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Algebra 1 View details

3. Inverse of Linear Functions

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Algebra 1 /

Chapter 3

Inverse of Linear Functions

Understanding the inverse of a function is crucial in mathematics. The lesson delves into the concept of function inverses, particularly focusing on linear functions. It emphasizes the idea that every function has an inverse relation. If this inverse relation also qualifies as a function, it's termed an 'inverse function'. The lesson provides a step-by-step guide on how to determine if two functions are inverses of each other, both algebraically and graphically. It highlights the significance of the line y=x, as the graphs of two inverse functions reflect across this line. The inverse essentially "undoes" the function, reversing its operations. This knowledge is pivotal for various applications, including real-world scenarios like currency conversion.

Lesson Settings & Tools

	12 Theory slides
	7 Exercises - Grade E - A
	Each lesson is meant to take 1-2 classroom sessions

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Slide of 12

Given a function

f,

sometimes it is possible to find another function that undoes

f .

This particular function is called the inverse function of

f .

This lesson will discuss how to verify whether two functions are inverse and will explain the procedure to find the inverse of a linear function.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Equation
Solving Multi-Step Equations
Literal Equation
Linear Function
Slope-Intercept Form
Graphing Linear Functions in Slope-Intercept Form
Function Notation

Challenge

Undoing a Function

Consider a function

f

that takes any real number and returns a number greater by two units.

What would be a function that undoes

f ?

Discussion

Inverse Relation

An inverse relation is a relation that interchanges the input and output values of a specific relation. In other words, it is the relation that undoes a certain relation. Consider the mapping diagram that shows a relation

p

that takes elements from set

A

and returns elements from set

B .

The relation that undoes

p

q .

In this case,

q

takes elements from set

B

and returns elements from set

A .

Relations can also be represented by a collection of ordered pairs. Here, the inverse of a relation is obtained by interchanging the elements of the ordered pairs of the original relation.

Relation p {(- 1, C), (0, E), (5, Z), (10, W), (20, Q)} Inverse Relation q {(C, - 1), (E, 0), (Z, 5), (W, 10), (Q, 20)}

Graphically, the inverse of a relation is a reflection across the line

y = x .

If the inverse relation of a function is also a function, then it is called an inverse function.

Discussion

Inverse of a Function

Every function has an inverse relation. If this inverse relation is also a function, then it is called an inverse function. In other words, the inverse of a function $f$ is another function $f^{- 1}$ such that they undo each other.

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

Also, if $x$ is the input of a function $f$ and $y$ its corresponding output, then $y$ is the input of $f^{- 1}$ and $x$ its corresponding output.

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

Example

Consider a function

f

and its inverse

f^{- 1} .

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

These functions will be shown to undo each other. To do so, it needs to be proven that

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

and that

f^{- 1} (f (x)) = x .

To start, the first equality will be proven. First, the definition of

f

will be used.

f (f^{- 1} (x)) = ? x ⇕ 2 f^{- 1} (x) - 3 = ? x

Now, in the above equation,

\frac{x + 3}{2}

will be substituted for

f^{- 1} (x) .

2 f^{- 1} (x) - 3 = ? x

Substitute

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

2 (\frac{x + 3}{2}) - 3 = ? x

▼

Simplify left-hand side

DenomMultFracToNumber

$2 \cdot \frac{a}{2} = a$

(x + 3) - 3 = ? x

RemovePar

Remove parentheses

x + 3 - 3 = ? x

SubTerm

Subtract term

x = x ✓

A similar procedure can be performed to show that

f^{- 1} (f (x)) = x .

	Definition of First Function	Substitute Second Function	Simplify
$f (f^{- 1} (x)) = ? x$	$2 f^{- 1} (x) - 3 = ? x$	$2 (\frac{x + 3}{2}) - 3 = ? x$	$x = x ✓$
$f^{- 1} (f (x)) = ? x$	$\frac{f ( x ) + 3}{2} = ? x$	$\frac{2 x - 3 + 3}{2} = ? x$	$x = x ✓$

Therefore, $f$ and $f^{- 1}$ undo each other. The graphs of these functions are each other's reflection across the line $y = x .$ This means that the points on the graph of $f^{- 1}$ are the reversed points on the graph of $f .$

$Graph of the linear function f(x)=2*x-3 with a point (a,b) on it and the graph of its inverse function f^{-1}(x)=x/2+3/2 with a point (b,a) on it are depicted. The line of symmetry x=y for these functions is also shown.$

In general, the inverse of a linear function is also a linear function.

