To find the inverse, swap x and y

∴ x-intercepts of relation = y-intercepts of inverse

The answer is C.

5 a

b

c

d

e

f

g

h

i

j

k

l

m

n

o

p

Exercise 5B — Functions and their inverses

1 a

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106

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b

c

d

e

f

g

h

2 a

Domain = {−4, −2, 0, 2, 3}

Range = {−2, 0, 1, 4, 6}

b

Domain = R Range = R c

Domain = R Range = R

d

Domain = R Range = [−9, ∞) e

Domain = R Range = [0, ∞) f

Domain = R\{0}

Range = R\{0}

g

Domain = R Range = [−16, ∞) h

Domain = R Range = R

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i

Domain = R Range = R⁺ j

Domain = R⁺ Range = R k

Domain = [−2, 2]

Range = [0, 2]

3 a

{(−2, −4), (0, −2)(1, 0)(4, 2) (6, 3)}

Domain = {−2, 0, 1, 4, 6}

Range = {−4, −2, 0, 2, 3}

b 3x + 4y = 12 The inverse is

3y + 4x = 12 or 4x + 3y = 12

Domain = R Range = R c f(x) = 5 − 2x The inverse is x = 5 − 2y −2y = x − 5 2y = 5 − x y = 5

2

−x

y = 1 1 2x 22

− +

Domain = R Range = R d f(x) = x²− 9 The inverse is x = y²− 9 y²= x + 9 y = ± x+ 9

The inverse is y = ± x+ 9

Domain = [−9, ∞) Range = R e f(x) = (x + 2)² The inverse is x = (y + 2)² y + 2 = ± x y = 2− ± x

Domain [0, ∞) Range R f f(x) = 4

x The inverse is x = 4

y y = 4 x

Domain = R\{0}

Range = R\{0}

g f(x) = x²+ 8x The inverse is x = y²+ 8y

x = y²+ 8y + 16 − 16 x = (y + 4)²− 16 x + 16 = (y + 4)² y + 4 = ± x+16 y = 4− ± x+16

Domain = [−16, ∞) Range = R h f(x) = ³

2 x The inverse is x = ³

2 y 2x = y³ y = ³2x f⁻¹ (x) = ³2x

Domain = R Range = R i f(x) = 2e^x The inverse is x = 2e^y 2

x = e^y

y = log

e 2

 x

  

f ⁻¹(x) = log

e 2

 x

  

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Domain = R⁺ Range = R j f(x) = loge (2x) The inverse is x = loge(2y) e^x= 2y y = 1

2 e x

f⁻¹(x) = 1 2

e x

Domain = R Range = R⁺ k f(x) = 4 x− ² x = 4 y− ² x² = 4 − y y² = 4 − x² y = ± 4 x− ²

Domain [0, 2]

Range [−2, 2]

4 a i

ii Domain = R Range = R iii Domain = R Range = R

b i

ii Domain = R⁻ Range = R⁺ iii Domain = R⁺ Range = R⁻ c i

ii Domain = R Range = [4, ∞]

iii Domain = [4, ∞) Range = R d i

ii Domain = R Range = R⁺ iii Domain = R⁺ Range = R e i

ii Domain = [−3, 3]

Range = [0, 3]

iii Domain = [0, 3]

Range = [−3, 3]

f i

ii Domain = (1, ∞) Range = R iii Domain = R Range = (1, ∞) g i

ii Domain = R Range = R iii Domain = R Range = R h i

ii Domain = R Range = [1, ∞) iii Domain = [1, ∞) Range = R i i

ii Domain = [−5, 5]

Range = [−5, 0]

iii Domain = [−5, 0]

Range = [−5, 5]

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109

j i

ii Domain = [2, 6]

Range = [−2, 0]

iii Domain = [−2, 0]

Range = [2, 6]

k i

ii Domain = (−3, ∞) Range = R iii Domain = R Range = (−3, ∞) l i

ii Domain = [−2, 3) Range = [0, 4) iii Domain = [0, 4) Range = [−2, 3) 5 Range of f(x) = (−∞, 0) The answer is D.

6 Domain of the inverse = (−∞, 0) The answer is A.

7 Range of the inverse = (0, ∞) The answer is E.

8 f (x) = − x

The inverse has the rule x = − y

−x = y y = (−x)² y = x² Function f(x) has Domain = R⁺ Range = R⁻

Hence the inverse function has Domain = R⁻

Range = R⁺ The answer is B.

