Slide 2.8- 1 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Jan 12, 2016
Slide 2.8- 1 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
OBJECTIVES
Inverse Functions
Learn the definition of an inverse function and a relation.
Learn to identify one-to-one functions.
Learn a procedure for finding an inverse function.
Learn to use inverse functions to find the range of a function.
Learn to apply inverse functions in the real world.
SECTION 2.8
1
2
3
4
5
Slide 2.8- 3 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 1Determining whether an Inverse Relation is a Function
Ray’s Music Mart has six employees. The first table lists the first names and the Social Security numbers of the employees, and the second table lists the first names and the ages of the employees.
a. Find the inverse of the function defined by the first table, and determine whether the inverse relation is a function.
b. Find the inverse of the function defined by the second table, and determine whether the inverse relation is a function.
Slide 2.8- 4 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 1Determining whether an Inverse Relation is a Function
Solution
Every y–value corresponds to exactly one x–value. Thus the inverse of the function defined in this table is a function.
Dwayne 590-56-4932
Sophia 599-23-1746
Desmonde 264-31-4958
Carl 432-77-6602
Anna 195-37-4165
Sal 543-71-8026
Slide 2.8- 5 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 1Determining whether an Inverse Relation is a Function
Solution continued
There is more than one x–value that corresponds to a y–value. For example, the age of 24 yields the names Dwayne and Anna. Thus the inverse of the function defined in this table is not a function.
Dwayne 24
Sophia 26
Desmonde 42
Carl 51
Anna 24
Sal 26
Slide 2.8- 6 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
DEFINITION OF A ONE-TO-ONE FUNCTION
A function is called a one-to-one function if each y-value in its range corresponds to only one x-value in its domain.
Slide 2.8- 7 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
A ONE-TO-ONE FUNCTION
Each y-value in the range corresponds to only one x-value in the domain.
Slide 2.8- 8 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
NOT A ONE-TO-ONE FUNCTION
The y-value y2 in the range corresponds to two x-values, x2 and x3, in the domain.
Slide 2.8- 9 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
NOT A FUNCTION
The x-value x2 in the domain corresponds to the two y-values, y2 and y3, in the range.
Slide 2.8- 10 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
HORIZONTAL- LINE TEST
A function f is one-to-one if no horizontal line intersects the graph of f in more than one point.
Slide 2.8- 11 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 2 Using the Horizontal-Line Test.
Use the horizontal-line test to determine which of the following functions are one-to-one.
a. f x 2x 5 b. g x x2 1 c. h x 2 x
Solution
No horizontal line intersects the graph of f in more than one point, therefore the function f is one-to-one.
Slide 2.8- 12 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 2 Using the Horizontal-Line Test.
Solution continued
There are many horizontal lines that intersect the graph of f in more than one point, therefore the function f is not one-to-one.
b. g x x2 1
Slide 2.8- 13 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 2 Using the Horizontal-Line Test.
Solution continued
No horizontal line intersects the graph of f in more than one point, therefore the function f is one-to-one.
c. h x 2 x
Slide 2.8- 14 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
DEFINITION OF f –1 FOR AONE-TO-ONE FUNCTION f
Let f represent a one-to-one function. The inverse of f is also a function, called the inverse function of f, and is denoted by f –1. If (x, y) is an ordered pair of f, then (y, x) is an ordered pair of f –1, and we write x = f –1(y). We have y = f (x) if and only if f –1(y) = x.
Slide 2.8- 15 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 3Relating the Values of a Function and Its Inverse
Assume that f is a one-to-one function.• If f (3) = 5, find f –1(5).• If f –1(–1) = 7, find f (7).
a. Let x = 3 and y = 5. Now 5 = f (3) if and only if f –1(5) = 3. Thus, f –1(5) = 3.
b. Let y = –1 and x = 7. Now, f –1(–1) = 7 if and only if f (7) = –1. Thus, f (7) = –1.
