UK Secondary (7-11)
Composite Functions
Lesson

The idea behind the composition of functions is best explained with an example.

Suppose we think about the function given by $f\left(x\right)=2x+1$f(x)=2x+1. We understand that the function takes values of $x$x in the domain and maps them to values, say $y=2x+1$y=2x+1, in the range.

Suppose however that this is only the first part of a two-stage treatment of $x$x. Suppose we now take these function values and map them using another function, say $g\left(x\right)=x^2$g(x)=x2. This means that the $y$y values given by $\left(2x+1\right)$(2x+1) become the squared values $\left(2x+1\right)^2$(2x+1)2. The diagram below captures the idea.

The function values $f\left(x\right)$f(x) have become the domain values of $g\left(x\right)$g(x). Thus we could describe the complete two-stage process by the expression $g\left(f\left(x\right)\right)$g(f(x)), sometimes written $g\left[f\left(x\right)\right]$g[f(x)] and spoken of as the "gof" of $x$x

Algebraically, we can write $g\left(f\left(x\right)\right)=g\left[2x+1\right]=\left(2x+1\right)^2$g(f(x))=g[2x+1]=(2x+1)2

Note that if we reversed the order of the two-stage processing, we would, in this instance, develop a different composite function.

Thus $f\left(g\left(x\right)\right)=f\left(x^2\right)=2\left(x^2\right)+1=2x^2+1$f(g(x))=f(x2)=2(x2)+1=2x2+1. This is known as the "fog" of $x$x.

## The domain of Composite functions

One very important point needs to be made here in terms of the domain of composite functions.

If we consider, say $g\left(f\left(x\right)\right)$g(f(x)), then range of the function $f\left(x\right)$f(x) (which is the function applied first) must be a subset of the domain of the function $g\left(x\right)$g(x) (the function that is applied second).

That is to say, the domain of $g\left(f\left(x\right)\right)$g(f(x)) must only consist of elements that can be mapped by both $f\left(x\right)$f(x) and then by $g\left(x\right)$g(x) without causing an issue in either function.

A similar situation applies for  $f\left(g\left(x\right)\right)$f(g(x)).

For example, if $f\left(x\right)=\sqrt{x}$f(x)=x and $g\left(x\right)=\frac{1}{x}$g(x)=1x, then both $f\left(g\left(x\right)\right)$f(g(x)) and $g\left(f\left(x\right)\right)$g(f(x)) have the restricted domain $x\in R^+$xR+. Lets explain why.

If we consider $g\left(f\left(x\right)\right)$g(f(x)), the range of $f\left(x\right)$f(x) includes non-negative real numbers, but we can't use all of these in $g\left(x\right)$g(x). The number zero needs to be deleted.

If we consider $f\left(g\left(x\right)\right)$f(g(x)), the range of $g\left(x\right)$g(x) includes all real numbers other than zero, but only positive reals can be considered for $f\left(x\right)$f(x)

Taking this last example, we have $f\left(g\left(x\right)\right)=f\left(\frac{1}{x}\right)=\sqrt{\frac{1}{x}}=\frac{1}{\sqrt{x}}$f(g(x))=f(1x)=1x=1x and $g\left(f\left(x\right)\right)=g\left(\sqrt{x}\right)=\frac{1}{\sqrt{x}}$g(f(x))=g(x)=1x and so in this instance the fog and the gof are equal.

The graph of the composition is shown here. Note that the composition is graphed for positive reals only.

#### Worked examples

##### Question 1

If $f\left(x\right)=4x+4$f(x)=4x+4,

1. find $f\left(2\right)$f(2).

2. find $f\left(-5\right)$f(5).

##### Question 2

Consider the functions $f\left(x\right)=-2x-3$f(x)=2x3 and $g\left(x\right)=-2x-6$g(x)=2x6.

1. Find $f\left(7\right)$f(7).

2. Hence, or otherwise, evaluate $g\left(f\left(7\right)\right)$g(f(7)).

3. Now find $g\left(7\right)$g(7).

4. Hence, evaluate $f\left(g\left(7\right)\right)$f(g(7)).

5. Is it true that $f\left(g\left(x\right)\right)=g\left(f\left(x\right)\right)$f(g(x))=g(f(x)) for all $x$x?

Yes

A

No

B

Yes

A

No

B

##### Question 3

Consider the functions $f\left(x\right)=-2x+6$f(x)=2x+6 and $g\left(x\right)=3x+1$g(x)=3x+1.

1. The function $r\left(x\right)$r(x) is defined as $r\left(x\right)=f\left(x^2\right)$r(x)=f(x2). Define $r\left(x\right)$r(x).

2. Using the results of the previous part, define $q\left(x\right)$q(x), which is $g\left(f\left(x^2\right)\right)$g(f(x2)).