.
• Mixing up even and odd: Even means same output, odd means opposite output.
• Looking at only part of the graph: Symmetry must hold for the entire graph.
• Ignoring the domain: The domain must include matching positive and negative inputs.

12. Quick check table

• Even: \(f(-x)=f(x)\), symmetry about the \(y\)-axis
• Odd: \(f(-x)=-f(x)\), rotational symmetry about the origin
• Neither: neither rule works

13. Final summary

To determine whether a function is even, odd, or neither, compute \(f(-x)\) and compare it with \(f(x)\) and \(-f(x)\). If \(f(-x)=f(x)\), the function is even and its graph is symmetric about the \(y\)-axis. If \(f(-x)=-f(x)\), the function is odd and its graph has rotational symmetry about the origin.

If neither relationship is true, then the function is neither even nor odd. Always simplify carefully and remember that symmetry must hold for the whole domain, not just a few values.

Monotonicity and Extrema are ideas that help us describe how a function behaves as we move from left to right on its graph.

When we study monotonicity, we ask: Where is the function increasing? and Where is it decreasing? When we study extrema, we ask: Where does the function reach high points or low points?

These ideas are important because they help us understand the overall shape of a graph, compare values of a function, and solve real-world problems involving maximum profit, minimum cost, highest height, lowest temperature, and more.

1. Increasing and Decreasing Functions

A function is increasing on an interval if its output values go up as the input values increase.

In simple words: as you move to the right, the graph goes up.

A function is decreasing on an interval if its output values go down as the input values increase.

In simple words: as you move to the right, the graph goes down.

A function can also be constant on an interval, meaning the output stays the same.

For example, if a function is increasing on the interval \((-2,3)\), that means for any two x-values in that interval, the larger x-value gives a larger y-value.

We often describe monotonicity using interval notation:

• Increasing on \((a,b)\)
• Decreasing on \((b,c)\)
• Constant on \((c,d)\)

2. What Does Monotonic Mean?

The word monotonic means a function keeps the same trend over an interval.

• If it keeps going upward, it is monotonic increasing.
• If it keeps going downward, it is monotonic decreasing.

A function does not have to be increasing everywhere or decreasing everywhere. Many functions change behavior at different x-values.

3. Local and Global Extrema

An extremum is a maximum or minimum value of a function. The plural is extrema.

There are two main types:

• Local (relative) maximum
• Local (relative) minimum

A local maximum is a point where the function is higher than nearby points.

A local minimum is a point where the function is lower than nearby points.

These are called “local” because we only compare the point to values close to it, not to the entire graph.

A global (absolute) maximum is the highest value of the function on its whole domain or on the interval being considered.

A global (absolute) minimum is the lowest value of the function on its whole domain or on the interval being considered.

Important: A local maximum is not always the highest point overall, and a local minimum is not always the lowest point overall.

4. Turning Points and Changes in Monotonicity

A graph often changes from increasing to decreasing, or from decreasing to increasing. These changes usually happen at turning points.

• If a function changes from increasing to decreasing, there is a local maximum.
• If a function changes from decreasing to increasing, there is a local minimum.

This is one of the most useful connections between monotonicity and extrema.

For example:

• Increasing then decreasing \(\rightarrow\) local maximum
• Decreasing then increasing \(\rightarrow\) local minimum

5. Reading Monotonicity and Extrema from a Graph

To analyze a graph, move from left to right and ask these questions:

1. Is the graph going up, down, or staying flat?
2. Where does the behavior change?
3. Where are the highest and lowest points?

When giving your answer:

• State the intervals where the function is increasing.
• State the intervals where the function is decreasing.
• Name any local maxima or minima.
• Name any global maximum or minimum if they exist.

6. Open and Closed Intervals

When we describe where a function is increasing or decreasing, we usually use open intervals, such as \((1,4)\), because the behavior is being described between the endpoints.

When we state a maximum or minimum point, we give the actual point, such as \((2,5)\), or say “the maximum value is 5 when \(x=2\).”

7. Extrema on a Restricted Interval

Sometimes a function may not have a highest or lowest value on its entire domain, but it does have one on a specific interval.

For example, the line \(f(x)=x\) has no global maximum or global minimum if all real numbers are allowed. It keeps increasing forever.

