HONORS ALGEBRA 2 : Problems on Inverse Functions

In each case, find f^-1(x).

Problem 1 :

Solution :

Replace f(x) by y.

Switch x and y.

Solve for y.

Replace y by f^-1(x).

Problem 2 :

Solution :

Replace f(x) by y.

Switch x and y.

Solve for y.

Replace y by f^-1(x).

Problem 3 :

Solution :

Replace f(x) by y.

Switch x and y.

Solve for y.

Replace y by f^-1(x).

Problem 4 :

Solution :

Replace f(x) by y.

Switch x and y.

Solve for y.

Replace y by f^-1(x).

Problem 5 :

Solution :

Replace f(x) by y.

Switch x and y.

Solve for y.

Replace y by f^-1(x).

Problem 6 :

f(x) = x² - 8

Solution :

f(x) = x² - 8

Replace f(x) by y.

y = x² - 8

Switch x and y.

x = y² - 8

Solve for y.

x + 8 = y²

Replace y by f^-1(x).

Problem 7 :

Solution :

Replace f(x) by y.

Switch x and y.

Solve for y.

Replace y by f^-1(x).

Problem 8 :

Solution :

Replace f(x) by y.

Switch x and y.

Solve for y.

Replace y by f^-1(x).

Problem 9 :

Solution :

Replace f(x) by y.

Switch x and y.

Solve for y.

Replace y by f^-1(x).

Problem 10 :

f(x) = 2^x - 1

Solution :

f(x) = 2^x - 1

Replace f(x) by y.

y = 2^x - 1

Switch x and y.

x = 2^y - 1

Solve for y.

x + 1 = 2^y

log₂(x + 1) = y

Replace y by f^-1(x).

f^-1(x) = log₂(x + 1)

Problem 11 :

f(x) = log₁₀x + 2

Solution :

f(x) = log₁₀x + 2

Replace f(x) by y.

y = log₁₀x + 2

Switch x and y.

x = log₁₀y + 2

Solve for y.

x - 2 = log₁₀y

10^{x - 2} = y

Replace y by f^-1(x).

f^-1(x) = 10^{x - 2}

Problem 12 :

If f(7) = -3 and f^-1(4) = 2, find the value of

Solution :

f(7) = -3 ----> (7, -3) is on f(x)

Then, (-3, 7) is on f^-1(x).

f^-1(-3) = 7

f^-1(4) = 2 ----> (4, 2) is on f^-1(x)

Then, (2, 4) is on f(x).

f(2) = 4

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