7-7 Inverse Relations & Functions

M11.D.1.1.3: Identify the domain, range, or inverse of a relation

Objectives

The Inverse of a Function

a. Find the inverse of relation m.

Relation m

x –1 0 1 2y –2 –1 –1 –2

Interchange the x and y columns.

Inverse of Relation m

x –2 –1 –1 –2y –1 0 1 2

Finding the Inverse of a Relation

b. Graph m and its inverse on the same graph.

Relation mReversing theOrdered Pairs Inverse of m

Continued

Find the inverse of y = x2 – 2.

y = x2 – 2

x = y2 – 2 Interchange x and y.

x + 2 = y2 Solve for y.

± x + 2 = y Find the square root of each side.

Interchanging x and y

The graph of y = –x2 – 2 is a parabola that opens downward with vertex (0, –2).

Graph y = –x2 – 2 and its inverse.

You can also find points on the graph of the inverse by reversing the coordinates of points on y = –x2 – 2.

The reflection of the parabola in the line x = y is the graph of the inverse.

Graphing a Relation and Its Inverse

Consider the function ƒ(x) = 2x + 2 .

a. Find the domain and range of ƒ.

Since the radicand cannot be negative, the domain is the set of numbers greater than or equal to –1.

Since the principal square root is nonnegative, the range is the set of nonnegative numbers.

b. Find ƒ –1

So, ƒ –1(x) = .x2 – 2

2

ƒ(x) = 2x + 2

y = 2x + 2 Rewrite the equation using y.

x = 2y + 2 Interchange x and y.

x2 = 2y + 2 Square both sides.

y =x2 – 2

2 Solve for y.

Finding an Inverse Function

(continued)

c. Find the domain and range of ƒ –1.

The domain of ƒ –1 equals the range of ƒ, which is the set of nonnegative numbers.

d. Is ƒ –1 a function? Explain.

For each x in the domain of ƒ–1, there is only one value of ƒ –1(x). So ƒ –1 is a function.

Note that the range of ƒ–1 is the same as the domain of ƒ.

Since x2 0, –1. Thus the range of ƒ–1 is the set of numbers

greater than or equal to –1.

x2 – 22

>– >–

Continued

The function d = 16t 2 models the distance d in feet that an

object falls in t seconds. Find the inverse function. Use the inverse to

estimate the time it takes an object to fall 50 feet.

d = 16t 2

t 2 = d

16Solve for t. Do not interchange variables.

t =d4 Quantity of time must be positive.

t =14 50 1.77

The time the object falls is 1.77 seconds.

Real-World Example

Vocabulary

If and are inverse functions, then f 1f

xxff ))(( 1

and

xxff ))(( 1

and (ƒ ° ƒ –1)(– 86) = – 86.

For the function ƒ(x) = x + 5, find (ƒ–1 ° ƒ)(652) and

(ƒ ° ƒ–1)(– 86).

12

Since ƒ is a linear function, so is ƒ –1.

Therefore ƒ –1 is a function.

So (ƒ –1 ° ƒ)(652) = 652

Composition of Inverse Functions

Homework

p 410 #1,2,5,6,14,15,23,24,31,32