SOLVED PROBLEMS ON INVERSE FUNCTIONS

Problems 1-8 : Find f-1(x).

Problem 1 :

f(x) = {(2, 3), (3, 5), (4, 7), (5, 9)}

Solution :

f-1(x) = {(3, 2), (5, 3), (7, 4), (9, 5)}

Problem 2 :

Solution :

Replace f(x) by y.

Switch x and y.

Solve for y interms of x.

Replace y by f-1(x).

Problem 3 :

f(x) = 3x - 5

Solution :

f(x) = 3√x - 5

y = 3√x - 5

x = 3√y - 5

x + 5 = 3√y

(x + 5)3 = (3√y)3

(x + 5)3 = y

f-1(x) = (x + 5)3

Problem 4 :

Solution :

y + 7 = (3 - x)2

y = (3 - x)2 - 7

Problem 5 :

f(x) = 4 - x2

Solution :

f(x) = 4 - x2

y = 4 - x2

x = 4 - y2

y2 = 4 - x

Problem 6 :

f(x) = x2 + 6x + 7

Solution :

f(x) = x2 + 6x + 7

y = x2 + 6x + 7

x = y2 + 6y + 7

x = y2 + 2(y)(3) + 32 - 32 + 7

x = (y + 3)2 - 32 + 7

x = (y + 3)2 - 9 + 7

x = (y + 3)2 - 2

x + 2 = (y + 3)2

(y + 3)= x + 2

Problem 7 :

f(x) = log2(x + 5)

Solution :

f(x) = log2(x + 5)

y = log2(x + 5)

x = log2(y + 5)

Convert the above equation to exponential.

2x = y + 5

2x - 5 = y

y = 2x - 5

f-1(x) = 2x - 5

Problem 8 :

f(x) = 10x - 3 + 2

Solution :

f(x) = 10x - 3 + 2

y = 10x - 3 + 2

x = 10y - 3 + 2

x - 2 = 10y - 3

Convert the above equation to logarithmic form.

log10(x - 2) = y - 3

log10(x - 2) + 3 = y

y = log10(x - 2) + 3

f-1(x) log10(x - 2) + 3

Problem 9 :

The function f(x) = a(-x3 - x + 2) has an inverse function such that f-1(6) = -2. Solve for a.

Solution :

f-1(6) = -2

-2 = f-1(6)

f(-2) = f[f-1(6)]

f(-2) = 6

a[-(-2)3 - (-2) + 2] = 6

a[-(-8) + 2 + 2] = 6

a(8 + 4) = 6

a(12) = 6

12a = 6

Problem 10 :

If the point P(-2, 5) is on the graph of y = f(x), then determine the coordinates of P' on the graph of

y = f-1(x - 3) + 1

Solution :

When P(-2, 5) is on the graph of y = f(x), let P'(a, b) is on the graph of y = f-1(x - 3) + 1.

Then, we have

5 = f(-2)

or

f(-2) = 5 ----(1)

b = f-1(a - 3) + 1

b - 1 = f-1(a - 3)

f(b - 1) = f[f-1(a - 3)]

f(b - 1) = a - 3 ----(2)

Comparing (1) and (2), we get

b - 1 = -2

b = -1

a - 3 = 5

a = 8

The coordinates of P' are

(a, b) = (8, -1)

Therefore, if the point P(-2, 5) is on the graph of y = f(x), then the coordinates of P' on the graph of y = f-1(x - 3) + 1 are (8, -1).

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