9 a f(x) = (x − 2)²− 3

domain [−2, ∞), range [−3, ∞) x = (y − 2)²− 3

x + 3 = (y − 2)² y − 2 = ± x+ 3 y = 2± x+ 3 f⁻¹: [−3, ∞) → R,

where f ⁻¹(x) = 2± x+ 3 b y = 3e^{x − 1}+ 2 domain R, range (2, ∞) x = 3e^{y − 1}+ 2

2

3

x− = e^{y − 1}

log 2

e 3 x−

 

 

  = y − 1

y = 2

1 log

e 3 x−

 

+   f⁻¹: (2, ∞) → R,

f⁻¹= 2

1 log

e 3 x−

 

+  

Exercise 5C — Inverse functions

1 a f(x) = 4x + 1

i f(x) is a straight line, so it is a one-to-one function. Its inverse is also a function.

ii

iii 4x + 1 = 1 1 4x− 4

From the CAS calculator, the point of intersection is

1 1

3 3

− − 

 

 

iv Domain = R Range = R v Domain = R Range = R b f(x) = 6x

i f(x) is a straight line, so it is a one-to-one function. Its inverse is also a function.

ii

iii 6x = 1 6x

From the CAS calculator, the point of intersection is (0, 0) iv Domain = R

Range = R v Domain = R Range = R c f(x) = 5

i f(x) is a horizontal line, so it is not a one-to-one function.

f⁻¹(x) does not exist.

d f(x) = x²+ 2

i f(x) is a parabola. It is not a one-to-one function as a horizontal line cuts the parabola twice.

f⁻¹(x) does not exist.

e f(x) = (x − 3)²

i f(x) is a parabola. It is not a one-to-one function as a horizontal line cuts the parabola twice.

f⁻¹(x) does not exist.

f f(x) = (x + 1)³

i f(x) is a cubic graph with a point of inflection at (−1, 0). It is a one-to-one function so its inverse is also a function.

ii

iii (x + 1)³= ³x− 1

From the CAS calculator, the point of intersection is (−2.3, −2.3).

iv Domain = R Range = R v Domain = R Range = R g f(x) = 2

x

i A hyperbola is a one-to-one function, so its inverse is also a function.

ii

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iii There is no point of intersection.

iv Domain = R\{0}

Range = R\{0}

v Domain = R\{0}

Range = R\{0}

h f(x) = − 16 x− ²

i f(x) is a semicircle [−4, 4] and is not a one-to-one function, so its inverse is not a function.

i f(x) = x²− 6x + 3

i f(x) is a parabola and is not a one-to-one function, so its inverse is not a function.

j f(x) = e^4x− 2

i f(x) is an exponential graph that is a one-to-one function. Its inverse exists as a function.

ii

iii e^4x− 2 = 1

log ( 2) 4 ^e x+

From the CAS calculator, the point of intersections are (0.2, 0.2) and (−1.99, −1.99)

iv Domain = R Range = (−2, ∞) v Domain = (−2, ∞) Range = R k f(x) = 2loge(x − 1)

i f(x) is a logarithmic graph that is a one-to-one function.

Its inverse exists as a function.

ii

iii 2log (_e x− = 1) ² 1

x

e +

There are no points of intersection.

iv Domain = (1, ∞) Range = R v Domain = R Range = (1, ∞)

2 a f(x) = 4x is a straight line. It is one-to-one, so f⁻¹(x) exists.

f(x) = 4x Its inverse is x = 4y y =

4 x

∴ f ⁻¹(x) = 4 x

b f(x) = 1₂ x − 2

The graph shows that f(x) is not one-to-one, so f⁻¹(x) does not exist.

c f(x) is a parabola reflected in the x-axis and translated 5 units up. It is not a one-to one function and so f⁻¹(x) does not exist.

d f(x) = (x − 1)² is a parabola. Not one-to-one, so f⁻¹(x) does not exist.

e f(x) = 2³

x is a cubic. One-to-one, so f⁻¹(x) exists.

The inverse is x = ³ y 2

2x = y³

∴ y = ³2x f⁻¹(x) = ³2x

f f(x) = x²+ 10x − 3 is a parabola. Not one-to-one, so f ⁻¹(x) does not exist.

g f(x) = x− is one-to-one, so f 2 ⁻¹(x) exists.

The inverse is x = y− 2

x² = y − 2

y = x²+ 2

∴ f ⁻¹(x) = x²+ 2, x ∈ [0, ∞) h f(x) = 16 x− ²

Semicircle. Not one-to-one, so f⁻¹(x) does not exist.

i f(x) = 2e^x+ 1

Exponential function. One-to-one, so f⁻¹(x) exists.