But on the interval \([0,4]\):

• The minimum value is \(0\) at \(x=0\).
• The maximum value is \(4\) at \(x=4\).

So the interval being considered is very important.

8. Worked Example 1: A Simple Table of Values

Suppose a function has these values:

$$ \begin{array}{c|ccccc} x & -2 & -1 & 0 & 1 & 2 \\ \hline f(x) & 5 & 3 & 1 & 2 & 4 \end{array} $$

Let us describe the behavior.

From \(x=-2\) to \(x=0\), the outputs go:

$$5 \to 3 \to 1$$

So the function is decreasing over that part.

From \(x=0\) to \(x=2\), the outputs go:

$$1 \to 2 \to 4$$

So the function is increasing over that part.

The lowest value shown is \(1\) at \(x=0\). Since the function changes from decreasing to increasing there, \((0,1)\) is a local minimum.

It is also the global minimum among the values in the table.

The highest value shown is \(5\) at \(x=-2\). This is the global maximum among the listed points.

Answer:

• Decreasing from \(x=-2\) to \(x=0\)
• Increasing from \(x=0\) to \(x=2\)
• Local minimum at \((0,1)\)
• Global minimum value \(1\)
• Global maximum value \(5\)

9. Worked Example 2: Quadratic Function

Consider the function

$$f(x)=x^2-4x+3$$

This is a quadratic, so its graph is a parabola. Since the coefficient of \(x^2\) is positive, the parabola opens upward.

That means the graph has a minimum at its vertex.

To find the x-coordinate of the vertex:

$$x=\frac{-b}{2a}=\frac{-(-4)}{2(1)}=\frac{4}{2}=2$$

Now find the y-value:

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

So the vertex is:

$$ (2,-1) $$

Now describe monotonicity.

Before \(x=2\), the graph goes down as we move right, so the function is decreasing on \(( -\infty, 2 )\).

After \(x=2\), the graph goes up as we move right, so the function is increasing on \(( 2, \infty )\).

Because the function changes from decreasing to increasing at \(x=2\), the point \((2,-1)\) is a local minimum.

It is also the global minimum, because the parabola opens upward and this is the lowest point on the entire graph.

There is no global maximum, because the function keeps increasing without bound.

Answer:

• Decreasing on \(( -\infty, 2 )\)
• Increasing on \(( 2, \infty )\)
• Local minimum at \((2,-1)\)
• Global minimum at \((2,-1)\)
• No global maximum

10. Worked Example 3: Cubic Function with Two Turning Points

Consider the function

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

Let us use its graph shape and key values to understand the behavior.

Evaluate the function at a few important x-values:

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

This cubic has a turning point near \((-1,2)\) and another near \((1,-2)\).

From the graph shape:

• The function increases until it reaches \((-1,2)\).
• Then it decreases from \(x=-1\) to \(x=1\).
• Then it increases again after \(x=1\).

So:

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

Since the function changes from increasing to decreasing at \(x=-1\), there is a local maximum at \((-1,2)\).

Since the function changes from decreasing to increasing at \(x=1\), there is a local minimum at \((1,-2)\).

However, this function has no global maximum and no global minimum, because cubic functions like this continue upward on one side and downward on the other.

Answer:

• Increasing on \(( -\infty, -1 )\) and \(( 1, \infty )\)
• Decreasing on \(( -1, 1 )\)
• Local maximum at \((-1,2)\)
• Local minimum at \((1,-2)\)
• No global maximum or global minimum

11. Worked Example 4: Extrema on a Closed Interval

Consider

$$f(x)=-x^2+4x+1 \quad \text{on } [0,5]$$

This parabola opens downward because the coefficient of \(x^2\) is negative. So its vertex will be a maximum point.

Find the vertex:

$$x=\frac{-b}{2a}=\frac{-4}{2(-1)}=2$$

Now find the function value there:

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

So the vertex is \((2,5)\), which is the highest point on the graph.

Because we are working on the closed interval \([0,5]\), we should also check the endpoints.

$$f(0)=-(0)^2+4(0)+1=1$$ $$f(5)=-(5)^2+4(5)+1=-25+20+1=-4$$

Now compare the values:

• \(f(0)=1\)
• \(f(2)=5\)
• \(f(5)=-4\)

The greatest value is \(5\), so the global maximum on \([0,5]\) is \(5\) at \(x=2\).

