Topics
8. Modeling with Functions
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United States of America
Grades 9-12

8.02 Combining functions

Lesson
Practice

Introduction

Just as we can perform operations on polynomials, we can also perform operations on different functions by adding, subtracting, or multiplying them provided we follow specific rules. We will use our tools to extend to different functions.

Combining functions

Operations with functions are defined using special notation:

  • Sum: \left(f+g\right)\left(x\right)=f\left(x\right)+g\left(x\right)

  • Difference: \left(f-g\right)\left(x\right)=f\left(x\right)-g\left(x\right)

  • Product: \left(f\cdot g\right)\left(x\right)=f\left(x\right)\cdot g\left(x\right)

With each operation, the domain of the new function becomes the intersection or overlap of the domains of the original functions.

In addition to the ways in which we can combine two functions, as above, 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

A function created when one function is substituted into another function

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.

Examples

Example 1

Consider the following pair of functions:

\begin{aligned} f\left(x\right) & = -5x+5\\\ g\left(x\right) & = 2x^2+3x-10 \end{aligned}

a

Find \left(f+g\right)\left(x\right)

Worked Solution
Create a strategy

To find \left(f+g\right)\left(x\right), we want to add the two functions together.

Apply the idea
\displaystyle \left(f+g \right)\left(x\right)\displaystyle =\displaystyle f\left(x\right)+g\left(x\right)
\displaystyle =\displaystyle \left(-5x+5\right)+\left(2x^2+3x-10\right)Substitute f\left(x\right), g\left(x\right)
\displaystyle =\displaystyle 2x^2-2x-5Combine like terms
b

Find \left(f-g\right)\left(x\right)

Worked Solution
Create a strategy

To find \left(f-g\right)\left(x\right) we want to subtract g\left(x\right) from f\left(x\right).

Apply the idea
\displaystyle \left(f-g \right)\left(x\right)\displaystyle =\displaystyle f\left(x\right)-g\left(x\right)
\displaystyle =\displaystyle \left(-5x+5\right)-\left(2x^2+3x-10\right)Substitute f\left(x\right), g\left(x\right)
\displaystyle =\displaystyle -2x^2-8x+15Distribute and combine like terms
c

Find \left(f \cdot g\right)\left(x\right)

Worked Solution
Create a strategy

To find \left(f \cdot g\right)\left(x\right) we want to find the product of the two functions.

Apply the idea
\displaystyle \left(f \cdot g \right)\left(x\right)\displaystyle =\displaystyle f\left(x\right) \cdot g\left(x\right)
\displaystyle =\displaystyle \left(-5x+5\right)\left(2x^2+3x-10\right)Substitute f\left(x\right), g\left(x\right)
\displaystyle =\displaystyle -5x\left(2x^2\right) -5x\left(3x\right) -5x\left(-10\right)+5\left(2x^2\right)+5\left(3x\right)+5\left(-10\right)Distribute the parentheses
\displaystyle =\displaystyle -10x^3 -15x^2 + 50x + 10x^2 + 15x - 50Evaluate the products
\displaystyle =\displaystyle -10x^3-5x^2+65x-50Combine like terms
d

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

Worked Solution
Create a strategy

To find \left(f \circ g\right)\left(x\right), we need to use g\left(x\right) as the input of f\left(x\right).

Apply the idea
\displaystyle \left(f \circ g\right)\left(x\right)\displaystyle =\displaystyle f\left(g\left(x\right)\right)Definition of function composition
\displaystyle =\displaystyle f\left(2x^2+3x-10\right)Substitute g\left(x\right)

The notation f\left(2x^2+3x-10\right) means we need to replace the independent variable in f\left(x\right) with 2x^2+3x-10.

\displaystyle f\left(x\right)\displaystyle =\displaystyle -5x+5Original function, f\left(x\right)
\displaystyle f\left(2x^2+3x-10\right)\displaystyle =\displaystyle -5\left(2x^2+3x-10\right)+5Substitute \left(2x^2+3x-10\right)
\displaystyle =\displaystyle -10x^2-15x+50+5Distribute -5
\displaystyle =\displaystyle -10x^2-15x+55Evaluate the addition

Therefore, \left(f\circ g\right)=-10x^2-15x+55.

Example 2

The projected birth rate of a city's population, in ten-thousands, is b\left(t\right) = 3t^2 and the projected death rate, in ten-thousands, is d\left(t\right) = t, for each year, t. The graph below shows the functions:

Population birth and death rates
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\text{Time}
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\text{Population (ten-thousands)}
a

Create a function equation to represent the net rate of population change, due to births and deaths, as a function of time t. Then, graph the function.

Worked Solution
Create a strategy

The net rate will be the difference between the birth and death rates, so we need to find n\left(t\right)=b\left(t\right) - d\left(t\right), of the population at time t and graph it.

Apply the idea

The net rate of population change is given by the equation n\left(t\right) = 3t^2 - t, where t represents the number of years and n represents the net population in ten-thousands.

Net population birth and death rates
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t
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n\left(t\right)
b

Find n\left(5\right) and explain what it tells us about the town's population.

