2
Inverse function
By: Ahmad Ghandehari
1- One to one functions.
Suppose the sets A and B are the subset of IR. ( 1R, is the set of real numbers ).
A function f with domain A and range B is called a one to one function, if none of two elements
of A have the same image.
And , every element of B is the image of only one element of A.
For each couple elements x1,x 2 belongs to A, if x1  x 2 then f (x1 )  f (x 2 ) or
f (x1 )  f (x 2 ) Then x1  x 2
Example:
Show that the function f, given by f ( x )  x 3  5 is one to one function.
Solution:
The domain of f is IR.
For each couple x1 , x 2 belongs to IR let f (x1 )  f (x 2 ) , so x13  5  x 32  5
x13  x 32  x1  x 2
Problem 1:
Show that the function f given by f (x)  x 3  5x  2 is one to one function.
Solution:
The domain of f is IR .
For each couple x1 , x 2 belongs to IR let f (x1 )  f (x 2 ) so
x13  5x1  2  x 32  5x 2  2
x13  x 32  5x1  5x 2  0
(x1  x 2 )( x12  x 22  x1.x 2 )  5( x1  x 2 )  0
(x1  x 2 )( x12  x 22  x1.x 2  5)  0
First, we show that
x12  x 22  x1.x 2  5 is positive.
x 2 2 x 22
x  x  x1.x 2  5  ( x1  )   x 22  5
2
4
2
1
2
2
 ( x1 
x2 2 3 2
)  x 2  50
2
4
So just x1  x 2  0  x1  x 2 .
1
Problem 2:
Show that the function f given by f ( x )  2 x  1  3 2 x  1 when x 
1
is one to one
2
function.
Solution:
1

For each couple x1 , x 2 belongs to the  ,    . Let f (x1 )  f (x 2 )
2

2x1  1  3 2x1  1  2x 2  1  3 2x 2  1
let ; 2x1  1  a  2x1  a  1
and
a 0
let ; 2x 2 1  b  2x 2  b  1
and
b0
So
a 11 3 a  b 11 3 b
Then
a 23 a  b23 b
Then
a  2  3 a b  2  3 b
a  b  3 a 3 b 0
( a  b )( a  b )  3( a  b )  0
( a  b )( a  b  3)  0
a  b 0 
We know that ( a  b  3) is positive, so
a  b  a b
Then 2x1  1  2x 2  1  2x1  2x 2  x1  x 2 .
2- Inverse function.
If f is a function given by f (1, a) , (2 , b) , (3, c)  and there exists a function g such
that g(a ,1) , (b , 2) , (c , 3)  , then we say that two functions f and g are inverse to
each other.
Now this question is propounds that, if in each function like f , we change the elements
of ordered pairs, then we will find a new function which is the inverse function of f ?
The answer is no, unless the function f is an one to one function.
3- Definition.
If f is a function given by f  (x , y) y  f (x) , in the event, there exists a function g
such that g  ( y , x) x  g( y) , then two functions f and g are inverse to each other.
In the former example we have proved that the function f ( x )  x 3  5 is one to one
function, now we want to find the equation of inverse function of f.
2
y  f (x)  x 3  5
x3  5  y  x3  y  5  x  3 y  5
now we substitute x to y and y to x, so y  3 x  5 . We call this function g
g( x )  3 x  5 .
Two functions f ( x )  x 3  5 and g( x )  3 x  5 are inverse to each other.
Attention: We show the inverse function of f, by symbol f 1 , so f ( x )  x 3  5 and
f 1 ( x )  3 x  5 .
Abstract: For to find the equation of inverse function, y  f ( x ) , we have to find x with
respect to y, then we substitute x to y and y to x.
Back to the two functions f ( x )  x 3  5 and f 1 ( x )  3 x  5 .
Point A (0, -5) is on the curve of f , then point A(-5 , 0) is on the curve of f 1 .
In generally if A(x, y) is on the curve of invertible function then A(y , x) is on the curve
of inverse function, we say that two points A and A are homologous to each other.
As we know two points A(x, y) and A(y , x) are symmetric to each other about the line
y  x , therefore, the curves of f and f 1 are symmetric to each other about the line y=x.
Problem 1:
Assume f ( x)  x 2  2x , x 1
a) Show that this function is invertible
b) Find the equation of f 1 ( x )
c) Draw the graphs of f and f 1 in one coordinate system.
Solation:
Part (a): We have to show that f is one to one function f ( x)  x 2  2x , x 1 or
x 1,  
For each couple x1 , x 2 1,  
let
f (x1 )  f (x 2 )
x12  2x1  x 22  2x 2
x12  x 22  2x1  2x 2  0
(x1  x 2 )( x1  x 2 )  2(x1  x 2 )  0
(x1  x 2 )( x1  x 2  2)  0
x1
, so x1  x 2  2 , so x1  x 2  2  0
3
therefore x1  x 2  0  x1  x 2 .
Part (b) : y  f ( x )  x 2  2x
x 1
y  x 2  2x  x 2  2x  y 
x 2  2x  1  y  1  ( x  1)2  y  1 ,
x 1
x  1  y  1  x  1  y  1 , interchanging x and y so y  1  x  1 or
f 1 ( x )  1  x  1
x
f (x)
1
1
2
3
0
3
4
8
x  1
x
1
0
3
8
f 1 ( x )
1
2
3
4
Problem 2:
Assume f ( x )  x 4  8x 2
and x 0 , 2
a) Show that f is one to one function
b) Find the equation of f 1 ( x ) .
Solution:
Part (a): f ( x )  x 4  8x 2 , 0  x  2
For each couple x1 , x 2  0 , 2
let
f (x1 )  f (x 2 ) , so
x14  8x12  x 42  8x 22
x14  x 42  8x12  8x 22  0
(x12  x 22 )( x12  x 22 )  8(x12  x 22 )  0
(x12  x 22 )( x12  x 22  8)  0
0  x1 , x 2  2 ; x12  x 22  8  0
so
x12  x 22  0  ( x1  x 2 )( x1  x 2 )  0
0  x1 , x 2  2 ; x1  x 2  0
so
x1  x 2  0  x1  x 2 so f is (1-1) function.
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Part (b):
y  f ( x )  x 4  8x 2
x 4  8x 2  y
x 4  8x 2  16  y  16
( x 2  4) 2  y  16
x 2  4   y  16
0  x  2  0  x2  4  x2  4  0
so
x 2  4   y  16  x 2  4  y  16 
x   4  y  16 , 0  x  2  x  4  y  16 .
So
y  4  x  16  f 1 ( x )  4  x  16
Problem 3:
We know that the function f given by f ( x )  x 3  3x 2  5 is a one to one function , so
find the equation of f 1 ( x ) .
Solution:
y  f ( x)  x 3  3x 2  3x  5
x 3  3x 2  3x  5  y
x 3  3x 2  3x  1  y  4
(x  1)3  y  4  x  1  3 y  4  x  1  3 y  4
x  y
so y  1  3 x  4  f 1 ( x )  1  3 x  4 .

