4 - Inverse Functions

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Before starting, you must check:
  1. Notes of Inverse Functions - Class 10
Remember the following points:
  1. The function obtained by inversing a function may not necessarily be a function.
  2. For the inverse of a function to be a function, the initial function should be a one to one onto function.
Here are more solutions from Algebra:

1. Find the inverse function of the following function:

a) f={(x,y): y=x and x = {1,2,3}}
Solution:
Given,
f={(1,1),(2,2),(3,3)}
So,
Inverse of function f = {(1,1),(2,2),(3,3)}
2a) If f={(1,2),(2,3),(4,5)}, find $f^{-1}$ and the domain and range of  $f^{-1}$.

Solution:
$f^{-1}$ = {(2,1),(3,2),(5,4)}
Domain of $f^{-1}$ = {2,3,5}
Range of $f^{-1}$ = {1,2,4}


3a) If f={(1,5),(2,4),(3,6)} and fog={(5,6),(4,5),(6,4)} then find $g^{-1}$.
Solution:
fog={(5,6),(4,5),(6,4)}
f = {(1,5),(2,4),(3,6)}
We know, in fog, f is the range and g is the domain.
So, g = {(5,3),(4,1),(6,2)}
And, $g^{-1}$ = {(1,4),(2,6),(3,5)}


4. Find the inverse functions of the following functions:

a) f(x) = 2x + 3
Solution:
Let, f(x) = y = 2x + 3
Interchanging the roles of x and y, we get,
x = 2y + 3
or, x - 3 = 2y
or, y = (x - 3)/2
So, $f^{-1} = \dfrac{x - 3}{2}$


c) f(x) = (x + 5)/3
Solution:
Let f(x) = y = (x + 5)/3
Interchanging the roles of x and y, we get,
x = (y + 5)/3
or, 3x = y + 5
or, 3x - 5 = y
or, y = 3x - 5
So, $f^{-1} = 3x - 5$


e) f:x - > (2x + 5)/6
Solution:
Let f(x) = y = (2x+5)/6
Interchanging the roles of x and y, we get,
x = (2y +5)/6
or, 6x = 2y + 5
or, 6x - 5 = 2y
or, y = (6x -5)/2
So, $f^{-1} = \dfrac{6x - 5}{2}$


g) f:x -> (2x +3)/(x -3); x ≠ 3
Solution:
Let f(x) = y = (2x + 3)/(x - 3)
Interchanging the roles of x and y, we get,
x = (2y + 3)/(y - 3)
or, x(y - 3) = 2y + 3
or, xy - 3x = 2y + 3
or, xy - 2y - 3x = 3
or, xy - 2y = 3 + 3x
or, y(x - 2) = 3x + 3
or, y = (3x + 3)/(x - 2)
So, $f^{-1} = \dfrac{3x + 3}{x - 2}$


i) f:x -> (4x + 7)/(5x + 3)
Solution:
Let f(x) = y = (4x + 7)/(5x +3)
Interchanging the roles of x and y, we get,
x = (4y +7)/(5y +3)
or, x(5y + 3) = 4y +7
or, 5xy + 3x = 4y + 7
or, 3x - 7 = 4y - 5xy
or, 3x - 7 = y(4 - 5x)
or, y(4 -5x) = 3x - 7
or, y = (3x - 7)/(4 - 5x)
So, $f^{-1} = \dfrac{3x - 7}{4 - 5x}$


5a) If f(x) = (x -5)/4 and f-¹(x) = 25, find the value of x.
Solution:
Let f(x) = y = (x -5)/4
Interchanging roles of x and y, we get,
x = (y -5)/4
or, 4x = y - 5
or, y = 4x + 5
So, f-¹(x) = 4x + 5
Given,
f-¹(x) = 25
or, 4x + 5 = 25
or, 4x = 25 - 5
or, 4x = 20
So, value of x = 5


6a) If f(x) = 8 - 3x, evaluate f-¹(-4) and ff(2).
Solution:
Let f(x) = y = 8 - 3x
Interchanging the roles of x and y, we get,
x = 8 - 3y
or, 3y + x = 8
or, 3y = 8 - x
or, y = (8-x)/3
So, f-¹(x) = (8 - x)/3
So, f-¹(-4) = {8 - (-4)}/3 = (8+4)/3 = 12/3 = 4
And,
ff(x) = fof(x) = f[f(x)]
= f(8 - 3x)
= 8 - 3(8 - 3x)
= 8 - 24 + 9x
= 9x - 16
So, ff(2) = 9×2 - 16 = 18 - 16 = 2