The least value is \(-4\), so the global minimum on \([0,5]\) is \(-4\) at \(x=5\).

For monotonicity:

• The function is increasing on \((0,2)\)
• The function is decreasing on \((2,5)\)

Answer:

• Increasing on \((0,2)\)
• Decreasing on \((2,5)\)
• Local and global maximum at \((2,5)\)
• Global minimum at \((5,-4)\)

12. Common Mistakes to Avoid

• Mixing up increasing and decreasing: Always read the graph from left to right.
• Confusing local and global extrema: Local means nearby; global means entire domain or interval.
• Forgetting endpoints on a closed interval: Endpoints can be the absolute maximum or minimum.
• Listing points instead of intervals for monotonicity: Say “increasing on \((1,4)\),” not “increasing at \(x=2\).”
• Assuming every function has a maximum and minimum: Some functions have neither.

13. How to Analyze Any Function

Here is a useful step-by-step method:

1. Look at the graph or function carefully.
2. Move from left to right.
3. Mark where the graph rises, falls, or stays flat.
4. Write the increasing and decreasing intervals.
5. Find where the graph changes direction.
6. Label local maxima and minima.
7. If a specific interval is given, compare important points and endpoints to find global extrema.

14. Quick Check Questions

Try these on your own:

• If a graph changes from decreasing to increasing at \(x=4\), what type of extremum occurs there?
• If a function is increasing on \(( -\infty, 1 )\) and decreasing on \((1, \infty)\), what happens at \(x=1\)?
• Can a function have a local maximum but no global maximum?
• Can the endpoint of a closed interval be a global minimum?

Answers:

• A local minimum
• A local maximum
• Yes
• Yes

15. Summary

Monotonicity describes whether a function is increasing or decreasing over an interval.

Extrema are the high and low points of a function. A local maximum or local minimum compares a point to nearby points, while a global maximum or global minimum compares it to the whole domain or interval.

A change from increasing to decreasing gives a local maximum, and a change from decreasing to increasing gives a local minimum.

When finding extrema on a closed interval, always check both turning points and endpoints.

Average Rate of Change tells us how much one quantity changes on average compared to how much another quantity changes.

When working with functions, the average rate of change compares the change in the output values to the change in the input values over an interval. It is one of the most important ideas in function theory because it connects tables, graphs, and formulas.

For a function \(f(x)\), the average rate of change from \(x=a\) to \(x=b\) is

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

This formula measures how much the function changes vertically compared to how much it changes horizontally.

Geometrically, this value is the slope of the secant line through the two points \((a, f(a))\) and \((b, f(b))\) on the graph.

This idea is very similar to the slope formula you already know:

$$m=\frac{y_2-y_1}{x_2-x_1}$$

In fact, average rate of change is just the slope formula applied to two points on a function.

Why it matters: Many functions are not linear, so they do not have one constant rate of change everywhere. Average rate of change helps us describe how the function behaves over a specific interval.

For example, a car trip may not happen at one constant speed the whole time. But you can still find the average speed over the trip. Average rate of change works the same way for functions.

Main idea:

• If the average rate of change is positive, the function increases on average over that interval.
• If it is negative, the function decreases on average over that interval.
• If it is zero, the function has the same output at both endpoints of the interval.

How to find average rate of change

1. Identify the interval, from \(x=a\) to \(x=b\).
2. Find the function values \(f(a)\) and \(f(b)\).
3. Substitute into the formula \(\frac{f(b)-f(a)}{b-a}\).
4. Simplify carefully.
5. Interpret the result in context, if a context is given.

Important note: Average rate of change is found over an interval, not at a single point. That is why we need two x-values.

Worked Example 1: Using a table

Suppose a function has the following values:

• \(f(1)=3\)
• \(f(5)=11\)

Find the average rate of change from \(x=1\) to \(x=5\).

Use the formula:

$$\frac{f(5)-f(1)}{5-1}=\frac{11-3}{4}=\frac{8}{4}=2$$

Answer: The average rate of change is \(2\).

Interpretation: On average, the function increases by 2 units in output for every 1 unit increase in input between \(x=1\) and \(x=5\).

Worked Example 2: Using an equation

Find the average rate of change of \(f(x)=x^2+1\) from \(x=2\) to \(x=6\).