The inverse is x = 2e^y+ 1

1

2 x− = e^y

y = 1

log^e 2 x−

 

 

 

∴ f ⁻¹(x) = log 1

e 2 x−

 

 

 , x > 1 j f(x) = 5 − e^{x − 2}

Exponential function, reflected in the x-axis and translated 2 units to the right and 5 units up.

Hence one-to-one, so f⁻¹(x) exists The inverse is x = 5 − e^{y − 2} e^{y − 2} = 5 − x y − 2 = loge(5 − x) y = loge(5 − x) + 2 ∴ f ⁻¹(x) = loge(5 − x) + 2, x < 5 k f(x) = 3e^−x− 2

Exponential function, so one-to-one, so f⁻¹(x) exists.

The inverse is x = 3e^−y− 2

2

3 x+ = e^−y

−y = 2

log^e 3 x+

 

 

 

∴ f ⁻¹(x) = log 2

e 3 x+

 

−  , x > −2

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l f(x) = log e(3x)

Logarithmic function. So one-to-one, so f⁻¹(x) exists.

The inverse is x = loge(3y)

e^x = 3y

∴ f ⁻¹(x) = 1 3

e , x x ∈ R m f(x) = 2loge(x − 4)

Logarithmic function. So one-to-one, so f⁻¹(x) exists.

The inverse is x = 2 loge(y − 4)

2

x = loge(y − 4)

y − 4 = ²

x

e ∴ f ⁻¹(x) = ²

x

e + 4, x ∈ R n f(x) = 1 + 2loge(x)

Logarithmic function. So one-to-one, so f⁻¹(x) exists.

The inverse is:

x = 1 + 2 loge(y)

1

2 x−

= loge(y)

∴ f ⁻¹(x) =

1 2 x

e

−

 

 

 , x ∈ R o f(x) = 3 − loge(2x + 3)

Logarithmic function. So one-to-one, so f⁻¹(x) exists.

The inverse is

x = 3 − loge(2y + 3) −x + 3 = loge(2y + 3) 2y + 3 = e^{(−x + 3)} 2y = e^{−x + 3}− 3 ∴ f ⁻¹(x) = ³ 3

2

e⁻x− , x ∈ R 3

Function Inverse of function

Domain Range Domain Range

dom f = ran f⁻¹ ran f = dom f⁻¹ dom f⁻¹ = ran f ran f⁻¹ = dom f

a R R⁻ R⁻ R

b [1, ∞) R R [1, ∞)

c [−3, 3] [0, 3] [0, 3] [−3, 3]

d R R⁺ R⁺ R

e R [−10, ∞) [−10, ∞) R

f R⁻ (0, ∞) (0, ∞) R⁻

g [−5, 5] [0, 8] [0, 8] [−5, 5]

h R⁺ R R R⁺

4 a The function can be divided into two one-to-one functions over the domains (−∞, 2] and [2, ∞). Any one-to-one function would need its domain to be contained in one of these.

The answer is B.

b Using the above logic, the answer is (−∞, 2] or [2, ∞) The answer is E.

5 a f(x) = 2 ₂ (x 3) +1

−

Domain restricted to (3, ∞) The answer is C.

b Domain restricted to [0, 3) The answer is D.

6 The graph of the function f(x) = x²+ 1 is as follows:

It is not a one-to-one function but could be made one-to-one by splitting the domain into either R⁺∪ {0} or R⁻∪ {0}, the right and left arms of the parabola respectively.

a The answer is D.

b The positive set above is R⁺∪ {0} and so [0, ∞) satisfies.

The answer is C.

7 a Select either the right half of the parabola or the left half.

These will both form one-to-one functions, So f⁻¹ exists.

The largest domain is (−∞, 0] or [0, ∞).

b Right half of parabola has domain [−3, ∞]. Left half has domain (−∞, −3]. Solution [−3, ∞).

c Divide the graph into three sections with vertical lines through x = −3 and x = −1. Each section is one-to-one, therefore has an inverse function f ⁻¹.

The largest domain is [−1, ∞).

d The function is one-to-one, with domain (0, ∞). So f ⁻¹ exists over the whole domain (0, ∞), or R⁺.

e The function is one-to-one, with domain R. So f ⁻¹ exists over the entire domain R.

f The function can be split into two one-to-one functions each with an inverse. So f ⁻¹ exists over the domains [−9, 0] or [0, 9].

g The function can be split into two one-to-one functions by a vertical line through x = 5. So f ⁻¹ exists over the domains [1, 5] or [5, 9].

h The y-axis divides f(x) into two one-to-one functions. So f ⁻¹ exists over (−∞, 0) or (0, ∞).

i The vertical line x = 5 divides f(x) into two one-to-one functions, so f ⁻¹ exists over the domain (−∞, 5) or (5, ∞).