7a) If f(4x+5) = 12x + 18 then find f-¹(x).
Solution:
Here,
f(4x + 5) = 12x + 18
or, f(4x + 5) = 12x + 15 + 3
or, f(4x + 5) = 3(4x + 5) + 3
or, f(x) = 3x + 3
Let f(x) = y = 3x + 3
Interchanging the roles of x and y, we get,
x = 3y + 3
or, x - 3 = 3y
or, (x - 3)/3 = y
So, f-¹(x) = (x -3)/3


8a) If f(x) = 2x +7, find f(x+2) and $f^{-}$(x+2).

Solution:
Given,
f(x) = 2x +7
Now,
f(x+2) = 2(x +2) + 7
= 2x + 4 + 7
= 2x + 11
And,
Let f(x) = y = 2x +11
Interchanging the roles of x and y, we get,
or, x = 2y + 11
or, x -11 = 2y
or, 2y = x -11
or, y = $\frac{x -11}{2}$
So, $f^{-1}$(x+2) = $\frac{x -11}{2}$


9a) If f(x) = 2x +1, g(x) = $\frac{x-5}{2}$ then find the value of $f^{-1}og^{-1}$(3).

Solution:

Given,
f(x) = 2x +1
Let, f(x) = y = 2x +1

or, y = 2x +1
Interchanging the roles of x and y, we get,

or, x = 2y +1
or, x -1 = 2y
or, 2y = x -1
or, y = $\frac{x -1}{2}$
So, $f^{-1}(x) = \frac{x-1}{2}$

And,
g(x) = $\frac{x-5}{2}$
Let, g(x) = y = $\frac{x-5}{2}$
or, y = $\frac{x-5}{2}$

Interchanging the roles of x and y, we get,

or, x = $\frac{y -5}{2}$
or, 2x = y -5
or, 2x +5 = y
So, $g^{-1}(x) = 2x+5$

Now,
$f^{-1}g{-1}(x) = f^{-1}[g^{-1}(x)]$
= $f^{-1}[2x +5]$
= $\frac{2x + 5-1}{2}$
= $\frac{2x+4}{2}$
= $\frac{2(x+2)}{2}$
= $x+2$
So,
$f^{-1}g{-1}(3) = (3)+2 = 5$


10a) If f(x) = 1 +2x and g(x) = $\frac{1}{1-x}$, where x ≠ 1, find the value of $g^{-1}\frac{1}{2}$ and fog(-1).

Solution:

Given,
f(x) = 1 + 2x
g(x) = $\frac{1}{1-x}$, x≠1

Let g(x) = y = $\frac{1}{1-x}$
or, y = $\frac{1}{1-x}$

Interchanging the roles of x and y, we get,

or, x = $\frac{1}{1 - y}$
or, x(1 - y) = 1
or, x - xy = 1
or, x -1 = xy
or, $\frac{x-1}{x}$ = y
So, $g^{-1}(x) = \frac{x-1}{x}$

And,
$g^{-1}(\frac{1}{2}) = \dfrac{ \frac{1}{2} - 1}{\frac{1}{2}}$
= $\dfrac{1 - 2}{2} × \dfrac{2}{1}$
= $-1$

Now,
fog(x) = f[g(x)]
= f[$\frac{1}{1-x}$]
= 1 + $2 × \frac{1}{1 - x}$
= 1 + $\frac{2}{1-x}$

Again,
fog(-1) = 1 + $\frac{2}{1-(-1)}$
= 1 + $\frac{2}{2}$
= 1 + 1
= 2

11a) If f(x) = 3x + 4; g(x) = 2(x +1), prove that fog = gof and find the value of $f^{-1}$(2).

Solution:

Given,
f(x) = 3x + 4
g(x) = 2(x + 1)

To prove: fog = gof


Taking LHS

fog(x) = f[g(x)]

= f[2(x +1)]

= f[2x + 2]

= 3(2x + 2) + 4

= 6x + 6 + 4

= 6x + 10


Taking RHS

gof(x) = g[f(x)]

= g[3x + 4]

= 2(3x + 4 + 1)

= 2(3x + 5)

= 6x + 10

Since, LHS = 6x + 10 = RHS
fog(x) = gof(x) #proved.


