Putting functions together to make new ones

To compose two functions is to put one inside the other. We already know that f(x) means to take x and follow the instructions in f(x) for what to do with it to get the corresponding y-value. But what if we replace x with another function?

Let's do this by a simple example. Let f(x) = x + 2 and g(x) = x².

Example 1

Given functions   $f(x) = x + 1$   and   $g(x) = 3x - 1,$   find the composite functions   $f(g(x))$   and   $g(f(x))$.

Solutions: We start with our two original functions

$$f(x) = x + 1 \;\;\;\; g(x) = 3x - 1$$

For f(g(x)), we're just plugging in the definition of g(x) wherever there is an x in the definition of f(x), like this.

$$f(g(x)) = f(3x - 1)$$

Now just put in the rest of f(x):

$$f(g(x)) = (3x - 1) + 1$$

and simplify (1 - 1 = 0) to get the result:

$$f(g(x)) = 3x$$

Example 2

Now let's look at some more complicated functions. If   $f(x) = x^2 + 1$   and   $g(x) = -2x - 1$, find the composite functions   $f(g(x)), \, g(f(x)), \, f(f(x)), \; and \; g(g(x)).$   The last two are sometimes called self-referential functions, or recursive functions. They can be very important, especially in computer programming.

Solutions: Here's   $f(g(x))$

$$ \begin{align} f(g(x)) &= (-2x - 1)^2 + 1 \\ &= (4x^2 + 4x + 1) + 1 \\ &= 4x^2 + 4x + 2 \end{align}$$

$g(f(x))$   goes like this:

$$ \begin{align} g(f(x)) &= -2 (x^2 + 1) - 1 \\ &= -2x^2 - 2 - 1 \\ &= -2x^2 - 3 \end{align}$$

Example 3

The number of widgets* (n) produced by a factory after t hours of operation is modeled by the function  $n(t) = 1000t - 8t^2$. If the cost (C, in dollars) of producing n widgets is $C(n) = 10,000 + 350n$, find the cost C(t) as a function of the time of factory operation.

Solution: What we're trying to do here is make C(t) from C(n) and n(t), which means finding the composition C(t) = C(n(t)).

If we take C(n) and plug the function n(t) in everywhere we see an n, we get:

$$ \begin{align} C(t) &= 10,000 + 350(1000t - 8t^2) \\ &= 10,000 + 350,000t - 2800t^2 \end{align}$$

Example 4

A conical water tank has a radius of 10 meters and a height of 12 m, as shown below. If water is flowing into the tank at a rate of 0.5 m³ per hour, find a function for the height of the water in the tank over time. That is, find h(t) from V(t) and the known volume of a cone.

Solution: Here's a picture of what's going on:

The volume of a cone is

Now the problem with that definition of a cone is that it is a function of two variables, r and h. Because we eventually want a function h(t), we'll need to 'get rid' of r, which means we'll need a piece of information relating r to h. Luckily we have one.

Notice that the cone and the inner cone of liquid are similar, exactly in the sense of the similar triangles you learned in geometry, so the ratio of their radii and heights are the same. So we have:

$$\frac{H}{R} = \frac{h}{r} \; \longrightarrow \; r = \frac{Rh}{h}$$

Rearranging gives us r(h),

$$r = \frac{10 h}{12} = \frac{5h}{6}$$

We'll need r², which is

$$r^2 = \frac{25 h^2}{36}$$

Practice problems

Here are some simple problems to practice putting one function inside of another. Practice these to get the hang of it. Given the functions f(x), g(x), h(x) and k(x) below, find each of the values or compositions indicated in the problems that follow.