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# 1.5 Inverse functions

```1.5 Inverse functions
Question
Is there any ways to express the statement: “x is an angle such
that cos x=1” in a short way?
In other words, if f is a function that turns x into y, then what is
THE THING THAT TURNS Y BACK INTO X?
It is called an Inverse function
Any function has an inverse?
f (x)  2 x
f (x)  x
f (x)  2
2
f :times 2
x  2x
f :power 2
x  x2
f
divided by 2

square root
?

x 2 x
x
x
The inverse is g( x ) 
x
2
The inverse is g( x )  x
only true when x  0.
No inverse
because there is no function
that can turn 2 into any number.
In this section, you’ll learn

What is an inverse function?

How to find an inverse function?

When does a function has an inverse?

How inverse trigonometric functions are defined?
Content

Definition of inverse functions

Find an inverse function

Condition for the existence of an inverse

Domain, range, and graph of inverse trigonometric
functions
What is an inverse function?
Given f is a function with domain A and range B.
The inverse function f 1 (if exists) has domain B and range A
and is defined as follows :
f 1  y   x

f ( x )  y for all y in B
Properties:
f 1  f ( x )  x
for all x  A
f  f 1 ( y )  y
for all y  B
This definition says that if f turns x int o y,
then f 1 will turn y back int o x.
It means that f 1 reverses the action of f .
Example: Find f 1 , if exists, given :
a) f  (0,3),(1,5),(3,9)
b) f ( x )  2 x  3
2x  3
c) f ( x ) 
x 1
Solution
a) The inverse of f simply reverses the ordered pairs, thus
f 1  (3,0),(5,1),(9,3)
You have to check to ensure that f 1 is a function, i.e.
each x has a unique y.
0
3
1
5
9
3
3
5
0
9
3
1
b) Let y  2 x  3, you have to solve this equation to find x in terms of y.
y  3  2x
y3
x
2
y3
1
So f ( y ) 
.
2
2x  3
c) Let y 
, you have to solve this equation to find x in terms of y.
x 1
yx  y  2 x  3
( y  2) x  3  y
3 y
x
2y
Notice that this expression must be defined and ensure that x  1.
So, y  2.
y3
1
Therefore, f ( y) 
for all y  2.
2y
DIY: Find f , if exists, given:
1
1) f  (0,3),(1,5),(3,9),(5,9)
A . f 1  (3,0),(5,1),(9,3),(9,5)
B . f 1 does not exist.
2) f ( x )  2  x
A . f 1 ( y )  (2  y)2
C . f 1 ( y )  (2  y)2 , y  2
B . f 1 ( y )  y  2, y  0
D . f 1 ( y )  y  2, y  2
Graph of an inverse
The graphs of f and f 1 (if exists) are symmetric with respect to
the line y=x.
The existence of an inverse
If f 1 exists, then its graph must satisfies the Vertical line test.
It implies that the graph of f must satisfies the HORIZONTAL LINE TEST.
The horizontal line test:
A function f has an inverse if and only if the graph of f intersects
any horizontal line at AT MOST ONE POINT.
f
f
f 1
f has an inverse.
f has no inverse.
Trigonometric functions
 Trigonometric functions DO NOT satisfy the Horizontal
line test.
 However, if we restrict the domain of these functions, the
inverse will exist.
The sine function
The inverse sin 1 x
sin 1 x  y  sin y  x and 

2
y

2
The cosine function
The inverse cos1 x
cos1 x  y  cos y  x and 0  y  
The tangent function
The inverse tan 1 x
tan 1 x  y  tan y  x and 

2
y

2
The cotangent function
The inverse cot 1 x
cot 1 x  y  cot y  x and 0  y  
Example: Evaluate

2
a. sin  

 2 


1
 1 
c. tan 

 3
b. cos 0
1
1
Solution

2
2
a. sin  
  y  sin y  
 2 
2


1
b. cos 0  y  cos y  0
1

2
y
and 0  y   so y 
 1 
1
c. tan 

y

tan
y


3
 3
1
and 
and 

2
y

2

2

2
so y  
.
so y 

6
.

4
.
 1
DIY : Find the domain of f ( x )  sin 
2
5

x

1

 .

Properties
Notice: These properties are all implied from two
properties of an inverse.
Can you write the similar properties for cosine and cotangent
functions?
Example: Evaluate

a. sin sin 1 0.5


b. sin sin 1 2

c. sin 1  sin 0.5
Solution
Applying the properties in the previous slide, we get:
because  1  0.5  1.


b. sin  sin 2  does not exist because 2 is outside [-1,1]
a. sin sin 1 0.5  0.5
1
c. sin 1  sin 0.5 =0.5 because 

2
 0.5 

2


DIY: Let f ( x )  sin 2 cos1 x  sin 1 x .
Find the domain of f and solve f ( x )  0.
Now, you’re able to

Find an inverse function

Determine if a function has an inverse

State domain, range, graph, and do calculation with
inverse trigonometric functions
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