y

x

Problem 4:
We know that the function f given by f ( x )  ( x  3) x  3x , x  0 is one to one
function, find the equation of f 1 ( x ) .
Solution:
y  f ( x )  ( x  3) x  3x , x  0
( x  3) x  3x  y  x x  3 x  3x  y
x x  3x  3 x  1  y  1  ( x  1)3  y  1
x 1  3 y 1 

x  1  3 y 1  x  1 3 y 1
5

2

interchanging x to y, so y  1  3 x  1

2


2
, f 1 ( x )  1  3 x  1 .
Notice:
If two curves of f and f 1 are intersect to each other, then intersection points usually are
on the line y  x , unless the function has homologous points, for instance
f (x )  x 3  1 which its inverse is f 1 ( x )  3 1  x
Points A(0 , 1) and A(1 , 0) belongs to f and f 1 so A(0 , 1) , and A(1 , 0) are two
intersection points of f and f 1 and none of each does not on the line y  x .
Inverse for the multiform function
A multiform function is invertible when
a) the intersection sets of range is empty
b) Each equation of function is invertible.
 2x , x  1
Example: Assume f ( x )  
x  1 , x  1
 y  2x , x  1 then y1  2 , 2 ,  
f (x)   1
y2  x  1 , x  1 then y1  2 , ( , 2)
( , 2) 2 ,     so
The intersection of two sets of range is empty.
y1  2x is one to one and
y2  x  1 is also one to one.
So the function f is invertible.
y1  2x  x 
y1
x
x
, y1  ; f11 ( x )  , x  2
2
2
2
y2  x  1  x  y2  1 , y2  x  1 ; f 21 (x)  x  1 , x  2
Some points:
1- If invertible function f is strictly increasing on [a, b], then the function f 1 is
strictly in creasing on [f(a) , f(b)].
2- If invertible function f is strictly decreasing on [a , b], then the function f 1 is
strictly decreasing on [f(b) , f(a)].
3- If in function f ( x ) 
ax  b
, ( c  0)
cx  d
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a  d  0 Then f 1 ( x )  f ( x )
4- If function f is invertible then
a): f 1 f ( x )  x , x belongs to the domain of f.


b): f f 1 (x)  x , x belongs to the range of f.

5- If function f is invertible function then we have f 1 f ( x )  
 
7
1
f ( x )