Again,

Let f(x) = y = 3x + 4

or, y = 3x + 4

Interchanging the roles of x and y

or, x = 3y + 4

or, x - 4 = 3y

or, 3y = x - 4

or, y = $\frac{x - 4}{3}$

$\therefore f^{-1}$(x) = $\frac{x -4}{3}$


Put x = 2 in $f^{-1}(x)$, we get,

$f^{-1}(2) = \dfrac{2-4}{3}$

= \dfrac{-2}{3}$

= - \dfrac{2}{3}$

Hence, the required value of $f^{-1}$(2) is -2/3.


12a) Given that function f(x) = 3x + a. If ff (6) = 10, find the value of a and $f^{-1}$(4).

Solution:

Given,
f(x) = 3x + a
We know,
fof (6) = f[f(x)]
or, 10 = f[3x + a]
or, 10 = 3(3x + a) + a
or, 10 = 9(6) + 3a + a
or, 10 = 9(6) + 4a
or, 10 = 54 + 4a
or, 4a = 10 - 54
or, 4a = - 44
or, 4 × a = 4 × (-11)
So, a = -11

Now,
f(x) = 3x + (-11) = 3x -11

Let y = f(x) = 3x - 11
or, y = 3x - 11
Interchanging the roles of x and y

or, x = 3y - 11
or, x + 11 = 3y
or, y = $\frac{x + 11}{3}$

$\therefore f^{-1}(x) = \frac{x +11}{3}$

And,
$f^{-1}$(4) = $\frac{4 +11}{3}$

$= \frac{15}{3}$
$= 5$
So, $f^{-1}$(-4) = 5.
13a) Given that f(x)=4x +7 and g(x) = 3x -5. If $fg^{-1}$ = 15, find the value of x.

Solution:
Given,
f(x) = 4x +7
g(x) = 3x -4

Let g(x) = y = 3x - 4
or, y = 3x - 4

Interchanging the roles of x and y, we get,

or, x = 3y - 4
or, x - 4 = 3y
or, y = $\frac{x - 4}{3}$
$\therefore g^{-1}(x) = \frac{x - 4}{3}$


Now,
$fg^{-1}$(x) = f[$g^{-1}$(x)]

or, 15 = f[$\frac{x -4}{3}$]

or, 15 = 4 × $\frac{x - 4}{3}$ + 7

or, 15 - 7 = $\frac{4x - 16}{3}$

or, 8×3 = 4x - 16
or, 24 + 16 = 4x
or, 40 = 4x
$\therefore$ x = 10
14a) If f(x) = 5 - $\frac{6}{x}$ then what is the value of x at $f(x) = - f^{-1}(-1)$?

Solution:
Here,
f(x) = 5 - $\frac{6}{x}$
Let y = f(x) = 5 - $\frac{6}{x}$

$or, y = 5 - $\frac{6}{x}$

Interchanging the roles of x and y, we get,

or, x = 5 - $\frac{6}{y}$
or, $\frac{6}{y}$ = (5 - x)

or, y = $\frac{6}{5-x}$
So, $f^{-1}$(x) = $\frac{6}{5-x}$


Also, $f^{-1}$(-1) = $\frac{6}{5-(-1)}$
= $\frac{6}{6}$
= 1
Given,
$f(x) = - f^{-1}(-1)$
$or, 5 - \frac{6}{x} = - (1)$

$or, \frac{5x -6}{x} = -1$

$or, 5x - 6 = -x$
$or, 5x + x = 6$
$or, 6x = 6$
$\therefore x = 1$
15a) If f(x) = 4x + 5 and g(x) = 8x + 9, prove that $f^{-1}$og(x) is a linear function.

Solution:
A function in the form of f(x) = mx + c, is said to be a linear function.

Given,
f(x) = 4x + 5
g(x) = 8x + 9
Let y = f(x) = 4x + 5
or, y = 4x + 5
Interchanging the roles of x and y, we get,
or, x = 4y + 5

or, x - 5 = 4y
or, y = $\frac{x - 5}{4}$
$\therefore f^{-1}(x) = \frac{x - 5}{4}$

Now,
$f^{-1}$og(x) = $f^{-1}$[g(x)]

$= f^{-1}[ 8x + 9]$
$= \frac{8x + 9 - 5}{4}$
$= \frac{8x + 4}{4}$
$= \frac{4(2x + 1)}{4}$
$= 2x +1$
Hence, $f^{-1}og(x)$ is a linear function. #proved.