Exercise 5D — Restricting functions 1 a

f is one-to-one over (−∞, 0] and [0, ∞). Largest possible domain for f ⁻¹ to exist is (−∞, 0] or [0, ∞).

b

f is one-to-one over (−∞, 0] and [0, ∞). The largest possible domain for f ⁻¹ to exist is (−∞, 0] or [0, ∞).

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c

f is one-to-one over (−∞, −3] and [−3, ∞). The largest possible domain for f ⁻¹ to exist is [−3, ∞).

d

f is one-to-one over (−∞, 0) and (0, ∞). The largest possible domain for f ⁻¹ to exist is (−∞, 0) or (0, ∞).

e

f is one-to-one over (−∞, 4) and (4, ∞). The largest possible domain for f ⁻¹ to exist is (−∞, 4) or (4, ∞).

f

f is one-to-one over (−∞, 2) and (2, ∞). The largest possible domain for f ⁻¹ to exist is R\{2}.

g

f is one-to-one over [−5, 0] and [0, 5]. The largest possible domain for f ⁻¹ to exist is [−5, 0] or [0, 5].

h

f is one-to-one over [−1, 0] and [0, 1]. The largest possible domain for f ⁻¹ to exist is [−1, 0] or [0, 1].

i

f is one-to-one over its domain [4, ∞). Hence the inverse f ⁻¹ exists over its entire domain [4, ∞).

j f(x) = x²− 2x + 5 = (x − 1)²+ 4

f is one-to-one over (−∞, 1] and [1, ∞). Hence the largest domain for the inverse f ⁻¹ to exist is (−∞, 1].

k

f is one-to-one over its entire domain. Hence f ⁻¹ exists over R.

l

f is one-to-one over its entire domain. Hence f ⁻¹ exists over (5, ∞).

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2 a Choose f(x) to have domain [0, ∞).

Range of f(x) = [3, ∞) ∴ dom f ⁻¹= [3, ∞) ran f ⁻¹= [0, ∞) y = x²+ 3

Rule for f ⁻¹: x = y²+ 3 ∴ y = ± x− 3

As ran f ⁻¹= [0, ∞), we choose the positive square root.

∴ f ⁻¹(x) = x− 3

Hence f ⁻¹: [3, ∞) → R, f ⁻¹(x) = x− 3 b Choose f(x) to have domain [0, ∞).

Range of f(x) = [−1, ∞) ∴ dom f ⁻¹ = [−1, ∞) ran f ⁻¹ = [0, ∞) y = 3x²− 1 Rule for f ⁻¹: x = 3y²− 1 ∴ y²= 1

3 x+

y = 1

3 x+

±

As ran f ⁻¹= [0, ∞), choose the positive square root.

∴ f ⁻¹(x) = 1 3 x+

Hence f ⁻¹: [−1, ∞) → R, f ⁻¹(x) = 1 3 x+ c f(x) has a domain [−3, ∞).

Range of f(x) = [−2, ∞) ∴ dom f ⁻¹= [−2, ∞) ran f ⁻¹= [−3, ∞) y = (x + 3)²− 2 Rule for f ⁻¹: x = (y + 3)²− 2 y + 3 = ± x+ 2 y = 3− ± x+ 2

As ran f ⁻¹= [−3, ∞), we choose the positive square root.

∴ f ⁻¹(x) = 3− + x+ 2

Hence f ⁻¹: [−2, ∞)→R, f ⁻¹(x) = 3− + x+ 2 d Choose f(x) to have domain (0, ∞).

Ran f = (−3, ∞) ∴ dom f ⁻¹= (−3, ∞) ran f ⁻¹= (0, ∞) y = 1₂

x − 3 Rule for f ⁻¹: x = 1₂

y − 3

x + 3 = 1₂ y y = 1

3

± x +

As ran f ⁻¹= (0, ∞), choose the positive square root.

∴ f ⁻¹(x) = 1 3 x+

Hence f ⁻¹: (−3, ∞) → R, f ⁻¹(x) = 1 3 x+ e Choose f(x) to have domain (4, ∞).

