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) = 3x3 + 7
(iii)             ƒ (x) = (-x + 9)3
(iv)             ƒ (x) =
 2x + 1 x - 1
,   x > 1

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
 2x 2
=
 8 - y 2
x =
 8 - y 2
As x = ƒ-1 (y)
ƒ-1 (y) =
 8 - y 2
To find ƒ-1 (x), replace y by x, we get
ƒ-1 (x) =
 8 - x 2
---------- (1)
(b)   ƒ-1 (-1)
Put x = -1 in equation (1), we get
ƒ-1 (-1) =
 8 - (-1) 2
ƒ-1 (-1) =
 8 + 1 2
ƒ-1 (-1) =
 9 2
Verification:
ƒ (ƒ-1 (x)) = ƒ (
 8 - x 2
)
ƒ (ƒ-1 (x)) = -2 (
 8 - 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)) =
 8 - (-2x + 8) 2
ƒ-1 (ƒ (x)) =
 8 + 2x - 8 2
ƒ-1 (ƒ (x)) =
 2x 2
ƒ-1 (ƒ (x)) = x
Hence vrified that:
ƒ (ƒ-1 (x)) = ƒ-1 (ƒ (x)) = x

(ii)             ƒ (x) = 3x3 + 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 = 3x3 + 7
By subtracting "7" on both sides
y - 7 = 3x3 + 7 - 7
3x3 = y - 7
Taking cubic root on both sides
33x3 = 3y - 7
x  33 = 3y - 7
By dividing "33" on both sides
 x  3√3 3√3
=
 3√y - 7 3√3
x =
3
 y - 7 3
As x = ƒ-1 (y)
ƒ-1 (y) =
3
 y - 7 3
To find ƒ-1 (x), replace y by x, we get
ƒ-1 (x) =
3
 x - 7 3
---------- (1)
(b)   ƒ-1 (-1)
Put x = -1 in equation (1), we get
ƒ-1 (-1) =
3
 -1 - 7 3
ƒ-1 (-1) =
3
 -8 3
Verification:
ƒ (ƒ-1 (x)) = ƒ (
3
 x - 7 3
)
ƒ (ƒ-1 (x)) = 3(
3
 x - 7 3
)3 + 7
ƒ (ƒ-1 (x)) = 3(
 x - 7 3
) + 7
ƒ (ƒ-1 (x)) = x - 7 + 7
ƒ (ƒ-1 (x)) = x
Now   ƒ-1 (ƒ (x)) = ƒ-1 (3x3 + 7)
ƒ-1 (ƒ (x)) =
3
 3x3 + 7 - 7 3
ƒ-1 (ƒ (x)) =
3
 3x3 3
ƒ-1 (ƒ (x)) = 3x3
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
3y = 3(-x + 9)3
y1/3 = -x + 9
By adding "x" on both sides
y1/3 + x = -x + x + 9
y1/3 + x = 9
By subtracting "y1/3" on both sides
y1/3 - y1/3 + x = 9 - y1/3
x = 9 - y1/3
As x = ƒ-1 (y)
ƒ-1 (y) = 9 - y1/3
To find ƒ-1 (x), replace y by x, we get
ƒ-1 (x) = 9 - x1/3       ---------- (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 - x1/3)
ƒ (ƒ-1 (x)) = (-(9 - x1/3) + 9)3
ƒ (ƒ-1 (x)) = (-9 + x1/3 + 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) =
 2x + 1 x - 1
,   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 =
 2x + 1 x - 1
By multiplaying "x - 1" on both sides
y(x - 1) =
 2x + 1 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
 -(1 + y) 2 - y
=
 (2 - y)(x) 2 - y
x =
 -(1 + y) 2 - y
As x = ƒ-1 (y)
ƒ-1 (y) =
 -(1 + y) 2 - y
To find ƒ-1 (x), replace y by x, we get
ƒ-1 (x) =
 -(1 + x) 2 - x
---------- (1)
(b)   ƒ-1 (-1)
Put x = -1 in equation (1), we get
ƒ-1 (-1) =
 -[1 + (-1)] 2 - (-1)
ƒ-1 (-1) =
 -(1 - 1) 2 + 1
ƒ-1 (-1) =
 -1 + 1 3
ƒ-1 (-1) =
 0 3
ƒ-1 (-1) = 0
Verification:
ƒ (ƒ-1 (x)) = ƒ (
 -(1 + x) 2 - x
)
ƒ (ƒ-1 (x)) =
2(
 -(1 + x) 2 - x
) + 1
-
 1 + x 2 - x
- 1
ƒ (ƒ-1 (x)) =
 -2 - 2x 2 - x
+ 1
-
 1 + x 2 - x
- 1
ƒ (ƒ-1 (x)) =
 -2 - 2x + 2 - x 2 - x
-
 1 + x 2 - x
- 1
ƒ (ƒ-1 (x)) =
 -2x - x 2 - x
-
 1 + x 2 - x
- 1
ƒ (ƒ-1 (x)) =
 -3x 2 - x
-
 1 + x 2 - x
- 1
ƒ (ƒ-1 (x)) =
 -3x 2 - x
 -1 - x - 2 + x 2 - x
ƒ (ƒ-1 (x)) =
 -3x 2 - x
 -1 - 2 2 - x
ƒ (ƒ-1 (x)) =
 -3x 2 - x
 -3 2 - x
ƒ (ƒ-1 (x)) =
 -3x -3
Now   ƒ-1 (ƒ (x)) = ƒ-1 (-2x + 8)
ƒ-1 (ƒ (x)) =
 8 - (-2x + 8) 2
ƒ-1 (ƒ (x)) =
 8 + 2x - 8 2
ƒ-1 (ƒ (x)) =
 2x 2
ƒ-1 (ƒ (x)) = x
Hence vrified that:
ƒ (ƒ-1 (x)) = ƒ-1 (ƒ (x)) = x