Example

Using Graphs to Determine if Two Functions Are Inverse Functions

Kriz enjoys playing video games with their friends. For their birthday they received a copy of Mathleaks: The Adventure, a math-based video game.

Kriz is very close to beating this level. They need to graph the functions given below and determine whether they are inverse functions.

f (x) = 4 x + 5 and g (x) = \frac{1}{4} x - \frac{5}{4}

Help Kriz beat the level and get one step closer beating the game by mastering math!

Answer

Graph:

Are $f$ and $g$ Inverse Functions? Yes.

Hint

Graph both functions on the same coordinate plane and see if they are each other's reflection across the line $y = x .$

Solution

Since $f$ and $g$ are linear functions written in slope-intercept form, they can both be graphed using their slope and $y -$ intercept.

Now, it can be seen whether the lines are each other's reflection across

y = x .

The lines are indeed each other's reflection across

y = x .

Therefore,

f

and

g

are inverse functions.

Pop Quiz

Determining if Two Functions Are Inverses by Looking at Their Graphs

It was discussed previously that if the graphs of two functions are each other's reflection across $y = x,$ then they are inverse functions. Therefore, it can be determined if two functions are inverse just by looking at their graphs. Determine whether the following linear functions are inverse by looking at the lines.

Do the lines shown correspond to inverse functions?

Example

Verifying Inverse Functions by Composition

If Paulina gets an A on a math test, her mother will buy her a new saxophone.

To get an A, Paulina must answer two questions correctly. Help her get the new saxophone!

a Given

f (x) = 5 x + 10

and

g (x) = \frac{1}{5} x - 2,

calculate

f (g (x))

and

g (f (x)),

and determine whether

f

and

g

are inverse functions.

Are

f

and

g

inverse functions?

b Given

h (x) = \frac{1}{2} x + 6

and

k (x) = 2 x - 6,

calculate

h (k (x))

and

k (h (x)),

and determine whether

h

and

k

are inverse functions.

Are

h

and

k

inverse functions?

Hint

a If

f (g (x)) = x

and

g (f (x)) = x,

then

f

and

g

are inverse functions.

b If

h (k (x)) = x

and

k (h (x)) = x,

then

h

and

k

are inverse functions.

Solution

a Consider the given functions.

f (x) = 5 x + 10 and g (x) = \frac{1}{5} x - 2

The composite function

f (g (x))

will be found first. To do so, the definition of

f

will be used.

f (g (x)) = 5 g (x) + 10

Next,

g (x) = \frac{1}{5} x - 2

can be substituted in the above equation.

f (g (x)) = 5 g (x) + 10

Substitute

$g (x) = \frac{1}{5} x - 2$

f (g (x)) = 5 (\frac{1}{5} x - 2) + 10

▼

Simplify right-hand side

Distr

Distribute $5$

f (g (x)) = 5 (\frac{1}{5} x) - 5 (2) + 10

AssociativePropMult

Associative Property of Multiplication

f (g (x)) = 5 (\frac{1}{5}) x - 5 (2) + 10

DenomMultFracToNumber

$5 \cdot \frac{a}{5} = a$

f (g (x)) = 1 x - 5 (2) + 10

Multiply

f (g (x)) = x - 10 + 10

AddTerms

Add terms

f (g (x)) = x

By following the same procedure, an expression for

g (f (x))

can be found.

	Definition of First Function	Substitute Second Function	Simplify
$f (g (x))$	$f (g (x)) = 5 g (x) + 10$	$f (g (x)) = 5 (\frac{1}{5} x - 2) + 10$	$f (g (x)) = x$
$g (f (x))$	$g (f (x)) = \frac{1}{5} f (x) - 2$	$g (f (x)) = \frac{1}{5} (5 x + 10) - 2$	$g (f (x)) = x$

It has been found that both $f (g (x))$ and $g (f (x))$ are equal to $x .$ Therefore, $f$ and $g$ are inverse functions.

b Consider the given functions.

h (x) = \frac{1}{2} x + 6 and k (x) = 2 x - 6

By following the same procedure as in Part A, the expressions for

h (k (x))

and

k (h (x))

will be found.