Ran f = (1, ∞) ∴ dom f ⁻¹= (1, ∞) ran f ⁻¹= (4, ∞)

y = 1 ₂ (x 4) +1

− Rule for f ⁻¹: x = 1 ₂

(y 4) +1

− x − 1 = 1 ₂

(y−4) (y − 4)²= 1

1 x− y − 4 = 1

1

± x

− y = 4 1

1

± x

−

As ran f ⁻¹= (4, ∞), choose the positive square root.

∴ f ⁻¹(x) = 1 4+ x 1

−

Hence f ⁻¹: (1, ∞) → R, f ⁻¹(x) = 1 4+ x 1

− f f(x) has a domain R\{2}.

∴ dom f ⁻¹= R\{0}

ran f ⁻¹= R\{2}

f(x) = 1 2:

x− find f ⁻¹(x) x = 1

2 y− y − 2 = 1

x y = 1

x+ 2 f ⁻¹(x) = 1 2

x+

Hence f ⁻¹: R\{0} → R, f ⁻¹(x) = 1 x+ 2 g Choose f(x) to have domain [0, 5].

Ran f = [−5, 0]

∴ dom f ⁻¹ = [−5, 0]

ran f ⁻¹ = [0, 5]

f(x) = − 25−x²: find f ⁻¹(x) x = − 25 y− ²

x² = 25 − y² y² = 25 − x² y = ± 25 x− ²

As ran f ⁻¹= [0, 5], choose the positive square root.

∴ f ⁻¹(x) = 25 x− ²

Hence f ⁻¹: [−5, 0] → R, f ⁻¹(x) = 25 x− ² h Choose f(x) to have domain [0, 1].

Ran f = [0, 1]

∴ dom f ⁻¹= [0, 1]

ran f ⁻¹= [0, 1]

f(x) = 1−x²: find f ⁻¹(x) x = 1 y− ²

x²= 1 − y² y²= 1 − x² y = ± −1 x²

As ran f ⁻¹= [0, 1], choose the positive square root.

∴ f ⁻¹(x) = 1 x− ²

Hence f ⁻¹: [0, 1] → R, f ⁻¹(x) = 1 x− ²

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i dom f = [4, ∞) ran f = [0, ∞) ∴ dom f ⁻¹= [0, ∞) ran f ⁻¹ = [4, ∞) f(x) = x− find f4: ⁻¹(x) x = y− 4

x²= y − 4 y = x²+ 4 f ⁻¹(x) = x²+ 4

Hence f ⁻¹: [0, ∞) → R, f ⁻¹(x) = x²+ 4 j f(x) = x²− 2x + 5

= (x − 1)²+ 4 Choose dom f = [1, ∞) Ran f = [4, ∞) ∴ dom f ⁻¹= [4, ∞) ran f ⁻¹= [1, ∞)

f(x) = (x − 1)²+ 4: find f ⁻¹(x) x = (y − 1)²+ 4

(y − 1)²= x − 4 y − 1 = ± x− 4 y = 1± x− 4

As ran f ⁻¹= [1, ∞), choose the positive square root.

f ⁻¹(x) = 1+ x− 4

Hence f ⁻¹: [4, ∞) → R, f ⁻¹(x) = 1+ x− 4 k dom f = R

ran f = (0, ∞) ∴ dom f ⁻¹= (0, ∞) ran f ⁻¹ = R f(x) = e^x+2: find f ⁻¹(x) x = e^{y + 2}

y + 2 = loge(x) y = loge(x) − 2 f⁻¹(x) = log e(x) − 2

Hence f ⁻¹: (0, ∞) → R, f ⁻¹(x) = loge(x) − 2 l dom f = (5, ∞)

ran f = R ∴ dom f ⁻¹ = R ran f ⁻¹ = (5, ∞)

f(x) = 2 log^e(x − 5): find f ⁻¹(x) x = 2 loge(y − 5)

y − 5 = ²

x

e y = 5 ²

x

+e f ⁻¹(x) = 5 ²

x

+e

Hence f ⁻¹: R → R, f ⁻¹(x) = 5 ²

x

+e 3 a

A function and its inverse can only intersect on the line y = x.

The answer is B.

b Domain of function [0, ∞) f(x) = x²: find rule for f ⁻¹(x).

x = y²

y = ± x f ⁻¹(x) = ± x

dom f = [0, ∞), then ran f ⁻¹= [0, ∞) so we take the positive square root. Hence f ⁻¹(x) = x , x ∈ [0, ∞) The answer is C.

c From solution for b above, [0, ∞].

