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Math Grade 9-12 Answer Key

Inverse Functions and Compositions

Finding inverses and evaluating composed functions

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Inverse Functions and Compositions

Finding inverses and evaluating composed functions

Math - Grade 9-12

Instructions: Read each problem carefully. Show your work in the space provided and write final answers clearly.
  1. 1

    Let f(x) = 3x - 7 and g(x) = x + 5. Find (f composed with g)(x), written as f(g(x)).

    Substitute the entire expression x + 5 into f(x) wherever x appears.

    f(g(x)) = f(x + 5) = 3(x + 5) - 7 = 3x + 15 - 7 = 3x + 8.
  2. 2

    Let f(x) = x squared + 2 and g(x) = 4x. Find g(f(x)).

    g(f(x)) = g(x squared + 2) = 4(x squared + 2) = 4x squared + 8.
  3. 3

    Let f(x) = 2x + 1 and g(x) = x squared - 3. Find f(g(4)).

    Work from the inside out by finding g(4) first.

    First, g(4) = 4 squared - 3 = 16 - 3 = 13. Then f(13) = 2(13) + 1 = 27, so f(g(4)) = 27.
  4. 4

    Find the inverse of f(x) = x - 9.

    Let y = x - 9. Swap x and y to get x = y - 9. Solve for y: y = x + 9. Therefore, f inverse of x is f^-1(x) = x + 9.
  5. 5

    Find the inverse of f(x) = 5x + 2.

    After switching x and y, undo addition before undoing multiplication.

    Let y = 5x + 2. Swap x and y to get x = 5y + 2. Solve for y: x - 2 = 5y, so y = (x - 2)/5. Therefore, f^-1(x) = (x - 2)/5.
  6. 6

    Find the inverse of f(x) = (x - 4)/3.

    Let y = (x - 4)/3. Swap x and y to get x = (y - 4)/3. Multiply both sides by 3 to get 3x = y - 4. Add 4 to get y = 3x + 4. Therefore, f^-1(x) = 3x + 4.
  7. 7

    Use composition to verify that f(x) = 2x - 6 and g(x) = (x + 6)/2 are inverse functions.

    To prove two functions are inverses, show that both f(g(x)) and g(f(x)) simplify to x.

    f(g(x)) = 2((x + 6)/2) - 6 = x + 6 - 6 = x. Also, g(f(x)) = ((2x - 6) + 6)/2 = 2x/2 = x. Since both compositions equal x, the functions are inverses.
  8. 8

    Let f(x) = x cubed. Find f^-1(x).

    Let y = x cubed. Swap x and y to get x = y cubed. Taking the cube root of both sides gives y = cube root of x. Therefore, f^-1(x) = cube root of x.
  9. 9

    Let f(x) = square root of (x + 1). State the domain of f and find f^-1(x).

    Remember that the range of the original function becomes the domain of the inverse function.

    For f(x) to be defined, x + 1 must be greater than or equal to 0, so the domain is x greater than or equal to -1. Let y = square root of (x + 1). Swap x and y to get x = square root of (y + 1). Square both sides to get x squared = y + 1, so y = x squared - 1. Since the range of f is y greater than or equal to 0, the domain of f^-1 is x greater than or equal to 0. Therefore, f^-1(x) = x squared - 1 for x greater than or equal to 0.
  10. 10

    Let f(x) = x squared with domain x greater than or equal to 0. Find f^-1(x).

    Let y = x squared. Swap x and y to get x = y squared. Since the original domain is x greater than or equal to 0, the inverse uses the nonnegative square root. Therefore, f^-1(x) = square root of x.
  11. 11

    Let f(x) = 1/(x - 2). Find f^-1(x).

    After switching x and y, clear the denominator first.

    Let y = 1/(x - 2). Swap x and y to get x = 1/(y - 2). Multiply both sides by y - 2 to get x(y - 2) = 1. Divide by x to get y - 2 = 1/x. Add 2 to get y = 1/x + 2. Therefore, f^-1(x) = 1/x + 2.
  12. 12

    Let f(x) = 4x - 1 and g(x) = (x + 1)/4. A student says f(g(x)) = x but g(f(x)) = 4x, so the functions are not inverses. Find the error and give the correct result.

    Substitute all of f(x) into g(x), including the plus 1 in the numerator.

    The error is in simplifying g(f(x)). The correct composition is g(f(x)) = ((4x - 1) + 1)/4 = 4x/4 = x. Also, f(g(x)) = 4((x + 1)/4) - 1 = x + 1 - 1 = x. Since both compositions equal x, the functions are inverses.
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