College Algebra Problems With Answers
sample 2 : Composite and Inverse Functions

College algebra problems and questions on composite and inverse functions are presented along with their solutions located at the bottom of the page.

  1. Let f(x) = √(x - 4) + 3.
    a) Find the inverse of f.
    b) Find the range of f -1.
  2. Let h(x) = (x - 1) / (-x + 3).
    a) Find the inverse of h.
    b) Find the range of h.
  3. Let f(x) = (x - 1)/(x + 5) and g(x) = 1/(x + 3).
    a) Find the composite function (f o g)(x).
    b) Find the domain of f o g.
  4. Function f is a function with inverse f -1. Function h is defined by h(x) = f(x) + k where k is a constant. Express the inverse function of h in terms of f -1 and k.
  5. Function f is a function with inverse f -1. Function h is defined by h(x) = A*f(x - h) + k where A, k and h are constants. Express the inverse function of h in terms of f -1, A, k and h.
  6. The graphs of functions f and g are shown below.
    a) Use the graph to find (f o g)(-4)
    b) Use the graph to find (g o f)(1)

    college algebra problem 4, function f(x) .

    college algebra problem 4, function g(x) .


  7. Functions f and h are defined by the tables
    x -3 -2 -1 0 1 2 3
    f(x) -6 -4 -2 1 2 6 16

    x 0 1 2 3 4 5 6
    h(x) 1 2 5 10 17 26 37

    Use the values in the tables to find
    a) (f o h)(1)
    b) (f o f)(0)
    c) (f o h)(5)
    d) (f o h-1)(5)
    e) (h o f-1)(6)

Answers to the Above Questions

    1. f -1(x) = (x - 3)3 + 4 , x≥3
    2. [4 , +infinity) : it is the domain of f

    1. h -1(x) = (-3x - 1) / (x + 1)
    2. (-infinity , -1) U (-1 , +infinity) : it is the domain of h-1

    1. (f o g)(x) = -(x + 2) / (5x + 16)
    2. domain of the composite function f o g : (-infinity , -16/5) U (-16/5 , -3) U (-3 , +infinity)

  1. h -1(x) = f -1(x - k)
  2. h -1(x) = f -1((x - k) / A) + h
    1. (f o g)(-4) = f(g(-4)) = f(2) = -2
    2. (g o f)(1) = g(f(1) = g(-3) = -1

    1. (f o h)(1) = f(h(1)) = f(2) = 6
    2. (f o f)(0) = f(f(0)) = f(1) = 2
    3. (f o h)(5) = f(h(5) = f(26) = undefined
    4. (f o h-1)(5) = f(h-1(5)) = f(2) = 6
    5. (h o f-1)(6) = h(f-1(6)) = h(2) = 5

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