The answer is D.

d Intersect at f(x) = f ⁻¹(x) ∴ x² = x x⁴ = x x⁴− x = 0 x(x³− 1) = 0 x = 0 or 1.

x = 0 ⇒ f(x) = 0 (0, 0) x = 1 ⇒ f(x) = 1 (1, 1) The answer is A.

4 a

b

c

d

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e

f

g

h

5 a f(x) = 3+ x− 1

Graph of f(x) = x translated 1 unit right and 3 units , up.

Hence f(x) is a one-to-one function over its domain.

∴ S = [1, ∞)

b Graph is sketched using dotted line on axes above.

f(x) = 3+ x− find f1: ⁻¹(x) x = 3+ y− 1 y− = x − 3 1 y = (x − 3)²+ 1 ran f = [3, ∞)

∴ dom f ⁻¹= [3, ∞)

Hence f ⁻¹: [3, ∞) → R, f ⁻¹(x) = (x − 3)²+ 1 6 a f(x) = (x − 4)²− 5

Graph is one-to-one over [−2, 4] and so a = 4.

b dom f = [−2, 4]

x = −2 ⇒ f(x) = 31 x = 4 ⇒ f(x) = −5 ∴ ran f = [−5, 31]

Hence

dom f ⁻¹= [−5, 31], ran f = [−2, 4]

f(x) = (x − 4)²− 5: find f ⁻¹(x) x = (y − 4)²− 5

x + 5 = (y − 4)² y − 4 = ± x+ 5 y = 4± x+ 5

As ran f ⁻¹= [−2, 4] we take the negative square root.

∴ y = 4− x+ 5 Hence f⁻¹: [−5, 31] → R, f⁻¹(x) = 4− x+ 5 7 a

From the graph dom f = R⁺ ∴ ran f ⁻¹= dom f = R⁺ f(x) = 1 :

− x find f⁻¹(x) x = 1

− y x² =

1 2

y

 

−

 

 

 

= 1 y ∴ y = 1₂

x

Hence f ⁻¹(x) = 1₂ x , x ∈ R⁻ Range is R⁺.

b f [f ⁻¹(x)]

= 2

f 1 x

  

  =

2

1 1 x

−

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Now 1₂ x = 1

x if x > 0 or 1 x

− if x < 0.

The domain of f ⁻¹(x) is R⁻ and so 1 x

− is taken.

f [f ⁻¹(x)] = 1 1 x

−−

= x for x ∈ R⁻ c f⁻¹[f(x)]

= f ⁻¹ 1 x

 

− 

 

= 1 ₂ 1

x

 

− 

 

= 1 1 x = 1 ×

1 x = x for x > 0.

d Both equate to x.

8 a

b f⁻¹ exists if f(x) is one-to-one. Choose either the left half, or the right half, that is, domain is (−∞, 0] or [0, ∞).

c

d Find f⁻¹(x), given f(x) = ₂1 2 x + x = ₂1

2 y + y²+ 2 = 1

x y²= 1

x− 2

y = 1 2

± x−

Choose the positive square root, as the graph shows f⁻¹(x) is positive.

∴ f ⁻¹(x) = 1 x− 2 Find f [f⁻¹(x)]

f [f⁻¹(x)] = f 1 2 x

 

−

 

 

 

= 1 ₂

1 2 2

x

 

− +

 

 

 

= 1

1 2 2

x

 − +

 

 

= 1 1 x = 1

1

× x

= x for x > 0 Find f⁻¹[f(x)]

f⁻¹[f(x)] = ¹ ₂1 f 2

x

−  

 

+

 

=

2

1 2

1 2 x

− +

= ² 2

1 2

1 x +

× −

= x²+ − 2 2 = x ² = x for x > 0.

Hence they are both equivalent and equal to x.

9 a

g(x) is an inverted parabola translated upwards 9 units.

For g⁻¹(x) to exist, we must restrict the domain of g so that g(x) is a one-to-one function. Choose the left or right half of the parabola, that is, domains (−∞, 0] or [0, ∞).

Given the domain is [b, 8], we have a section of the right half of the parabola. The smallest value of b is 0.

b Find g⁻¹, given g(x) = 9 − x² x = 9 − y²

y²= 9 − x y = ± 9 x− Now Dom g = [0, 8]

x = 0 ⇒ g(x) = 9 x = 8 ⇒ g(x) = −55 ∴ Ran g = [−55, 9]

Hence

dom g⁻¹= [−55, 9]

ran g⁻¹= [0, 8]

As ran g⁻¹ is positive, we take the positive square root.

Hence g⁻¹: [−55, 9] → R, g⁻¹(x) = 9 x− c ran g⁻¹= dom g = [0, 8].

