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  33
33
=
3y - 7
33
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

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