Step 1: Find the function values.

$$f(2)=2^2+1=5$$ $$f(6)=6^2+1=37$$

Step 2: Apply the formula.

$$\frac{f(6)-f(2)}{6-2}=\frac{37-5}{4}=\frac{32}{4}=8$$

Answer: The average rate of change is \(8\).

What this means: Even though \(f(x)=x^2+1\) is not linear, from \(x=2\) to \(x=6\) it increases at an average rate of 8 units of output per 1 unit of input.

Worked Example 3: A nonlinear function with a negative result

Find the average rate of change of \(f(x)=-x^2+4x\) from \(x=1\) to \(x=3\).

First find the function values:

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

Now use the formula:

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

Answer: The average rate of change is \(0\).

Interpretation: The function starts and ends at the same y-value on this interval, so the secant line is horizontal.

This is a good reminder that a curved graph can rise and fall within an interval, yet still have an average rate of change of zero if the endpoint outputs are equal.

Worked Example 4: Interpreting a real-world situation

A ball is thrown upward, and its height in metres after \(t\) seconds is given by

$$h(t)=-5t^2+20t+1$$

Find the average rate of change from \(t=1\) to \(t=3\).

Step 1: Find the heights.

$$h(1)=-5(1)^2+20(1)+1=-5+20+1=16$$ $$h(3)=-5(3)^2+20(3)+1=-45+60+1=16$$

Step 2: Use the formula.

$$\frac{h(3)-h(1)}{3-1}=\frac{16-16}{2}=0$$

Answer: The average rate of change is \(0\) metres per second.

Interpretation: Between 1 second and 3 seconds, the ball has the same height at both times. That does not mean the ball stood still the whole time. It means the overall change in height across that interval was zero.

Connecting average rate of change to the graph

On a graph, choose two points on the function. Draw the line through them. This is the secant line.

The slope of that secant line is the average rate of change over that interval.

• A steep upward secant line means a large positive average rate of change.
• A steep downward secant line means a large negative average rate of change.
• A horizontal secant line means the average rate of change is zero.

Average rate of change vs. constant rate of change

For a linear function, the average rate of change is the same on every interval because the graph is a straight line.

For a nonlinear function, the average rate of change can be different on different intervals because the graph curves.

Example: For \(f(x)=x^2\), compare two intervals.

$$\text{From }x=1\text{ to }x=2:\quad \frac{4-1}{2-1}=3$$ $$\text{From }x=2\text{ to }x=3:\quad \frac{9-4}{3-2}=5$$

The average rate of change changes because \(x^2\) is nonlinear.

Common mistakes to avoid

• Mixing up the order: If you use \(f(b)-f(a)\), then the denominator must be \(b-a\). Keep the same order in both parts.
• Using x-values instead of function values: Do not write \(\frac{b-a}{b-a}\). The numerator must be the change in outputs.
• Forgetting parentheses: When substituting negative numbers, use parentheses carefully.
• Thinking it describes the whole graph: Average rate of change only describes the function on the chosen interval.

Quick check

Find the average rate of change of \(f(x)=2x^2-3\) from \(x=-1\) to \(x=2\).

First find the outputs:

$$f(-1)=2(-1)^2-3=2-3=-1$$ $$f(2)=2(2)^2-3=8-3=5$$

Now calculate:

$$\frac{f(2)-f(-1)}{2-(-1)}=\frac{5-(-1)}{3}=\frac{6}{3}=2$$

So the average rate of change is \(2\).

Summary

The average rate of change of a function over an interval from \(x=a\) to \(x=b\) is

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

It tells you how much the output changes on average for each 1-unit change in the input.

It is also the slope of the secant line connecting the two points on the graph.

This concept is especially useful for nonlinear functions, where the rate of change is not constant. Understanding average rate of change prepares you for later topics such as limits and derivatives.

Geometric Function Transformations are changes made to the graph of a function that move it, stretch it, compress it, or flip it.

These transformations are important because many complicated graphs can be understood by starting with a simple parent function and then applying a few changes.

For example, the graph of \(y=x^2\) is a basic parabola. From that one graph, you can create many others such as \(y=(x-3)^2+2\), \(y=-2(x+1)^2\), or \(y=\frac{1}{2}(x-4)^2-5\).

In this lesson, you will learn how to recognize and apply translations, dilations, and reflections to a general function \(y=f(x)\).

