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) = ³√x - 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)² = 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.

2^x = y + 5

2^x - 5 = y

y = 2^x - 5

f^-1(x) = 2^x - 5

Problem 8 :

f(x) = 10^{x - 3} + 2

Solution :

f(x) = 10^{x - 3} + 2

y = 10^{x - 3} + 2

x = 10^{y - 3} + 2

x - 2 = 10^{y - 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

Comparing (1) and (2), we get

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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