I n v e r s e f u n c t i o n s M M 1 2 - 5

117

10 g(x) = 3+ 4 x− ² defines a circle, radius 2 units, translated 3 units upwards.

The two maximal domains are [−2, 0] and [0, 2]

Choose the domain [−2, 0] range = [3, 5]

∴ dom g⁻¹= [3, 5]

ran g⁻¹= [−2, 0]

Find the rule for g⁻¹(x):

x = 3+ 4 y− ² x − 3 = 4 y− ² 4 − y²= (x − 3)² y² = 4 − (x − 3)² y = ± 4 (− −x 3)²

As ran g⁻¹ is negative, choose the negative square root.

Hence g⁻¹: [3, 5] → R, g⁻¹(x) = − 4 (− −x 3)²

Now choose the domain [0, 2] range = [3, 5]

∴ dom g⁻¹= [3, 5]

ran g⁻¹= [0, 2]

Find the rule for g⁻¹(x): y = ± 4 (− −x 3) ,² as before.

As ran g⁻¹ is positive, choose the positive square root.

Hence g⁻¹: [3, 5] → R, g⁻¹(x) = 4 (− −x 3)²

The two inverse functions g⁻¹ are:

g⁻¹: [3, 5] → R, g⁻¹(x) = − 4 (− −x 3)² g⁻¹: [3, 5] → R, g⁻¹(x) = 4 (− −x 3) .² Chapter review

Short answer 1 a

b

2 a Bottom half of a semicircle, radius 6.

Domain = [−6, 6]

Range = [−6, 0]

b Exponential graph, dilated by a factor of 2 parallel to the y-axis, translated 2 units to the right. The x-axis is an asymptote.

At x = 0, y = 2e⁻² ∴ y-intercept (0, 2e⁻²)

Domain = R Range = R⁺ 3 a

b

c

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I n v e r s e f u n c t i o n s

4 a f(x) = 3 − loge(x + 4) is a log function reflected in the x-axis and translated (−4, 3). Hence its domain is (−4, ∞).

As the log function is one-to-one, f(x) is defined over its entire domain.

Domain = (−4, ∞)

b g(x) = (x − 2)² is a parabola. A vertical line at x = 2 divides g(x) into two one-to-one functions. Hence g(x) is one-to-one over (−∞, 2] or [2, ∞).

c f(x) = 1₂ 5−x

This graph shows that f(x) is one-to-one over R⁻ or R⁺. 5 a f(x) = 2x − 1

i

f(x) is a 1:1 function.

iii Let f(x) = y

inverse → swap x and y x = 2y − 1

x + 1 = 2y y = 1

2 x+

b f(x) = 2(x − 1)³+ 1 i

f(x) is a 1:1 function iii Let f(x) = y

inverse → swap x and y x = 2(y − 1)³+ 1 x − 1 = 2(y − 1)³ (y − 1)³ = 1

2 x−

y − 1 = ³ 1 2 x−

y = ³ 1 1

2 x− + c f(x) = x²+ x

i

f(x) is a many: 1 function ii cusps: ⇒ (−1, 0) and (0, 0) restrict domain to [0, ∞)

iii For the domain [0, ∞), f(x) = x²+ x Let f(x) = y

inverse → swap x and y x = y²+ y x = y²+ y + 1

4 − 1 4 x = (y + 1

2)²− 1 4 x + 1

4 = (y +1 2)² (y + 1

2)² = x + 1 4 y + 1

2 = 1

x 4

± +

y = 1 1

4 2

± x+ −

y = 1 1

4 2

x+ − as dom f(x) = [0, ∞) d f(x) = 2− − x 3

i

f(x) is a 1:1 function iii Let f(x) = y

inverse → swap x and y x = 2− − y 3 x + 3 = 2 y− 2 − y = (x + 3)² y = 2 − (x + 3)² = 2 − (x²+ 6x + 9) = −x²− 6x − 7

The whole parabola is not needed, ran f(x) = [−3, ∞) ∴ domain of inverse = [−3, ∞)

∴ y = −x² – 6x − 7, x ∈ [−3, ∞) e f(x) = 3e^x+ 1

i

f(x) is a 1:1 function iii Let f(x) = y

inverse → swap x and y

I n v e r s e f u n c t i o n s M M 1 2 - 5

119

x = 3e^y+ 1 x − 1 = 3e^y e^y = 1

3 x−

y = 1

log^e 3 x−

 

 