Worked Solution
Create a strategy

We can use the graph to find the population, n, when t=5.

Apply the idea

When t=5 on the graph, n=70. This means the city's net change in population is projected to be an additional 700\,000 people in 5 years.

Reflect and check

We also could have substituted t=5 into the equation and evaluated to find n.

Example 3

The price of a stock at the end of the nth trading day is given by S\left(n\right) = \dfrac{1}{n+1}. A trader decides to purchase stocks, and the number of stocks they own at the end of the nth trading day is approximated by B\left(n\right) = \left(n + 1\right)^2.

a

Find the equation for the value, V\left(n\right), of the trader’s stocks at day n.

Worked Solution
Create a strategy

The value of the stocks at day n can be calculated by multiplying the price of the stock by the number of stocks. So, we need to find V\left(n\right)= \left(S \cdot B \right)\left(n\right).

Apply the idea
\displaystyle V\left(n\right)\displaystyle =\displaystyle S\left(n\right)\cdot B\left(n\right)Original equation
\displaystyle =\displaystyle \dfrac{1}{n+1}\cdot\left(n + 1\right)^2Substitution
\displaystyle =\displaystyle n+1Multiplicative inverse

Since \dfrac{n+1}{n+1}=1 only when n\neq -1, we cannot include n=-1 the domain. But since n represents days, n=-1 would not make sense in context. Therefore, V\left(n\right)=n+1 for n\geq 0.

b

Sketch the graph of V\left(n\right).

Worked Solution
Apply the idea
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n
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V
c

Explain what the new function tells us about the stocks.

Worked Solution
Create a strategy

Consider the meaning of the variables and their units and the patterns of change in the function.

Apply the idea

For every n days, the value of the total stock owned by the trader will increase at a constant rate of 1 dollar a day.

Example 4

A cylindrical tank initially contains 200 \text{ in}^3 of grain and starts being filled at a constant rate of 40 \text{ 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.

Worked Solution
Create a strategy

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 of grain

  • h\left(g\right) represents the height of the grain in terms of g

  • 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).

Apply the idea
\displaystyle V\displaystyle =\displaystyle \pi r^2 hVolume of a cylinder
\displaystyle g\displaystyle =\displaystyle \pi \left(12\right)^2 \cdot h\left(g\right)Substituting V=g, r=12, and h=h\left(g\right)
\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 the volume of grain g.

b

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

Worked Solution
Create a strategy

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

Apply the idea
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.

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.

Worked Solution
Create a strategy

\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}.

Apply the idea
\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
Reflect and check

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)}{144\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.

Worked Solution
Create a strategy

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 \circ g\right)\left(t\right). This represents height as a function of time.

Apply the idea

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

e

If the barrel can hold 10\,000 \text{ in}^3 of grain, determine the domains of g\left(t\right), h\left(g\right) and A\left(t\right).

Worked Solution
Create a strategy

The lower boundary of the domain of g\left(t\right) is 0 as the time starts at 0 seconds. This means the lower boundary of the domain of A\left(t\right) is also 0 seconds.

To calculate the upper boundaries, we can use the fact that the barrel can hold a maximum of 10\, 000 \text{ in}^3 of grain. The time it takes to fill the barrel will be the upper boundary of both g\left(t\right) and A\left(t\right).

As g is the input for h\left(g\right), the range of g\left(t\right) will be the domain of h\left(g\right). So, the lower boundary of h\left(g\right) will be the amount of grain in the barrel initially, and the upper amount will be the maximum amount of grain the barrel can hold.

Apply the idea

The lower boundary of h\left(g\right) is 200 \text{ in}^3 as this is how much is in the barrel initially, and the upper boundary is 10\, 000 \text{ in}^3 as this is the maximum amount of grain the barrel can hold.

Calculating the total amount of time needed to fill the barrel:

\displaystyle g\left(t\right)\displaystyle =\displaystyle 40t+200
\displaystyle 10\,000\displaystyle =\displaystyle 40t+200Substitute g\left(t\right)=10\,000
\displaystyle 9800\displaystyle =\displaystyle 40tSubtract 200 from both sides
\displaystyle 245\displaystyle =\displaystyle tDivide both sides by 40

This means the barrel will be completely full after 245 seconds. The domain of both g\left(t\right) and A\left(t\right) is \left[0, 245 \right].

  • Domain of g\left(t\right): \left[0, 245\right]
  • Domain of h\left(g\right): \left[200, 10\,000\right]
  • Domain of A\left(t\right): \left[0, 245\right]
Idea summary

Operations with functions are defined using special notation:

  • Sum: \left(f+g\right)\left(x\right)=f\left(x\right)+g\left(x\right)

  • Difference: \left(f-g\right)\left(x\right)=f\left(x\right)-g\left(x\right)

  • Product: \left(f\cdot g\right)\left(x\right)=f\left(x\right)\cdot g\left(x\right)

  • Composition:\left(f \circ g\right)\left(x\right)= f\left(g\left(x\right)\right)

Outcomes

F.BF.A.1.B

Combine standard function types using arithmetic operations.

F.BF.A.1.C (+)

Compose functions.

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