Suppose you are given the two functions  f (x) = 2x + 3 and g(x) = –x2 + 5. Composition means that you can plug g(x) into  f (x). This is written as "( f o g)(x)", which is pronounced as "f-compose-g of x". And "(  f o g)(x)" means " f (g(x))". That is, you plug something in for x, then you plug that value into g, simplify, and then plug the result into  f. The process here is just like what we saw on the previous page, except that now we will be using formulas to find values, rather than just reading the values from lists of points.

Given f(x) = 2x + 3 and g(x) = –x2 + 5, find (g o f )(1).

When I work with function composition, I usually convert "( f o g)(x)" to the more intuitive "  f (g(x))" form. This is not required, but I certainly find it helpful. In this case, I get:

(g o f )(1) = g( f(1))

This means that, working from right to left (or from the inside out), I am plugging x = 1 into f(x), evaluating f(x), and then plugging the result into g(x). I can do the calculations bit by bit, like this: Since f(1) = 2(1) + 3 = 2 + 3 = 5, and since g(5) = –(5)2 + 5 = –25 + 5 = –20, then (g o f )(1) = g( f(1)) = g(5) = –20. Doing the calculations all together (which will be useful later on when we're doing things symbolically), it looks like this:

(g o f )(1) = g( f (1))    = g(2(   ) + 3)   ... setting up to insert the original input    = g(2(1) + 3)    = g(2 + 3)    = g(5)    = –(   )2 + 5    ... setting up to insert the new input    = –(5)2 + 5    = –25 + 5    = –20

Note how I wrote each function's rule clearly, leaving open parentheses for where the input (x or whatever) would go. This is a useful technique. Whichever method you use (bit-by-bit or all-in-one), the answer is:

(g o f )(1) = g( f (1)) = –20

I just computed (g o f )(1); the composition can also work in the other order:

Given f(x) = 2x + 3 and g(x) = –x2 + 5, find ( f o g)(1).

First, I'll convert this to the more intuitive form, and then I'll simplify:

( f o g)(1) = f (g(1))

Working bit-by-bit, since g(1) = –(1)2 + 5 = –1 + 5 = 4, and since f(4) = 2(4) + 3 = 8 + 3 = 11, then ( f o g)(1) = f (g(1)) = f(4) = 11. On the other hand, working all-in-one (right to left, or from the inside out), I get this:

( f o g)(1) = f (g(1))     = f (–(    )2 + 5)    ... setting up to insert the original input     = f (–(1)2 + 5)     = f (–1 + 5)     = f (4)     = 2(   ) + 3     ... setting up to insert the new input     = 2(4) + 3     = 8 + 3     = 11

Either way, the answer is:   Copyright © Elizabeth Stapel 2002-2011 All Rights Reserved

( f o g)(1) = f (g(1)) = 11

A verbal note:  "f o g" is not pronounced as "fogg" and "g o f " is not pronounced as "goff". They are pronounced as "f-compose-g" and "g-compose-f ", respectively. Don't make yourself sound ignorant by pronouncing these wrongly!

As you have seen above, you can plug one function into another. You can also plug a function into itself:

Given f(x) = 2x + 3 and g(x) = –x2 + 5, find ( f o f )(1).

( f o f )(1) = f ( f (1))     = f (2(   ) + 3)    ... setting up to insert the original input     = f (2(1) + 3)     = f (2 + 3)     = f (5)     = 2(   ) + 3     ... setting up to insert the new  input     = 2(5) + 3     = 10 + 3     = 13

Given f(x) = 2x + 3 and g(x) = –x2 + 5, find (g o g)(1).

(g o g)(1) = g(g(1))     = g(–(    )2 + 5)    ... setting up to insert the original input     = g(–(1)2 + 5)

    = g(–1 + 5)     = g(4)     = –(   )2 + 5    ... setting up to insert the new  input     = –(4)2 + 5     = –16 + 5     = –11

In each of these cases, I wrote out the steps carefully, using parentheses to indicate where my input was going with respect to the formula. If it helps you to do the steps separately, then calculate g(1) outside of the other g(x) as a separate step. That is, do the calculations bit-by-bit, first finding g(1) = 4, and then plugging 4 into g(x) to get g(4) = –11.

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:   Copyright © Elizabeth Stapel 2002-2011 All Rights Reserved

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:

( 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

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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: Copyright © Elizabeth Stapel 2002-2011 All Rights

Reserved

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).

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,  find (f + g)(x), (f – g)(x), (f×g)(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

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

= –15x2 + 2x + 8

Given f(x) = 2x, g(x) = x + 4, and h(x) = 5 – x3,  find (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.