 

f f(x) = 2 3(x 2)−1

− i

f(x) is a 1:1 function iii Let f(x) = y

inverse → swap x and y

x = 2 1

3(y 2)−

− x + 1 = 2

3(y−2) 3(y − 2) = 2

1 x+ y − 2 = 2

3(x+1) y = 2

3(x 1)+2 + g f(x) = 1 ₂

(x 3) +1 + i

f(x) is a many:1 function

ii asymptote: x = −3 ∴ domain f(x) = (−3, ∞) iii Let f(x) = y

inverse → swap x and y x = 1 ₂

(y 3) +1 + x − 1 = 1 ₂

(y+3) (y + 3)²= 1

1 x− y + 3 = 1 1

± x

− y + 3 = 1

1

x− (discard negative because dom f(x) = (−3, ∞))

y = 1 1 3

x −

−

6 a f(x) = loge(x − 2) + 1 is a log graph with an asymptote at x = 2.

x-intercept (y = 0):

loge(x − 2) + 1 = 0

loge(x − 2) = −1 x − 2 = e⁻¹ x = 2 + e⁻¹ y-intercept (x = 0):

There is no y-intercept.

Domain = (2, ∞) Range = R

b f(x) = loge(x − 2) + 1: find f ⁻¹(x) x = loge(y − 2) + 1

y − 2 = e^{x − 1} y = 2 + e ^{x − 1} Inverse function f⁻¹(x) = 2 + e^{x − 1}, x ∈ R c f ⁻¹(m) = 3

2 + e^{m − 1}= 3 e^{m − 1}= 1 e^{m − 1}= e⁰ m − 1 = 0 m = 1 d

f ⁻¹: Domain R; Range (2, ∞) e Mark (3, 1) on graph of f(x) 7 g(x) = x²+ 5x − 1

=

5 2 25 2 4 1

x+  − −

 

 

=

5 2 29

2 4

x 

+ −

 

 

g(x) is a parabola. A vertical line at x = 5

− divides g(x) into 2 two one-to-one functions. Hence g(x) is one-to-one over (−∞, 5

− ] 2 ∴ b = 5

− 2

8 a a = 1 for horizontal asymptote using (0, 0) 1 + be⁰= 0

1 + b = 0 b = −1

b range h(x) = 1₂ 0, 1 e

 − 

 

 

c let h(x) = y for inverse interchange x and y x = 1 – e^−y

e^−y= 1 − x −y = loge(1 − x) y = −loge(1 − x) h⁻¹: 0, 1 1₂

e

 − 

 

  → R, h⁻¹(x) = −loge(1 − x)

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I n v e r s e f u n c t i o n s

d

9 f(x) = e^2x− 1

a let f(x) = y for inverse interchange x and y x = e^2y− 1

e^2y= x + 1 2y = loge(x + 1) y = 1

log ( 1) 2 ^e x+ f ⁻¹: (−1, ∞) → R, f ⁻¹(x) = 1

log ( 1) 2 ^e x+ b

c f(−f ⁻¹(2x)) =

2 1log (2 1)

2 ^{e x} 1

e

 

− + − = e^{−log (2e x+1)}− 1 = e^{log (2e x+1)−1}− 1

= 1 1

2x 1− +

= 1 (2 1)

2 1

x x

− +

+

= 2

2 1

x x

− + Multiple choice

1 Relation has coordinates (−4, 0) and (0, 2)

∴ Inverse relation has coordinates (0, −4) and (2, 0) The answer is B.

2

The answer is E.

3 f(x) has an asymptote at y = 1.

∴ f ⁻¹(x) will have an asymptote at x = 1. The mirror image of f(x) through y = x is graph C.

∴C

4 x²+ (y + 1)²= 16 is a circle.

∴ Many-to-many relation. Therefore the inverse will also be a many-to-many relation

∴ C

5 The x-intercept of f ⁻¹(x) = y-intercept of f(x) f(x) = 3− − x 3

f(0) = 3 3− ≈ −1.27 ∴ D

6 The y-intercept of f ⁻¹(x) = x-intercept of f(x).

f(x) = 2 (x 2)+3

− 0 = 2

2 3

x +

− −3 = 2 2 x− x − 2 = 2 3

−

x = 2 2

− + 3 = 4

3 ∴ A.

7 f(x) = 2 2 3

x +

−

asymptotes: x = 2 and y = 3

∴ For f ⁻¹(x), asymptotes: x = 3 and y = 2 ∴ C

8 y = 4₂

x : find the inverse x = 4₂

In document Maths Quest Methods Solutions y11 (Page 111-126)