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2. Radical Functions
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United States of AmericaTennessee
Integrated Math 3

2.03 Composition of functions

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Lesson

Concept summary

We can also create a composite function using an operation that combines two functions f and g and produces a function h such that h\left(x\right)=g\left(f\left(x\right)\right), where the function g is applied to the result of applying the function f to x

The output, or function values, of the function f\left(x\right) have become the input, or x-values, of the function g\left(x\right). We introduce a new symbol \circ to represent this new function.

Composite function

g\left(f\left(x\right)\right)=\left(g\circ f\right)\left(x\right)

In a composition of functions, the inner function is evaluated first, followed by the outer function. For example in the composition g\left(f\left(x\right)\right), the function f is applied first, followed by the function g. This means that \left(g \circ f\right)\left(x\right) is not necessarily equal to \left(f \circ g\right)\left(x\right).

The domain of \left(g \circ f\right)\left(x\right) is restricted to all x-values in the domain of f whose range values, f\left(x\right), are in the domain of g.

Worked examples

Example 1

The table shows some of the outputs of the functions f\left(x\right) and g\left(x\right).

Use the table to evaluate the following:

xf\left(x\right)g\left(x\right)
0-28
157
2124
319-1
426-8
533-17
a

\left(g \circ f\right)\left(1\right)

Approach

To evaluate \left(g \circ f\right)\left(1\right) we want to find g\left(f\left(x\right)\right).

We can see, from the table, that f\left(1\right)=5. We now want to find g\left(f\left(1\right)\right)=g\left(5\right).

Solution

From the table we can see that g\left(5\right)=-17.

Therefore \left(g \circ f\right)\left(1\right)=-17.

b

\left(f \circ g\right)\left(2\right)

Approach

To evaluate \left(f \circ g\right)\left(2\right) we want to find f\left(g\left(x\right)\right).

We can see, from the table, that g\left(2\right)=4. We now want to find f\left(g\left(2\right)\right)=f\left(4\right).

Solution

From the table we can see that f\left(4\right)=26.

Therefore \left(f \circ g\right)\left(2\right)=26.

Example 2

A cylindrical tank initially contains 200 in^3 of grain and starts being filled at a constant rate of 40 in^3 per second.

The radius of the tank is 12 inches. Let g be the amount of grain in the container after t seconds.

a

State the function for h\left(g\right), the height of the grain in the container, in terms of g.

Approach

As the tank fills with grain, the amount of grain takes the shape of a cylinder which has a volume given by V=\pi r^2h.

We know that g represents the volume, h\left(g\right) represents the height of the grain in terms of g and r is given to be 12 inches. Substituting these values into the volume of a cylinder, V=\pi r^2h, we can form an equation relating g and h\left(g\right).

Solution

\displaystyle g\displaystyle =\displaystyle \pi r^2 \cdot h\left(g\right)Setting up an equation
\displaystyle g\displaystyle =\displaystyle \pi\left(12^2\right) \cdot h\left(g\right)Substituting r=12
\displaystyle g\displaystyle =\displaystyle 144 \pi \cdot h\left(g\right)Evaluating the square
\displaystyle \dfrac{g}{144\pi}\displaystyle =\displaystyle h\left(g\right)Divide both sides by 144\pi
\displaystyle h\left(g\right)\displaystyle =\displaystyle \dfrac{g}{144\pi}Symmetric property of equality

The function h\left(g\right)=\dfrac{g}{144\pi} represents the height of the grain in the container, in terms of g.

b

State the function for g\left(t\right), the amount of grain in the tank after t seconds.

Approach

We know that initially, t=0, there are 200 in^3 of grain in the tank. Each second that passes, 40 in^3 is added.

t0123
g\left(t\right)200240280320

Creating a table of values, we can see that we have a linear equation where the amount of grain is equal to 200 plus 40 for every second that passes.

Solution

The function g\left(t\right)=40t+200

c

The function A\left(t\right) is defined as A\left(t\right)=\left( h \circ g \right)\left(t\right). Form an equation for A\left(t\right) in terms of t.

Approach

\left(h \circ g\right)\left(t\right) is the same as h\left(g\left(t\right)\right), so want to substitute g\left(t\right)=40t+200 into the function h\left(g\right)=\dfrac{g}{144\pi}.

Solution

\displaystyle A\left(t\right)\displaystyle =\displaystyle \left( h \circ g \right)\left(t\right)
\displaystyle =\displaystyle h\left(g\left(t\right)\right)Definition of \left( h \circ g \right)\left(t\right)
\displaystyle =\displaystyle h\left(40t+200\right)Substitute g\left(t\right)
\displaystyle =\displaystyle \dfrac{40t+200}{144\pi}Substitute h\left(g\right)
\displaystyle =\displaystyle \dfrac{5t+25}{18\pi}Simplifying the quotient

Reflection

In the working above we substituted g\left(t\right)=40t+200 into the function for h. We can also obtain the same answer by first substituting h\left(g\right)=\dfrac{g}{144\pi} into h\left(g\left(t\right)\right):

\displaystyle A\left(t\right)\displaystyle =\displaystyle h\left(g\left(t\right)\right)
\displaystyle =\displaystyle \dfrac{g\left(t\right)}{16\pi}Substitute h\left(g\right)
\displaystyle =\displaystyle \dfrac{40t+200}{144\pi}Substitute g\left(t\right)
\displaystyle =\displaystyle \dfrac{5t+25}{18\pi}Simplifying the quotient
d

Explain what A\left(t \right) represents.

Approach

g\left(t\right) represents the amount of grain in the container after t seconds, and h\left(g\right) represents the height of grain in terms of the amount of grain. Composing the two gives us \left(h \cdot g\right)\left(t\right). This represents height as a function of time.

Solution

A\left(t \right) represents the height of the grain in the container, in inches, after t seconds.

Outcomes

M3.F.BF.A.1

Build a function that describes a relationship between two quantities.*

M3.F.BF.A.1.A

Combine standard function types using composition.

M3.MP1

Make sense of problems and persevere in solving them.

M3.MP3

Construct viable arguments and critique the reasoning of others.

M3.MP6

Attend to precision.

M3.MP8

Look for and express regularity in repeated reasoning.

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