Return to the Purplemath home page

 The Purplemath Forums
Helping students gain understanding
and self-confidence in algebra

powered by FreeFind


Return to the Lessons Index  | Do the Lessons in Order  |  Get "Purplemath on CD" for offline use  |  Print-friendly page

Operations on Functions

First you learned (back in grammar school) that you can add, subtract, multiply, and divide numbers. Then you learned that you can add, subtract, multiply, and divide polynomials. Now you will learn that you can also add, subtract, multiply, and divide functions. Performing these operations on functions is no more complicated than the notation itself. For instance, when they give you the formulas for two functions and tell you to find the sum, all they're telling you to do is add the two formulas. There's nothing more to this topic than that, other than perhaps some simplification of the expressions involved.

  • Given f(x) = 3x + 2 and g(x) = 4 5x,
    (f + g)(x), (f g)(x), (fg)(x), and (f / g)(x).
  • To find the answers, all I have to do is apply the operations (plus, minus, times, and divide) that they tell me to, in the order that they tell me to.

      (f + g)(x) = f(x) + g(x) = [3x + 2] + [4 5x] = 3x 5x + 2 + 4 = 2x + 6

      (f g)(x) = f(x) g(x) = [3x + 2] [4 5x] = 3x + 5x + 2 4 = 8x 2

      (fg)(x) = [f(x)][g(x)] = (3x + 2)(4 5x) = 12x + 8 15x2 10x

          = 15x2 + 2x + 8

      (f/g)(x) = [f(x)]/[g(x)] = [3x + 2]/[4 - 5x]

  • Given f(x) = 2x, g(x) = x + 4, and h(x) = 5 x3,
    (f + g)(2), (h g)(2), (f h)(2), and (h / g)(2).
  • To find the answers, I can either work symbolically (like in the previous example) and then evaluate, or I can find the values of the functions at x = 2 and then work from there. It's probably simpler in this case to evaluate first, so:

      f(2) = 2(2) = 4

      g(2) = (2) + 4 = 6

      h(2) = 5 (2)3 = 5 8 = 3

    Now I can evaluate the listed expressions:

      (f + g)(2) = f(2) + g(2) = 4 + 6 = 10

      (h g)(2) = h(2) g(2) = 3 6 = 9

      (f h)(2) = f(2) h(2) = (4)(3) = 12

      (h / g)(2) = h(2) g(2) = 3 6 = 0.5

If you work symbolically first, and plug in the x-value only at the end, you'll still get the same results. Either way will work. Evaluating first is usually easier, but the choice is up to you.

You can use the Mathway widget below to practice operations on functions. Try the entered exercise, or type in your own exercise. Then click "Answer" to compare your answer to Mathway's. (Or skip the widget and continue with the lesson.)

(Clicking on "View Steps" on the widget's answer screen will take you to the Mathway site, where you can register for a free seven-day trial of the software.)

  • Given f(x) = 3x2 x + 4, find the simplified form of the following expression, and evaluate at h = 0:
        [f(x + h) - f(x)] / h 
      Copyright Elizabeth Stapel 2005-2011 All Rights Reserved
  • This isn't really a functions-operations question, but something like this often arises in the functions-operations context. The simplest way for me to proceed with this exercise is to work in pieces, simplifying as I go; then I'll put everything together and simplify at the end.

    For the first part of the numerator, I need to plug the expression "x + h" in for every "x" in the formula for the function, using what I've learned about function notation, and then simplify:

      f(x + h) = 3(x + h)2 (x + h) + 4
                   = 3(x2 + 2xh + h2) x h + 4

                   = 3x2 + 6xh + 3h2 x h + 4

    The expression for the second part of the numerator is just the function itself:

      f(x) = 3x2 x + 4

    Now I'll subtract and simplify:

      f(x + h) f(x) = [3x2 + 6xh + 3h22 x h + 4] [3x2 x + 4]  
                             = 3x2 3x2 + 6xh + 3h2 x + x h + 4 4
                             = 6xh + 3h2 h

    All that remains is to divide by the denominator; factoring lets me simplify:

      (6xh + 3h^2 - h) / h = 6x + 3h - 1

    Now I'm supposed to evaluate at h = 0, so:

      6x + 3(0)  1 = 6x  1

      Simplified form:  6x + 3h  1

      Value at h = 0:  6x  1

That's pretty much all there is to "operations on functions" until you get to function composition. Don't let the notation for this topic worry you; it means nothing more than exactly what it says: add, subtract, multiply, or divide; then simplify and evaluate as necessary. Don't overthink this. It really is this simple.

Oh, and that last example? They put that in there so you can "practice" stuff you'll be doing in calculus. You likely won't remember this by the time you get to calculus, but you'll follow a very similar process for finding something called "derivatives".

Top  |  Return to Index

Cite this article as:

Stapel, Elizabeth. "Operations on Functions." Purplemath. Available from Accessed


  Linking to this site
  Printing pages
  School licensing

Reviews of
Internet Sites:
   Free Help
   Et Cetera

The "Homework

Study Skills Survey

Tutoring from Purplemath
Find a local math tutor

This lesson may be printed out for your personal use.

Content copyright protected by Copyscape website plagiarism search

  Copyright 2005-2012  Elizabeth Stapel   |   About   |   Terms of Use


 Feedback   |   Error?