Lesson 10

Combining Functions

  • Let’s make some new functions using other functions.

10.1: Notice and Wonder: Are Book Sales Improving?

What do you notice? What do you wonder?

\(t\) (years since 2010) number of books sold
in the US (millions)		 population of
the US (millions)
0 2,530 309.35
1 2,400 311.64
2 2,730 313.99
3 2,720 316.23
4 2,700 318.62
5 2,710 321.04
6 2,700 323.41

 

10.2: How Many Books Can One Person Have?

The table shows the values of two functions, \(P\) and \(B\), where \(P(t)\) is the population of the US, in millions, \(t\) years after 2010, and \(B(t)\) is the number of books sold per year, in millions, \(t\) years after 2010.

\(t\) (years since 2010) \(B(t)\) (millions) \(P(t)\) (millions) \(R(t)\)
0 2,530 309.35                  
1 2,400 311.64  
2 2,730 313.99  
3 2,720 316.23  
4 2,700 318.62  
5 2,710 321.04  
6 2,700 323.41  

  1. Plot the values of \(B\) as a function of \(t\). What does the plot tell you about book sales?

    Empty coordinate plane. Horizontal axis from 0 to 6, time in years since 2010. Vertical axis from 2,350 to 2750, number of books sold in millions.
  2. How many books were sold per person in 2010 and 2016? What do these values tell you about book sales?
  3. Define a new function \(R\) by \(R(t) = \frac{B(t)}{P(t)}\). Complete the table and then graph the values of \(R(t)\). What do the values of \(R\) tell you?

10.3: Adding Functions

  1. Here are the graphs of two functions, \(E\) and \(L\). Define a new function \(S\) by adding \(E\) and \(L\), so \(S(x) = E(x) + L(x)\). On the same axes, sketch what you think the graph of \(S\) looks like.

    2 functions, E and L, graphed on coordinate plane. Axes from negative 6 to 6. E is a decreasing exponential with y-intercept of 1. L is a line with positive slope and y-intercept of 0.
  2. Sketch the graph of the sum of \(E\) and each of the following functions.
    2 functions, E and M, graphed on coordinate plane. Axes from negative 6 to 6. E is a decreasing exponential with y-intercept of 1. M is a horizontal line with y-intercept of negative 1.
    2 functions, E and N, graphed on coordinate plane. Axes from negative 6 to 6. E is a decreasing exponential with y-intercept of 1. N is a line with negative slope and y-intercept of 0.
    2 functions, E and P, on coordinate plane. Axes from negative 6 to 6. E is a decreasing exponential with y-intercept of 1. P is a parabola that opens up with y-intercept of 0.
    2 functions, E and Q, on coordinate plane. Axes from negative 6 to 6. E is a decreasing exponential with y-intercept of 1. Q is a parabola that opens down with y-intercept of 0.


Here are the graphs of two functions, \(U\) and \(V\). Define a new function \(W\) by multiplying \(U\) and \(V\), so \(W(x) = U(x) V(x)\). On the same axes, sketch what you think the graph of \(W\) looks like.

Intersecting Lines U and V. Line U, y intercept =2, slope =-1. Line V, y intercept =-1, slope =1.

 

Summary

We can add, subtract, multiply, and divide functions to get new functions. For example, the cost in dollars of producing \(n\) cups of lemonade at a lemonade stand is \(C(n) = 5 + 0.8n\). The revenue (amount of money collected) from selling \(n\) cups is \(R(n) = 2n\) dollars. The profit \(P(n)\) from selling \(n\) cups is the revenue minus the cost, so

\(\displaystyle P(n) = R(n) - C(n) = 2n - (5 + 0.8n) = 1.2n - 5\)

Here are the graphs of \(C\), \(R\), and \(P\). Can you see how each value on \(P\) is the result of the difference between the corresponding points on \(R\) and \(C\)?

The average profit per cup, \(A(n)\), from selling \(n\) cups, is the quotient of the profit and the number of cups, so

\(\displaystyle A(n) = \frac{P(n)}{n} = \frac{1.2n - 5}{n} = 1.2 - \frac5n\)

Graph of three lines.

Here are the graphs of \(P\) and \(A\). Can you see how the value of \(A(n)\) is the result of the quotient of \(P(n)\) and \(n\)? Why does it make sense that both functions are negative when \(n<4\frac16\) and positive when \(n>4\frac16\)?

Graph of 2 lines.

Since \(n\) can only be positive, \(P(n)\) and \(A(n)\) always have the same sign for a given \(n\) value. Notice that for the average profit to be positive, the seller has to sell at least 5 cups (since \(4\frac16\) is not in the domain, we must round up). It is also true that for a large number of cups, the average profit is close to \$1.20 per cup.