Solution
We know that y = f (x) if and only if f –1(y) = x.
Slide 2.8- 16 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
INVERSE FUNCTION PROPERTY
Let f denote a one-to-one function. Then
f o f 1 x f f 1 x x
for every x in the domain of f –1.
1.
f 1 o f x f 1 f x x
for every x in the domain of f .
2.
Slide 2.8- 17 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
UNIQUE INVERSE FUNCTION PROPERTY
Let f denote a one-to-one function. Then if g is any function such that
g = f –1. That is, g is the inverse function of f.
f g x x for every x in the domain of g and
g f x x for every x in the domain of f, then
Slide 2.8- 18 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 4 Verifying Inverse Functions
Verify that the following pairs of functions are inverses of each other:
f x 2x 3 and g x x 3
2.
Solution
Form the composition of f and g.
f og x f g x fx 3
2
2x 3
2
3 x 3 3
x
Slide 2.8- 19 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 4 Verifying Inverse Functions
Solution continued
Now find
go f x g f x g 2x 3
2x 3 3
2x
g f x .
Since f g x g f x x, we conclude that
f and g are inverses of each other.
Slide 2.8- 20 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
SYMMETRY PROPERTY OFTHE GRAPHS OF f AND f –1
The graph of the function f and the graph of f –1 are symmetric with respect to the line y = x.
Slide 2.8- 21 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 5 Finding the Graph of f –1 from the Graph of f
The graph of the function f is shown. Sketch the graph of the f –1.
Solution
Slide 2.8- 22 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
PROCEDURE FOR FINDING f –1
Step 1 Replace f (x) by y in the equation for f (x).
Step 2 Interchange x and y.Step 3 Solve the equation in Step 2 for y.Step 4 Replace y with f –1(x).
Slide 2.8- 23 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 6 Finding the Inverse Function
Find the inverse of the one-to-one function
f x x 1
x 2, x 2.
Solution
y x 1
x 2Step 1
x y 1
y 2Step 2
x y 2 y 1
xy 2y y 1Step 3
Slide 2.8- 24 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 6 Finding the Inverse Function
Solution continued
f 1 x 2x 1
x 1, x 1Step 4
Step 3(cont.)
xy 2y 2x y y 1 2x y
xy y 2x 1
y x 1 2x 1
y 2x 1
x 1
Slide 2.8- 25 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 7 Finding the Domain and Range
Find the domain and the range of the function
f x x 1
x 2, x 2.
Solution
Domain of f, all real numbers x such that x ≠ 2, in interval notation (–∞, 2) U (2, –∞).
Range of f is the domain of f –1.
f 1 x 2x 1
x 1, x 1
Range of f is (–∞, 1) U (1, –∞).
Slide 2.8- 26 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 8 Finding an Inverse Function
Find the inverse of g(x) = x2 – 1, x ≥ 0.Solution
Step 1 y = x2 – 1, x ≥ 0
Step 2 x = y2 – 1, y ≥ 0
Step 4 g 1 x x 1
Step 3
y x 1, y 0
Since y ≥ 0, reject y x 1.
x 1 y2 , y 0
Slide 2.8- 27 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 8 Finding an Inverse Function
Solution continued
Here are the graphs of g and g –1.
Slide 2.8- 28 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 9 Water Pressure on Underwater Devices
The formula for finding the water pressure p (in pounds per square inch), at a depth d (in feet)
p 15d
33.
pressure gauge on a diving bell breaks and shows a reading of 1800 psi. Determine how far below the surface the bell was when the gauge failed.
below the surface is Suppose the
p 15d
33.
Solution
The depth is given by the inverse of
Slide 2.8- 29 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 9 Water Pressure on Underwater Devices
Solution continued
p 15d
3333p 15d
d 33p
15
Solve the equation for d.
Let p = 1800.
d 33 1800
15d 3960
The device was 3960 feet below the surface when the gauge failed.