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Composition of Functions:
     Composing Functions with Functions (page 3 of 5)

Sections: Composing functions that are sets of point, Composing functions at points, Composing functions with other functions, Word problems using composition, Inverse functions and composition

You can also evaluate compositions symbolically. It is simpler to evaluate a composition at a point because you can simplify as you go, since you'll always just be plugging in numbers and simplifying. Evaluating a symbolic compositon, where you're first plugging x into some function and then plugging that function into some other function, can be much messier. But the process works just as the at-a-number composition does, and using parentheses to be carefully explicit at each step will be even more helpful.

• Given f(x) = 2x + 3 and g(x) = –x² + 5, find ( f o g)(x).

In this case, I am not trying to find a certain numerical value. Instead, I am trying to find the formula that results from plugging the formula for g(x) into the formula for f(x). I will write the formulas at each step, using parentheses to indicate where the inputs should go:

( f o g)(x) = f (g(x))
    = f (–x² + 5)
    = 2(             ) + 3     ... setting up to insert the input formula
    = 2(–x² + 5) + 3
    = –2x² + 10 + 3
    = –2x² + 13

If you plug in "1" for the x in the above, you will get ( f o g)(1) = –2(1)² + 13 = –2 + 13 = 11, which is the same answer we got before. Previously, we'd plugged a number into g(x), found a new value, plugged that value into f(x), and simplified the result. This time, we plugged a formula into f(x), simplified the formula, plugged the same number in as before, and simplified the result. The final numerical answers were the same. If you've done the symbolic composition (the composition with the formulas) correctly, you'll get the same values either way, regardless of the value you pick for x. This can be a handy way of checking your work.

Here's another symbolic example:

• Given f(x) = 2x + 3 and g(x) = –x² + 5, find (g o f )(x).

(g o f )(x) = g( f(x))
    = g(2x + 3)
    = –(           )² + 5    ... setting up to insert the input
    = –(2x + 3)² + 5
    = –(4x² + 12x + 9) + 5
    = –4x² – 12x – 9 + 5
    = –4x² – 12x – 4

There is something you should note from these two symbolic examples. Look at the results I got:

( f o g)(x) = –2x² + 13
(g o f )(x) = –4x² – 12x – 4

That is, ( f o g)(x) is not the same as (g o f )(x). This is true in general; you should assume that the compositions ( f o g)(x) and (g o f )(x) are going to be different. In particular, composition is not the same thing as multiplication. The open dot "o" is not the same as a multiplication dot "•", nor does it mean the same thing. While the following is true:

f(x) • g(x) = g(x) • f(x)            [always true for multiplication]

...you cannot say that:

( f o g)(x) = (g o f )(x)           [generally false for composition]

That is, you cannot reverse the order in composition and expect to end up with the correct result. Composition is not flexible like multiplication, and is an entirely different process. Do not try to multiply functions when you are supposed to be plugging them into each other.

• Given f(x) = 2x + 3 and g(x) = –x² + 5, find ( f o f )(x).

( f o f )(x) = f ( f (x))
    = f (2x + 3)
    = 2(           ) + 3    ... setting up to insert the input
    = 2(2x + 3) + 3
    = 4x + 6 + 3
    = 4x + 9

• Given f(x) = 2x + 3 and g(x) = –x² + 5, find (g o g)(x).

(g o g)(x) = g(g(x))
    = –(             )² + 5     ... setting up to insert the input
    = –(–x² + 5)² + 5
    = –(x⁴ – 10x² + 25) + 5
    = –x⁴ + 10x² – 25 + 5
    = –x⁴ + 10x² – 20

Sometimes you have to be careful with the domain and range of the composite function.

• Given f (x) = sqrt(x) and g(x) = x – 2, find the domains of ( f o g)(x) and (g o f )(x).

Since  f (x) involves a square root, the inputs have to be non-negative. This means that the domain (the set of x-values) for f (x) is "all x > 0". Then, in (g o f )(x), where I'm plugging x first into f (x) = sqrt(x), the domain is at least restricted to "all x > 0". Let's see what the two compositions look like:   Copyright © Elizabeth Stapel 1999-2009 All Rights Reserved

( f o g)(x) =  f (g(x))
    =  f (x – 2)
    = sqrt(         )
    = sqrt(x – 2)

The domain for the square root is all inputs that make "x – 2" non-negative. That is, all x such that x – 2 > 0. Solving this for x, I get that the domain of ( f o g)(x) is "all x > 2".

Now to do the other composition:

(g o f )(x) = g( f (x))
    = g( sqrt(x) )
    = (              ) – 2
    = ( sqrt(x) ) – 2
    = sqrt(x) – 2

The domain for this is all inputs that make the square root defined. Since there is only "x" inside the square root, then the domain of (g o f )(x) is "all x > 0".

If your initial functions are just plain old polynomials, then their domains are "all x", and so will be the domain of the composition. It's pretty much only if your dealing with denominators (where you can't divide by zero) or square roots (where you can't have a negative) that the domain ever becomes an issue.

Usually composition is used to combine two functions. But sometimes you are asked to go backwards. That is, they will give you a function, and they'll ask you to come up with the two original functions that they composed. For example:

• Given h(x) = (x + 1)² + 2(x + 1) – 3, determine two functions f (x) and g(x) which, when composed, generate h(x).

This is asking you to notice patterns and to figure out what is "inside" something else. In this case, this looks similar to the quadratic x² + 2x – 3, except that, instead of squaring x, they're squaring x + 1. In other words, this is a quadratic into which they've plugged x + 1. So let's make g(x) = x + 1, and then plug this function into f (x) = x² + 2x – 3:

( f o g)(x) = f (g(x))
    = f (x + 1)
    = (         )² + 2(         ) – 3
    = (x + 1)² + 2(x + 1) – 3

Then h(x) may be stated as the composition of  f (x) = x² + 2x – 3 and g(x) = x + 1.

• Given h(x) = sqrt(4x + 1), determine two functions f (x) and g(x) which, when composed, generate h(x).

Since the square root is "on" (or "around") the "4x + 1", then the 4x + 1 is put inside the square root. I need to take x, do "4x + 1" to it, and then take the square root of the result:

g(x) = 4x + 1,  f(x) = sqrt(x), and h(x) = ( f o g)(x).

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