Exercise 2D

1 a i y ℝ c i y ℝ

ii Let y = f(x) ii Let y = f(x)

y = 2x + 3 y = 4 − 3x
y 3 4 y
x x
2 3
x3 4 x
f−1(x) = f−1(x) =
2 3

iii The domain of f−1(x) is x  ℝ iii The domain of f−1(x) is x  ℝ

The range of f−1(x) is y  ℝ The range of f−1(x) is y  ℝ

iv iv

d i y ℝ
b i y ℝ
ii Let y = f(x)
ii Let y = f(x) y = x3 − 7
x5 x  3 y 7
y
2 f−1(x) = 3
x7
x  2y  5
f−1(x) = 2x − 5 iii The domain of f−1(x) is x  ℝ
The range of f−1(x) is y  ℝ
iii The domain of f−1(x) is x  ℝ
The range of f−1(x) is y  ℝ iv
iv

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2 a Range of f is f ( x )  ℝ 3
Let y = f(x)
y = 10 − x
x = 10 − y
−1
f (x) = 10 − x, {x ∈ ℝ}

b Range of f is f ( x )  ℝ
Let y = g(x)
x
y
5
x  5y
−1
g (x) = 5x, {x ∈ ℝ}
g : x| → 4 − x, {x ∈ ℝ, x > 0}
c Range of f is f ( x )  0 g has range {g(x) ∈ ℝ, g(x) < 4}
Let y = h(x)
3 The inverse function is g 1 ( x )  4  x
y
x Now {Range g}  {Domain g 1}
3
x and {Domain g}  {Range g}
y
Hence, g−1(x) = 4 − x, {x ∈ ℝ, x < 4}
3
h−1(x) = , {x  0}
x Although g( x) and g 1 ( x) have identical
d Range of f is f ( x )  ℝ equations, their domains and hence ranges
Let y = k(x) are different, and so are not identical.
y=x−8
x=y+8 4 a i Maximum value of g when x  3
k−1(x) = y + 8, {x ∈ ℝ} 1
Hence {g(x) ∈ ℝ, 0 < g(x) ≤ }
3

1
ii g 1 ( x ) 
x

iii Domain g–1 = Range g

1
 Domain g−1 :{x ∈ ℝ, 0 < x ≤ }
3
–1
Range g = Domain g
 Range g−1(x):{g−1(x)∈ℝ, g−1(x) ≥ 3}

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4 a iv 4 c i g  x    as x  2
Hence {g(x) ∈ ℝ, g(x) > 0}

3 2y 3
ii Letting y  x
x2 y
2x  3
Hence g 1 ( x ) 
x

iii Domain g–1 = Range g

 Domain g−1 :{x ∈ ℝ, x > 0}
Range g–1 = Domain g
 Range g−1(x):{g−1(x)∈ℝ, g−1(x) > 2}

b i Minimum value of g  x   1 iv
when x  0
Hence {g(x) ∈ ℝ, g(x) ≥ −1}

y 1
ii Letting y  2 x  1  x 
2
x 1
Hence g 1 ( x) 
2

iii Domain g–1 = Range g

 Domain g−1 :{x ∈ ℝ, x ≥ −1}
Range g–1 = Domain g
g ( x)  ,
1

 Range g 1 ( x) :  
 g 1 ( x)  0 d i Minimum value of g  x   2
when x  7
iv Hence {g(x) ∈ ℝ, g(x) ≥ 2}

ii Letting y  x  3  x  y 2  3
Hence g 1 ( x )  x 2  3

iii Domain g–1 = Range g

 Domain g−1 :{x ∈ ℝ, x ≥ 2}
Range g–1 = Domain g
 Range g−1(x):{g−1(x)∈ℝ, g−1(x) ≥ 7}

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4 d iv 4 f iii Domain g–1 = Range g
 Domain g−1 :{x ∈ ℝ, x ≥ 0}

Range g–1 = Domain g

g 1 ( x)  , 
 Range g 1 ( x) :  
 g 1 ( x)  2 
iv

e i 22  2  6
Hence {g(x) ∈ ℝ, g(x) > 6}

ii Letting y = x2 + 2
y − 2 = x2
x = y2
Hence g−1(x) = x2

5 t(x) = x2 − 6x + 5, {x ∈ ℝ, x ≥ 5}
iii Domain g–1 = Range g
 Domain g−1 :{x ∈ ℝ, x > 6}
Range g–1 = Domain g Let y  x 2  6 x  5
g 1 ( x)  , y  ( x  3)2  9  5 (completing the square)
1 
 Range g ( x) :  
g 1 ( x)  2 y  ( x  3)2  4

iv This has a minimum point at (3,  4)

For the domain x  5, t( x) is a

one-to-one function so we can find
an inverse function.