	Definition of First Function	Substitute Second Function	Simplify
$h (k (x))$	$h (k (x)) = \frac{1}{2} k (x) + 6$	$h (k (x)) = \frac{1}{2} (2 x - 6) + 6$	$h (k (x)) = x + 3$
$k (h (x))$	$k (h (x)) = 2 h (x) - 6$	$k (h (x)) = 2 (\frac{1}{2} x + 6) - 6$	$k (h (x)) = x + 6$

It has been found that neither $h (k (x))$ nor $k (h (x))$ is equal to $x .$ Therefore, $h$ and $k$ are not inverse functions.

Pop Quiz

Determining if Two Functions Are Inverse by Composition

For the functions $f$ and $g,$ use a composition to determine whether they are inverse functions.

determine if two functions are inverse by compounding

Discussion

Finding the Inverse of a Function Algebraically

A function can be represented by a table of values, a graph, a mapping diagram, or a function rule, among other ways. Depending on how the function is presented, finding its inverse can be done in different ways. When a function rule is given, finding the inverse algebraically is advantageous. Consider the following example function.

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

There is a series of steps to follow in order to find the inverse function

f^{- 1} (x) .

Replace

f (x)

With

y

To begin, since

f (x) = y

describes the input-output relationship of the function, replace

f (x)

with

y

in the function rule.

f (x) = \frac{2 x - 1}{3} \to y = \frac{2 x - 1}{3}

Switch

x

and

y

Because the inverse of a function reverses

x

and

y,

the variables can be switched. Notice that every other piece in the function rule remains the same.

y = \frac{2 x - 1}{3} switch x = \frac{2 y - 1}{3}

Solve for

y

Solve the resulting equation from the previous step for

y .

This will involve using the inverse operations.

x = \frac{2 y - 1}{3}

▼

Solve for

y

MultEqn

$LHS \cdot 3 = RHS \cdot 3$

3 x = 2 y - 1

AddEqn

$LHS + 1 = RHS + 1$

3 x + 1 = 2 y

DivEqn

$LHS / 2 = RHS / 2$

\frac{3 x + 1}{2} = y

RearrangeEqn

Rearrange equation

y = \frac{3 x + 1}{2}

Replace

y

With

f^{- 1} (x)

Just as

f (x) = y

shows the input-output relationship of

f,

so does

f^{- 1} (x) = y .

Therefore, replacing

y

with

f^{- 1} (x)

gives the rule for the inverse of

f .

y = \frac{3 x + 1}{2} \to f^{- 1} (x) = \frac{3 x + 1}{2}

Notice that in

f,

the input is multiplied by

2,

decreased by

1,

and divided by

3 .

From the rule of

f^{- 1},

it can be seen that

x

undergoes the inverse of these operations in the reverse order. Specifically,

x

is multiplied by

3,

increased by

1,

and divided by

2 .

Example

Finding the Inverse of a Function

Zosia lives in Honolulu, Hawaii.

She is obsessed with math. While watching a surfer catch some gnarly waves, she came up with a function that models the distance of the surfer from the shore.

f (x) = - \frac{4}{3} x + 24

Here,

f (x)

is the distance of the surfer from the shore after

x

seconds of riding a wave. Given the distance from shore, Zosia now wants to finds the function that gives the time, in seconds, the surfer has been riding a wave. To do so, she needs to find

f^{- 1},

the inverse of

f .

Help Zosia find this function! Write the answer in slope-intercept form.

Hint

Start by replacing $f (x)$ with $y .$ After that, switch the $x -$ and $y -$ variables. Then, solve the obtained equation for $y .$ Finally, replace $y$ with $f^{- 1} (x) .$

Solution

To find the inverse of the function, there are four steps to follow.

Replace $f (x)$ with $y .$
Switch $x$ and $y .$
Solve the obtained equation for $y .$
Replace $y$ with $f^{- 1} (x) .$

These steps will be done one at a time.

