Inverse Functions
Consider the functions
Supplemental Problems for Section 1.3 f ( x )
 x

2 and g ( x )
 x

2 x y
 f ( x )
 x

2 x y
 g ( x )
 x

2
-2 0
-1
0
1
1
2
3
2 4
Each function “undoes” what the other computes. We say that f and g are inverses of each other.
Note: The domain of f is the range of g and the domain of g is the range of f .
Definition: A function g is the inverse of the function f is f ( g ( x ))
 x and g ( f ( x ))
 x where x is known as the identity function.
Notation: The inverse of f is denoted as f

1
. Note, in general, f

1
( x )

1 f ( x )
.
Example 1: Show that g ( x )
 x

2 is the inverse of f ( x )
 x

2 .
1
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2
Note: The graph of f

1
is the reflection of the graph of f across the line y
 x .
Reflective Property of Inverse Functions: The graph of f contains the point ( a , b ) if and only if the graph of f

1
contains the point ( b , a ) .
For example, consider the graph of f ( x )
 x

2 and its inverse f

1
( x )
 x

2 .

Example 1: Sketch the graph of the inverse of the following function
3
Example 2: If f ( x )

12 x

5 , find f

1
(

6 )
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4
Example 3: If f ( x )
 x x

5

8
, find f

1
(

7 )
Example 4: If f ( x )

3

5 x

4 e x
, find f

1
( 7 )
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5
Note: N ot all functions have an inverse.
Consider f ( x )
 x
2
Fact: A function f has an inverse if and only if it is one to one , which says each y value in its range has a unique value x in its domain assigned to it.
For example, y

4 f ( x )
 x
2
is not one-to-one. Consider the value in the range of
. Then for both x
 
2 and x

2 , f ( x )
 x
2
given by x
 
2
 f (

2 )

(

2 )
2 
4 x

2
 f ( 2 )

( 2 )
2 
4
The range value of y

4 has two x values, x
 
2 and x

2 , in the domain that the function f ( x )
 x
2
assigns to it. To be one-to-one, the each range value can only have one value x in the domain assigned to it. This leads to the following test for determining if a function has an inverse
Horizontal Line Test
A function is one-to-one and hence has an inverse if and only if every horizontal line intersects the graph at most once.

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However, an inverse can be defined for a function without an inverse in some cases by restricting its domain.
For example, consider f ( x )
 x
2
defined for the domain x

0 .

Example 5: Determine whether the function 2 t
2 
4 is one-to-one and hence as an inverse.
Solution: Using Maple plot(2*t^2 + 4, t = -2..2);
7
Example 6: Determine whether the function 2 t
2 
4 , t

0 is one-to-one and hence as an inverse.
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Solution:
> plot(2*t^2 + 4, t = 0..4);
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Example 7: Determine whether the function cos( x ) , 0
 x
  is one-to-one and hence as an inverse.
Solution: > plot(cos(x), x = 0..Pi);
8
Example 8: Determine whether the function 8 sin( x )

3 cos( 10 x ) is one-to-one and hence as an
█ inverse.
Solution: > plot(8*sin(x)-3*cos(10*x), x = -2*Pi..2*Pi);
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9
Procedure for Finding an Inverse Function
1. Solve for x as a function of y .
2. Interchange x and y . The resulting equation is f

1
( x ) .
Example 8: Find the inverse of the function f ( x )

5 x

3 .
Solution:
Example 9: Find the inverse of the function f ( x )

6
 x
2
, x

0 .
Solution:
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Example 10: Find the inverse of the function f ( x )
 x
1

3
.
Solution:
Example 10: Find the inverse of the function f ( x )

12 x
3 
4 .
Solution: We start by assigning y
 f ( x )

12 x
3 
4 and solve the equation for x using the following procedure.
12 x
3 
4
 y
10
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12 x
3  y

4 x
3  y

4
12 x

3 y

4
12
Switching x and y gives the inverse function f

1
( x )

3 x

4
.
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Example 10: Find the inverse of the function f ( x )

2 e
11 e x x 
5

12
.
Solution: We start by assigning y
 f ( x )

2 e
11 e x x 
5

12
and solve the equation for x using the following procedure.
2 e x
11 e x

5

12
 y (Set function equal to y )
11
2 e x 
5
 y ( 11 e x 
12 ) (Multiply both sides by 11 e x 
12 )
2 e x 
5

11 ye x 
12 y (Distribut e the y )
2 e x 
11 ye x 
12 y

5 (Isolate the e x on left hand side of equation )
( 2

11 y ) e x 
12 y

5 (Factor e x
) e x 
12
2
 y

11
5 y
(Divide both sides by 2

11 y ) ln e x  ln


12 y
2


11
5 y

 (Take ln of both sides ) x
 ln


12
2
 y

5
11 y

 (Use fact that ln e x  x ln e
 x ( 1 )
 x to solve for x )
Switching x and y gives the inverse function f

1
( x )
 ln
12 x
2

5

11 x
.

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