Fsc Part 2 Mathematics (Complete Solutions)
Q2
For the real valued functions f defined below. Find
(a) ƒ^{1} (x)  (b) ƒ^{1} (1) 
and verify ƒ (ƒ^{1} (x)) = ƒ^{1} (ƒ (x)) = x 
(i) ƒ (x) = 2x + 8

(ii) ƒ (x) = 3x^{3} + 7


(iii) ƒ (x) = (x + 9)^{3}


Solution
(i) ƒ (x) = 2x + 8
(a) ƒ^{1} (x)
Let ƒ (x) = y
So that y is the image of x under ƒ
Now solve this equation for x as follow:
y = 2x + 8
By adding "2x" on both sides
y + 2x = 2x + 2x + 8
y + 2x = 8
By subtracting "y" on both sides
y  y + 2x = 8  y
2x = 8  y
By dividing "2" on both sides

= 

x  = 

As x = ƒ^{1} (y)
ƒ^{1} (y)  = 

To find ƒ^{1} (x), replace y by x, we get
ƒ^{1} (x)  = 

 (1) 
(b) ƒ^{1} (1)
Put x = 1 in equation (1), we get
ƒ^{1} (1)  = 

ƒ^{1} (1)  = 

ƒ^{1} (1)  = 

Verification:
ƒ (ƒ^{1} (x))  =  ƒ  ( 

) 
ƒ (ƒ^{1} (x))  =  2  ( 

) + 8 
ƒ (ƒ^{1} (x)) = (8  x) + 8
ƒ (ƒ^{1} (x)) = 8 + x + 8
ƒ (ƒ^{1} (x)) = x
Now ƒ^{1} (ƒ (x)) = ƒ^{1} (2x + 8)
ƒ^{1} (ƒ (x))  = 

ƒ^{1} (ƒ (x))  = 

ƒ^{1} (ƒ (x))  = 

ƒ^{1} (ƒ (x)) = x
Hence vrified that:
ƒ (ƒ^{1} (x)) = ƒ^{1} (ƒ (x)) = x
(ii) ƒ (x) = 3x^{3} + 7
(a) ƒ^{1} (x)
Let ƒ (x) = y
So that y is the image of x under ƒ
Now solve this equation for x as follow:
y = 3x^{3} + 7
By subtracting "7" on both sides
y  7 = 3x^{3} + 7  7
3x^{3} = y  7
Taking cubic root on both sides
^{3}√3x^{3} = ^{3}√y  7
x ^{3}√3 = ^{3}√y  7
By dividing "^{3}√3" on both sides

= 

x = 

As x = ƒ^{1} (y)
ƒ^{1} (y) = 

To find ƒ^{1} (x), replace y by x, we get
ƒ^{1} (x) = 

 (1) 
(b) ƒ^{1} (1)
Put x = 1 in equation (1), we get
ƒ^{1} (1) = 

ƒ^{1} (1) = 

Verification:
ƒ (ƒ^{1} (x)) = ƒ ( 

) 
ƒ (ƒ^{1} (x)) = 3( 

)^{3} + 7 
ƒ (ƒ^{1} (x)) = 3( 

) + 7 
ƒ (ƒ^{1} (x)) = x  7 + 7
ƒ (ƒ^{1} (x)) = x
Now ƒ^{1} (ƒ (x)) = ƒ^{1} (3x^{3} + 7)
ƒ^{1} (ƒ (x)) = 

ƒ^{1} (ƒ (x)) = 

ƒ^{1} (ƒ (x)) = ^{3}√x^{3}
ƒ^{1} (ƒ (x)) = x
Hence vrified that:
ƒ (ƒ^{1} (x)) = ƒ^{1} (ƒ (x)) = x
(iii) ƒ (x) = (x + 9)^{3}
(a) ƒ^{1} (x)
Let ƒ (x) = y
So that y is the image of x under ƒ
Now solve this equation for x as follow:
y = (x + 9)^{3}
Taking cubic root on both sides
^{3}√y = ^{3}√(x + 9)^{3}
y^{1/3} = x + 9
By adding "x" on both sides
y^{1/3} + x = x + x + 9
y^{1/3} + x = 9
By subtracting "y^{1/3}" on both sides
y^{1/3}  y^{1/3} + x = 9  y^{1/3}
x = 9  y^{1/3}
As x = ƒ^{1} (y)
ƒ^{1} (y) = 9  y^{1/3}
To find ƒ^{1} (x), replace y by x, we get
ƒ^{1} (x) = 9  x^{1/3}  (1)
(b) ƒ^{1} (1)
Put x = 1 in equation (1), we get
ƒ^{1} (1) = 9  (1)^{1/3}
Verification:
ƒ (ƒ^{1} (x)) = ƒ (9  x^{1/3})
ƒ (ƒ^{1} (x)) = ((9  x^{1/3}) + 9)^{3}
ƒ (ƒ^{1} (x)) = (9 + x^{1/3} + 9)^{3}
ƒ (ƒ^{1} (x)) = x^{(1/3)}^{(3)}
ƒ (ƒ^{1} (x)) = x
Now ƒ^{1} (ƒ (x)) = ƒ^{1} (x + 9)^{3}
ƒ^{1} (ƒ (x)) = 9  ((x + 9)^{3})^{1/3}
ƒ^{1} (ƒ (x)) = 9  (x + 9)^{(3)}^{(1/3)}
ƒ^{1} (ƒ (x)) = 9 + x  9
ƒ^{1} (ƒ (x)) = x
Hence vrified that:
ƒ (ƒ^{1} (x)) = ƒ^{1} (ƒ (x)) = x
(iv) ƒ (x) = 

, x > 1 
(a) ƒ^{1} (x)
Let ƒ (x) = y
So that y is the image of x under ƒ
Now solve this equation for x as follow:
y = 

By multiplaying "x  1" on both sides
y(x  1) = 

(x  1) 
yx  y = 2x + 1
By subtracting "1" on both sides
yx  y  1 = 2x + 1  1
yx  y  1 = 2x
By subtracting "yx" on both sides
1  y = 2x  yx
(1 + y) = 2x  yx
By taking "" and "x" common
(1 + y) = (2  y)(x)
By dividing "2  y" on both sides

= 

x = 

As x = ƒ^{1} (y)
ƒ^{1} (y)  = 

To find ƒ^{1} (x), replace y by x, we get
ƒ^{1} (x)  = 

 (1) 
(b) ƒ^{1} (1)
Put x = 1 in equation (1), we get
ƒ^{1} (1)  = 

ƒ^{1} (1)  = 

ƒ^{1} (1)  = 

ƒ^{1} (1)  = 

ƒ^{1} (1) = 0
Verification:
ƒ (ƒ^{1} (x))  =  ƒ  ( 

) 
ƒ (ƒ^{1} (x))  = 

ƒ (ƒ^{1} (x))  = 

ƒ (ƒ^{1} (x))  = 

ƒ (ƒ^{1} (x))  = 

ƒ (ƒ^{1} (x))  = 

ƒ (ƒ^{1} (x))  = 

ƒ (ƒ^{1} (x))  = 

ƒ (ƒ^{1} (x))  = 

ƒ (ƒ^{1} (x))  = 

Now ƒ^{1} (ƒ (x)) = ƒ^{1} (2x + 8)
ƒ^{1} (ƒ (x))  = 

ƒ^{1} (ƒ (x))  = 

ƒ^{1} (ƒ (x))  = 

ƒ^{1} (ƒ (x)) = x
Hence vrified that:
ƒ (ƒ^{1} (x)) = ƒ^{1} (ƒ (x)) = x