1. Start with the parent function

A parent function is the original, simplest version of a function family.

• Linear: \(f(x)=x\)
• Quadratic: \(f(x)=x^2\)
• Absolute value: \(f(x)=|x|\)
• Square root: \(f(x)=\sqrt{x}\)
• Cubic: \(f(x)=x^3\)

When we transform a function, we compare the new equation to the parent form \(y=f(x)\).

A very useful general form is

$$y=a\,f(b(x-h))+k$$

Each part of this equation affects the graph in a different way:

• \(h\): horizontal translation
• \(k\): vertical translation
• \(a\): vertical stretch/compression and possible reflection
• \(b\): horizontal stretch/compression and possible reflection

2. Vertical translations

A vertical translation moves a graph up or down.

If

$$y=f(x)+k$$

then:

• if \(k>0\), the graph moves up \(k\) units
• if \(k<0\), the graph moves down \(|k|\) units

This change affects the output, so it is added outside the function.

Example: starting with \(y=x^2\)

• \(y=x^2+4\) moves the parabola up 4 units
• \(y=x^2-3\) moves the parabola down 3 units

3. Horizontal translations

A horizontal translation moves a graph left or right.

If

$$y=f(x-h)$$

then:

• if \(h>0\), the graph moves right \(h\) units
• if \(h<0\), the graph moves left \(|h|\) units

This often feels backward at first. The reason is that the change is happening to the input \(x\), not the output.

Example: starting with \(y=|x|\)

• \(y=|x-2|\) moves the graph right 2 units
• \(y=|x+5|\) moves the graph left 5 units

Important note: horizontal changes inside the function work in the opposite direction of what the sign seems to suggest.

4. Vertical stretches and compressions

A vertical dilation changes how tall or steep a graph looks.

If

$$y=a f(x)$$

then:

• if \(|a|>1\), the graph is vertically stretched
• if \(0<|a|<1\), the graph is vertically compressed

Every \(y\)-value is multiplied by \(a\).

Examples with \(y=x^2\):

• \(y=3x^2\): vertical stretch by factor 3
• \(y=\frac{1}{2}x^2\): vertical compression by factor \(\frac{1}{2}\)

A stretch makes the graph appear narrower for parabolas, while a compression makes it appear wider.

5. Horizontal stretches and compressions

A horizontal dilation changes how wide a graph looks.

If

$$y=f(bx)$$

then:

• if \(|b|>1\), the graph is horizontally compressed by factor \(\frac{1}{|b|}\)
• if \(0<|b|<1\), the graph is horizontally stretched by factor \(\frac{1}{|b|}\)

Again, this can seem reversed because the change is inside the function.

Examples with \(y=|x|\):

• \(y=|2x|\): horizontal compression by factor \(\frac{1}{2}\)
• \(y=|\frac{1}{3}x|\): horizontal stretch by factor 3

6. Reflections

A reflection flips a graph across an axis.

Reflection in the \(x\)-axis happens when the output is multiplied by \(-1\):

$$y=-f(x)$$

This changes every \(y\)-value to its opposite.

Reflection in the \(y\)-axis happens when the input is multiplied by \(-1\):

$$y=f(-x)$$

This changes every \(x\)-value to its opposite.

Examples:

• \(y=-x^2\): reflection of \(y=x^2\) in the \(x\)-axis
• \(y=|-x|\): reflection of \(y=|x|\) in the \(y\)-axis, but this looks the same because \(|-x|=|x|\)

7. The full transformation form

Most transformed functions can be written as

$$y=a\,f(b(x-h))+k$$

Here is what each part means:

• \(h\): move right \(h\) if positive, left if negative
• \(k\): move up \(k\) if positive, down if negative
• \(a\): vertical stretch/compression by \(|a|\), and reflect in the \(x\)-axis if \(a<0\)
• \(b\): horizontal compression/stretch by \(\frac{1}{|b|}\), and reflect in the \(y\)-axis if \(b<0\)

Order idea: when reading the equation, it is helpful to identify the pieces first. When imagining the graph from the parent function, many students find it easier to think of inside changes and outside changes separately.