16a) Let f(x) = $\frac{3x +5}{x -2}$ and g(x) = 7 be two functions. Prove that $f^{-1}$og(x) is a constant function.

Solution:

A function in the form of f(x) = c where c is a constant number is said to be a constant function.

Here,
f(x) = $\frac{3x +5}{x -2}$
g(x) = 7
Let y = f(x) = $\frac{3x + 5}{x -2}$

or, y = $\frac{3x +5}{x -2}$

Interchanging the roles of x and y, we get,

$or, x = \frac{3y + 5}{y -2}$

$or, x(y - 2) = 3y  + 5$

$or, xy - 2x = 3y +5$
$or, xy - 3y - 2x = 5$
$or, y(x - 3) = 2x +5$
$or, y = \frac{2x +5}{x -3}$

$\therefore f^{-1}(x) = \frac{2x +5}{x -3}$


Now,
$f^{-1}og(x) = f^{-1}[g(x)]$

$= f^{-1}[9]$
$= \frac{2×9+5}{9-3}$
$= \frac{18+5}{6}$
$= \frac{23}{6}$
Hence, $f^{-1}og(x)$ is a constant function. #proved.


17a) Let, f(x) = 2x +3 and g(x) =x²+2x +1 be two functions. Prove that $f^{-1}$og(x) is a quadratic function.

Solution:
Given,
f(x) = 2x + 3
g(x) = x² +2x +1 = (x +1)²

Let y = f(x) = 2x + 3
or, y = 2x + 3
Interchanging the roles of x and y, we get,
or, x = 2y + 3

or, x - 3 = 2y
or, y = $\frac{x -3}{2}$
$\therefore f^{-1}(x) = \frac{x -3}{2}$


Now,
$f^{-1}og(x) = f^{-1}[g(x)]$

$= f^{-1}[(x +1)²]$
$= 2(x +1)² + 3$
It is the equation of a parabola which is definitely a quadratic function.

$= 2(x²+2x+1)+3$
$= 2x² + 4x + 2 + 3$
$= 2x² + 4x + 5$ #proved


18 a) Given that f(x) = x³ + x² + 1 and g(x) = x -1. Prove that $fog^{-1}$ is a cubic function.

Solution:

Let y = g(x) = x - 1
or, y = x -1
Interchanging the roles of x and y, we get,
or, x = y - 1
or, y = x + 1
$\therefore g^{-1}(x) = x + 1$

Now,
$fog^{-1}(x)$= $f[g^{-1}(x)]$

$= f[x + 1]$
$= (x +1)^3 + (x +1)^2 + 1$

$= x^3 + 3x^2 + 3x + 1 + x^2 + 2x + 1 + 1$

$= x^3 + 4x^2 + 5x + 3$
Since, the above function has its highest degree equal to 3, it is proved that $fog^{-1}(x)$ is a cubic function.


19 a) Given that f(x) = 3x + 5 and g(x) = $\dfrac{3x +2}{4}. What value of x makes $fog^{-1}(x)$ an identity function?

Solution:

Let y = g(x) = $\dfrac{3x +2}{4}$
or, y = $\dfrac{3x +2}{4}$
Interchanging the roles of x and y, we get,

$or, x = \dfrac{3y +2}{4}$
$or, 4x = 3y +2$
$or, 4x - 2 = 3y$
$\therefore y = \dfrac{4x -2}{3}$


Given,
$fog^{-1}(x) = x$
$or, f[g^{-1}(x)] = x$
$or, f[ \frac{4x -2}{3} ] = x$

$or, 3 × \dfrac{4x -2}{3} + 5 = x$

$or, 4x - 2 + 5 = x$
$or, 4x - x + 3 = 0$
$or, 3x = - 3$
$\therefore x = -1$

About Readmore Optional Mathematics Book 10

Author: D. R. Simkhada
Editor: I. R. Simkhada

Readmore Publishers and Distributors
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Phone: 4672071, 5187211, 5187226
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About this page:


Inverse Functions - Solved Exercises | Class 10 Readmore Optional Mathematics is a collection of the solutions related to exercises of Inverse Functions from the Algebra chapter for Nepal's Secondary Education Examination (SEE) appearing students. 

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