Step $1$

Here, the variable

y

will be substituted for

f (x)

in the given function.

f (x) = - \frac{4}{3} x + 24 \to y = - \frac{4}{3} x + 24

Step $2$

In this step, the variables

x

and

y

must be switched.

y = - \frac{4}{3} x + 24 switch x = - \frac{4}{3} y + 24

Step $3$

Here, the equation obtained in the previous step will be solved for

y .

x = - \frac{4}{3} y + 24

▼

Solve for

y

SubEqn

$LHS - 24 = RHS - 24$

x - 24 = - \frac{4}{3} y

MultEqn

$LHS \cdot (- \frac{3}{4}) = RHS \cdot (- \frac{3}{4})$

(x - 24) (- \frac{3}{4}) = y

Distr

Distribute $(- \frac{3}{4})$

x (- \frac{3}{4}) - 24 (- \frac{3}{4}) = y

CommutativePropMult

Commutative Property of Multiplication

- \frac{3}{4} x - 24 (- \frac{3}{4}) = y

MultPosNeg

$a (- b) = - a \cdot b$

- \frac{3}{4} x - (- 24 (\frac{3}{4})) = y

MoveLeftFacToNum

$a \cdot \frac{b}{c} = \frac{a \cdot b}{c}$

- \frac{3}{4} x - (- \frac{7 2}{4}) = y

CalcQuot

Calculate quotient

- \frac{3}{4} x - (- 18) = y

SubNeg

$a - (- b) = a + b$

- \frac{3}{4} x + 18 = y

RearrangeEqn

Rearrange equation

y = - \frac{3}{4} x + 18

Step $4$

Finally,

f^{- 1} (x)

will be substituted for the

y -

variable in the equation obtained in the previous step.

y = - \frac{3}{4} x + 18 ↓ f^{- 1} (x) = - \frac{3}{4} x + 18

Pop Quiz

Finding the Inverse Function

Find the inverse of the given function. Write the answer in slope-intercept form — in the format $y = m x + b .$ If the slope and $y -$ intercept are not integers, express them as decimal numbers. If necessary, round them to two decimal places.

finding the inverse of a linear function

Closure

Undoing a Function

In this lesson, it has been seen that the function that undoes a function

f

is its inverse

f^{- 1} .

Consider the function given in the challenge presented at the beginning of the lesson.

Find the inverse function of

f (x) = x + 2 .

Hint

Start by replacing $f (x)$ with $y .$ After that, switch the $x -$ and $y -$ variables. Then, solve the obtained equation for $y,$ and finally replace $y$ with $f^{- 1} (x) .$

Solution

To find the inverse of the function, there are four steps to follow.

Replace $f (x)$ with $y .$
Switch $x$ and $y .$
Solve the obtained equation for $y .$
Replace $y$ with $f^{- 1} (x) .$

These steps will be done one at a time.

Step $1$

Here, the variable

y

will be substituted for

f (x)

in the given function.

f (x) = x + 2 substitute y = x + 2

Step $2$

In this step, the variables

x

and

y

must be switched.

y = x + 2 switch x = y + 2

Step $3$

Here, the equation obtained in the previous step will be solved for

y .

x = y + 2 \Leftrightarrow y = x - 2

Step $4$

Finally,

f^{- 1} (x)

will be substituted for the

y -

variable in the equation obtained in the previous step.

y = x - 2 substitute f^{- 1} (x) = x - 2

It has been found that the inverse function of

f (x) = x + 2,

and therefore the function that undoes it, is

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

Level 1

Level 2

Level 3

Inverse of Linear Functions

Exercises

Let

f (x)

be a function and

f^{- 1} (x)

its inverse. The following information is given.

f (x) f^{- 1} (x) = \frac{1}{a} x + 5 = 3 x + b

Find the values of

a

and

b .

To find the values of a and b, we will determine the function rule for f^(-1)(x) by using the equation of f(x). Then we will compare the obtained equation for f^(- 1)(x) with the given equation.

Determining f^(-1)(x)

We know that f^(- 1)(x) and f(x) are inverse functions. Therefore, to find the function rule for f^(- 1)(x), we will substitute y for f(x), swap the x- and y-variables, and then solve for y.

Lastly, we can replace y with f^(-1)(x) to complete the equation of the inverse function. f^(-1)(x)=ax-5a

Compare

Notice that we now have two equations for f^(-1)(x), the given equation and the one that we wrote. ccc Given Equation & & Obtained Equation [0.8em] & & f^(-1)(x)=ax-5a f^(-1)(x)= 3x+ b & & ⇕ & & f^(-1)(x)= ax+( - 5a) Let's start by comparing the linear terms of the equations. In the given equation, the coefficient of this term is 3. In the equation we found, the corresponding coefficient is a. Therefore, we know that a= 3. Let's substitute this value into our equation to find the value of b.

We can see that b= - 15. With this information, we can formulate our final answer. a= 3 and b= - 15