• Inside: horizontal changes and possible reflection in the \(y\)-axis
• Outside: vertical changes and possible reflection in the \(x\)-axis

8. A table of common effects

• \(y=f(x)+k\): up \(k\)
• \(y=f(x)-k\): down \(k\)
• \(y=f(x-h)\): right \(h\)
• \(y=f(x+h)\): left \(h\)
• \(y=af(x)\): vertical scale factor \(|a|\)
• \(y=f(bx)\): horizontal scale factor \(\frac{1}{|b|}\)
• \(y=-f(x)\): reflect in the \(x\)-axis
• \(y=f(-x)\): reflect in the \(y\)-axis

9. Worked Example 1: Translation only

Describe the transformations from \(y=x^2\) to

$$y=(x-3)^2+4$$

Step 1: Compare with \(y=f(x-h)+k\).

Here, \(h=3\) and \(k=4\).

Step 2: Interpret the changes.

• \((x-3)\) means shift right 3
• \(+4\) means shift up 4

Result: The parabola moves right 3 units and up 4 units.

The vertex, which was originally at \((0,0)\), moves to \((3,4)\).

10. Worked Example 2: Reflection and vertical stretch

Describe the transformations from \(y=|x|\) to

$$y=-2|x|$$

Step 1: Identify the outside multiplier.

The factor is \(a=-2\).

Step 2: Interpret the sign and size.

• The negative sign reflects the graph in the \(x\)-axis
• The factor 2 gives a vertical stretch by 2

Result: The \(V\)-shaped graph flips upside down and becomes steeper.

11. Worked Example 3: Horizontal compression and translation

Describe the transformations from \(y=\sqrt{x}\) to

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

Step 1: Look inside the function.

The expression inside is \(2(x-1)\).

Step 2: Identify each transformation.

• \((x-1)\) means shift right 1
• The factor 2 inside means a horizontal compression by factor \(\frac{1}{2}\)

Result: The square root graph starts at \((1,0)\) instead of \((0,0)\), and it grows more quickly because it has been compressed horizontally.

12. Worked Example 4: Multiple transformations

Describe the transformations from \(y=x^3\) to

$$y=-\frac{1}{2}(x+4)^3-1$$

Step 1: Identify the parent function.

The parent function is \(f(x)=x^3\).

Step 2: Compare with \(y=a f(x-h)+k\).

We have:

• \(a=-\frac{1}{2}\)
• \(h=-4\) because \((x+4)=(x-(-4))\)
• \(k=-1\)

Step 3: Interpret each one.

• \(x+4\) means shift left 4
• \(-\frac{1}{2}\) means reflect in the \(x\)-axis
• \(\frac{1}{2}\) means a vertical compression by factor \(\frac{1}{2}\)
• \(-1\) means shift down 1

Result: The cubic graph moves left 4, is reflected across the \(x\)-axis, compressed vertically, and then moved down 1.

13. How to graph a transformed function using key points

A good way to graph transformations is to start with key points on the parent function.

For example, for \(y=x^2\), useful points are:

• \((-2,4)\)
• \((-1,1)\)
• \((0,0)\)
• \((1,1)\)
• \((2,4)\)

If the function becomes \(y=(x-1)^2+3\), then each point moves right 1 and up 3.

So the new points are:

• \((-1,7)\)
• \((0,4)\)
• \((1,3)\)
• \((2,4)\)
• \((3,7)\)

This method is especially helpful when several transformations happen together.

14. Common mistakes to avoid

• Mixing up left and right: \(x-h\) moves right, while \(x+h\) moves left.
• Mixing up horizontal and vertical scale factors: outside multipliers affect vertical change; inside multipliers affect horizontal change.
• Forgetting the reciprocal in horizontal dilations: \(f(2x)\) compresses by \(\frac{1}{2}\), not 2.
• Missing reflections: a negative outside reflects in the \(x\)-axis; a negative inside reflects in the \(y\)-axis.

15. Strategy for analyzing any transformed function

1. Identify the parent function.
2. Look for changes inside the function.
3. Look for changes outside the function.
4. Decide whether there are translations, dilations, or reflections.
5. If needed, track key points from the parent graph.

16. Brief Summary

Function transformations let you build new graphs from familiar parent functions.

Translations move a graph, dilations change its size, and reflections flip it across an axis.

The general form \(y=a\,f(b(x-h))+k\) helps you read all of these changes clearly.

If you remember that inside changes affect horizontal movement in the opposite direction and outside changes affect vertical movement directly, you will be able to analyze most transformed functions with confidence.