Make y the subject:

y  ( x  3)2  4
y  4  ( x  3)2
f i Minimum value of g  x   0 y  4  x 3
when x  2
y 4 3  x
Hence {g(x) ∈ ℝ, g(x) ≥ 0}

ii Letting y  x3  8  x  3 y  8
Hence g−1(x) = 3
x 8

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5 (continued) 6 c Domain of m−1(x): {x ∈ ℝ, x > 5}

Domain t–1 = Range t 5

7 a As x  2,
 Domain g−1 :{x ∈ ℝ, x ≥ 0} x20
Hence, t−1(x) = x  4  3, {x ∈ ℝ, x ≥ 0} and hence h( x )  

b To find h1 (3) we can find what element

of the domain gets mapped to 3

6 a m(x) = x2 + 4x + 9, {x ∈ ℝ, x > a} Suppose h (a )  3 for some a such that

Let y  x 2  4 x  9 a2
2a  1
y  ( x  2) 2  4  9 Then 3
a2
y  ( x  2) 2  5 2a  1  3a  6
This has a minimum value of (2, 5) 7a
1
So h (3)  7

2x 1
c Let y  and find x as a
x2
function of y
y ( x  2)  2 x  1
yx  2 y  2 x  1
yx  2 x  2 y  1
x( y  2)  2 y  1
For m( x) to have an inverse it must 2 y 1
x
be one-to-one. Hence the least value y2
of a is  2 2x  1
So h 1 ( x)  , x  , x  2
x2
b Changing the subject of the formula:
y  ( x  2) 2  5
y  5  ( x  2) 2
y 5  x  2
y 5  2  x
Hence m 1 ( x )  x  5  2

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7 d If an element b is mapped to itself,  3 x 
then h(b )  b 9 st(x) = s  
 x 
2b  1 3
b  3 x
b2  x  1
2b  1  b (b  2) 
3
2b  1  b 2  2b  3 x  x
x 
x
0  b 2  4b  1
 3 
4  16  4 4  20 st(x) = t  
b   x 1 
2 2

 3  x31 
42 5
  2 5 3
 x 1 
2

 3 x  3 3
x 1 
The elements 2  5 and 2  5 get
 3
x 1
mapped to themselves by the function. x

8 a nm(x) = n  2 x  3 The functions s(x) and t(x) are the

2x  3  3 inverse of each other as st(x) = ts(x) = x

2
x 10 a Let y  2x 2  3

 x 3 The domain of f 1 ( x ) is the range of f ( x).

b mn(x) = m  
 2  f(x) = 2x2 −3, {x ∈ ℝ, x < 0}
 x 3 has range f(x) > −3
 2 3
 2  x 3
Letting y  2 x 2  3  x  
x 2
The functions m(x) and n(x) are the We need to consider the domain of
inverse of each other as f ( x) to determine if either
mn(x) = nm(x) = x.
x3 x3
f 1  x    or f 1  x   
2 2
2
f(x) = 2x − 3 has domain {x ∈ ℝ, x < 0}
Hence f 1 ( x ) must be the negative square root
x3
f 1 ( x )   , {x ∈ ℝ, x > −3}
2

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10 b If f( a )  f 1 ( a) then a is negative 11 d Let y = g(x)
(see graph). y = ln(x − 4)
ey = x − 4
Solve f(a)  a x = ey + 4
2a 2  3  a g−1(x) = ex + 4
2a 2  a  3  0 Range of g(x) is g(x) ∈ℝ,
so domain of g−1(x) is { x  ℝ}
(2a  3)(a  1)  0
3 e g−1(x) = 11
a  , 1 ex + 4 = 11
2
Therefore a  1 ex = 7
x = ln7
x = 1.95

3( x  2) 2
12 a f(x) = 
x  x  20 x  4
2

3( x  2) 2
= 
( x  5)( x  4) x  4
3( x  2) 2( x  5)
= 
( x  5)( x  4) ( x  5)( x  4)
3 x  6  2 x  10
=
( x  5)( x  4)
11 a Range of f(x) is f(x) > −5 x4
=
( x  5)( x  4)
b Let y = f(x) 1
y = ex − 5 = , x4
x5
ex = y + 5
x = ln(y + 5) b The range of f is
−1
f (x) = ln(x + 5) 1
Range of f(x) is f(x) > −5, {f(x) ∈ ℝ, f(x) < }
9
so domain of f−1(x) is {x ∈ ℝ, x > −5}
c Let y = f(x)
c
1
y=
x5
yx + 5y = 1
yx = 1 − 5y
1 5y
x=
y
1
x = 5
y
1
f−1(x) =  5
x
The domain of f−1(x) is
1
{x ∈ ℝ, x > and x ≠ 0}
9

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