1. Copyright © Cengage Learning. All rights reserved.1
Functions and
Their Graphs
Copyright © Cengage Learning. All rights reserved.

2. Copyright © Cengage Learning. All rights reserved.
1.9
INVERSE FUNCTIONS
Copyright © Cengage Learning. All rights reserved.

3. What You Should Learn
Find inverse functions informally and verify that two functions are inverse functions of each other.
Use graphs of functions to determine whether functions have inverse functions.
Use the Horizontal Line Test to determine if functions are one-to-one.
Find inverse functions algebraically.

4. Inverse Functions

5. Inverse Functions
Recall that a function can be represented by a set of ordered pairs.
For instance, the function f (x) = x + 4 from the set A = {1, 2, 3, 4} to the set B = {5, 6, 7, 8} can be written as follows.
f (x) = x + 4: {(1, 5), (2, 6), (3, 7), (4, 8)}
In this case, by interchanging the first and second coordinates of each of these ordered pairs, you can form the inverse function of f, which is denoted by f –1.
It is a function from the set B to the set A, and can be written as follows.
f –1(x) = x – 4: {(5, 1), (6, 2), (7, 3), (8, 4)}
Note that the domain of f is equal to the range of f –1, and vice versa, as shown in Figure 1.93.
Also note that the functions f and f –1 have the effect of “undoing” each other.
In other words, when you form the composition of f with f –1 or the composition of f –1 with f, you obtain the identity function.
f (f –1(x)) = f (x – 4) = (x – 4) + 4 = x
f –1(f (x)) = f –1(x + 4) = (x + 4) – 4 = x

6. var('x,y')
f = 4*x; show(f);
sol = solve(f(x=y)==x,y); show(sol);
g = sol[0].rhs(); show(g)
p1 = plot(f,(x,-6,4),legend_label='f',color='red');
p2 = plot(g,(x,-6,4),legend_label='g',color='blue');
show(p1+p2,gridlines=true,figsize=(3,3))

7. Do not be confused by the use of –1 to denote the inverse function f –1.
In this text, whenever f –1 is written, it always refers to the inverse function of the function f and not to the reciprocal of f (x).
If the function g is the inverse function of the function f, it must also be true that the function f is the inverse function of the function g.
For this reason, you can say that the functions f and g are inverse functions of each other.

8. My example: var('x,y') f = 3*x-2; show(f);
sol = solve(f(x=y)==x,y); show(sol);
g = sol[0].rhs(); show(g)
p1 = plot(f,(x,-6,4),legend_label='f',color='red');
p2 = plot(g,(x,-6,4),legend_label='g',color='blue');
show(p1+p2,gridlines=true,figsize=(3,3))

9. The Graph of an Inverse Function

10. The Graph of an Inverse Function
The graphs of a function f and its inverse function f –1 are related to each other in the following way.
If the point (a, b) lies on the graph of f, then the point (b, a) must lie on the graph of f –1, and vice versa.
This means that the graph of f –1 is a reflection of the graph of f in the line y = x, as shown in Figure 1.94.

12. var('x,y')
f = x^2; show(f);
sol = solve(f(x=y)==x,y); show(sol);
g = sol[1].rhs(); show(g)
p1 = plot(f,(x,0,9),legend_label='f',color='red');
p2 = plot(g,(x,0,9),legend_label='g',color='blue');
p3 = plot(x,(x,0,9),legend_label='x',color='green');
p = p1+p2+p3; p.ymax(9);
show(p,gridlines=true,figsize=(3,3))

13. One-to-One Functions

14. One-to-One Functions
The reflective property of the graphs of inverse functions gives you a nice geometric test for determining whether a function has an inverse function.
If no horizontal line intersects the graph of f at more than one point, then no y-value is matched with more than one x-value.
This is the essential characteristic of what are called one-to-one functions.

15. One-to-One Functions Consider the function given by f (x) = x2.
var('x,y')
f = x^2; show(f);
sol = solve(f(x=y)==x,y); show(sol);
g = sol[1].rhs(); show(g)
p1 = plot(f,(x,-3,3),legend_label='f',color='red');
p2 = plot(g,(x,0,3),legend_label='g',color='blue');
p3 = plot(x,(x,0,3),legend_label='x',color='green');
p = p1+p2+p3; p.ymax(9); p.ymin(-1);
show(p,gridlines=true,figsize=(3,3))
One-to-One Functions
Consider the function given by f (x) = x2.
The table on the left is a table of values for f (x) = x2.
The table of values on the right is made up
by interchanging the columns of the first table.
The table on the right does not represent a function because the input x = 4 is matched with two different outputs: y = –2 and y = 2.
So, f (x) = x2 is not one-to-one and does not have an inverse function.

16. Example 5(a) – Applying the Horizontal Line Test
The graph of the function given by f (x) = x3 – 1 is shown in Figure 1.97.
Because no horizontal line intersects
the graph of f at more than one point,
you can conclude that f is a one-to-one
function and does have an inverse function.
Figure 1.97

17. Example 5(b) – Applying the Horizontal Line Test
cont’d
The graph of the function given by f (x) = x2 – 1 is shown in Figure 1.98.
Because it is possible to find a horizontal line that intersects the graph of f at more than one point, you can conclude that f is not a one-to-one function and does not have an inverse function.
Figure 1.98

18. Finding Inverse Functions Algebraically

19. Finding Inverse Functions Algebraically
For simple functions, you can find inverse functions by inspection.
For more complicated functions, however, it is best to use the following guidelines.
The key step in these guidelines is Step 3—interchanging the roles of x and y.
This step corresponds to the fact that inverse functions have ordered pairs with the coordinates reversed.

20. var('x,y')
f = (5-3*x)/2; show(f);
sol = solve(f(x=y)==x,y); show(sol);
g = sol[0].rhs(); show(g)
p1 = plot(f,(x,-6,8),legend_label='f',color='red');
p2 = plot(g,(x,-6,8),legend_label='g',color='blue');
show(p1+p2,gridlines=true,figsize=(3,3))

22. reset(); var('x,y');
f = (x^5-3)/2; show(f);
sol = solve(f(x=y)==x,y); show(sol);
g = sol[4].rhs(); show(g)
p1 = plot(f,(x,-9,9),legend_label='f',color='red');
p2 = plot(g,(x,-9,9),legend_label='g',color='blue');
p3 = plot(x,(x,-9,9),legend_label='x',color='green');
p = p1+p2+p3; p.ymin(-10); p.ymax(10);
show(p,gridlines=true,figsize=(3,3))

23. reset(); var('x,y');
f = (2*x+3)/(x-1); show(f);
sol = solve(f(x=y)==x,y); show(sol);
g = sol[0].rhs(); show(g)
p1 = plot(f,(x,-9,9),legend_label='f',color='red');
p2 = plot(g,(x,-9,9),legend_label='g',color='blue');
p3 = plot(x,(x,-9,9),legend_label='x',color='green');
p = p1+p2+p3; p.ymin(-10); p.ymax(10);
show(p,gridlines=true,figsize=(3,3))