WORKSHEET ON INVERSE FUNCTIONS

Problem 1 :

Suppose f(x) = x² - 1, with the domain of f being the set of positive numbers. 

(a) Evaluate f⁻¹(8)

(b) Evaluate [f(8)]⁻¹

Problem 2 :

Suppose f(x) = 4x + 6, evaluate f⁻¹(5).

Problem 3 :

Suppose f(x) = 7x - 5, evaluate f⁻¹(-3).

Problem 4 :

Suppose f(x) = 2 + (x - 5)/(x + 6), with the domain of f being the set of positive numbers. 

(a) Evaluate f ⁻¹(4)

(b) Evaluate [f(4)]⁻¹

Problem 5 :

Suppose h(x) = 3x² - 4, where the domain of h is the set of positive numbers, find h⁻¹(x).

1. Answer :

(a) Evaluate f⁻¹(8) :

f(x) = x² - 1

Replace f(x) by y. 

y  =  x² - 1

Interchange x and y. 

x = y² - 1

Solve for y. 

x + 1 = y²

Taking square on both sides, 

y = ±√(x + 1)

Replace y by f^-1(x). 

f⁻¹(x) = ±√(x + 1)

f⁻¹(8) = ±√(8 + 1)

= ±3

(b) Evaluate [f(8)]⁻¹

[f(8)]^-1 = 1/f(8)

= 1/(8² - 1)

= 1/(64 - 1)

= 1/63

2. Answer :

f(x) = 4x + 6

Replace f(x) by y. 

y = 4x + 6

Interchange x and y. 

x = 4y + 6

Solve for y. 

x - 6 = 4y

y = (x - 6)/4

Replace y by f^-1(x).

f^-1(x) = (x - 6)/4

f⁻¹(5) = (5 - 6)/4

= -1/4

3. Answer : 

f(x) = 7x - 5

Replace f(x) by y. 

y = 7x - 5

Interchange x and y. 

x = 7y - 5

Solve for y. 

x + 5 = 7y

y = (x + 5)/7

Replace y by f^-1(x).

f^-1(x) = (x + 5)/7

f⁻¹(-3) = (-3 + 5)/7

= 2/7

4. Answer :

f(x) = 2 + (x - 5)/(x + 6)

f(x) = [2(x + 6) + (x - 5)]/(x + 6)

f(x) = [2x + 12 + x - 5]/(x + 6)

f(x) = [3x + 7]/(x + 6)

f(x) = (3x + 7)/(x + 6)

(a) Evaluate f ⁻¹(4) :

f(x) = (3x + 7)/(x + 6)

Replace f(x) by y. 

y = (3x + 7)/(x + 6)

Interchange x and y. 

x = (3y + 7)/(y + 6)

Solve for y. 

x(y + 6) = 3y + 7

xy + 6x = 3y + 7

xy - 3y = 7 - 6x

y(x - 3) = 7 - 6x

y = (7 - 6x)/(x - 3)

Replace y by f^-1(x).

f^-1(x) = (7 - 6x)/(x - 3)

f⁻¹(4) = (7 - 24)/(4 - 3)

= -17/1

= -17

(b) Evaluate [f(4)]⁻¹.

[f(x)]⁻¹ = 1/f(x)

[f(x)]⁻¹ = (x + 6)/(3x + 7)

[f(4)]⁻¹ = (4 + 6)/(12 + 7)

= 10/19

5. Answer :

h(x) = 3x² - 4

Replace h(x) by y. 

y = 3x² - 4

Interchange x and y. 

x = 3y² - 4

Solve for y.  

x + 4 = 3y²

(x + 4)/3 = y²

Taking square on both sides, 

y = ±√[(x + 4)/3]

Replace y by h^-1(x). 

h⁻¹(x) = ±√[(x + 4)/3]

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