MAT 142: Composition of Functions

Reminders

At our last meeting, we discussed each of the following:

The limit of a function as a generalization of function evaluation.
Reading limits from a graph, including one-sided limits.
Evaluating limits algebraically via direct substitution and the factor-and-reduce technique, when possible.
The slope of the secant line as an average rate of change.
The derivative as the limit of the difference quotient.

Try the following warm-up problems.

Problem 1: Consider \(f\left(x\right) = 8 - x^2\). Find \(f\left(0\right)\), \(f\left(5\right)\), \(f\left(-2\right)\), \(f\left(x + h\right)\), and \(f\left(x^2 - 5\right)\).

Problem 2: Consider \(\displaystyle{g\left(x\right) = \frac{x^2 + x}{x + 1}}\). Find \(g\left(0\right)\), \(g\left(5\right)\), \(g\left(-2\right)\), \(g\left(x + h\right)\), and \(g\left(2 - x\right)\).

Motivating Application

A forest fire leaves behind an area of burned grass in an expanding circular pattern. The radius of the circle is increasing with time according to

\[r\left(t\right) = 2t + 1\]

where \(t\) is in minutes and \(r\) is in feet.

Express the area burned as a function of time \(t\). Find the total area burned after 10 minutes.

Before we work through this, consider the following:

What is the formula for the burned area in terms of the radius?
Is the radius truly the independent variable for the area function?
What is the radius of the burn after 5 minutes? What is the area burned at that point?

We’ll return to this application at the end of today’s class.

Objectives

Today we introduce a powerful new way to combine functions.

After today’s class meeting, you should be able to:

Understand what it means to compose two functions.
Construct the algebraic definition of a composed function \(\left(f \circ g\right)\left(x\right)\).
Evaluate a composed function at a given input value.
Read composed function values from graphs.
Recognize when composition arises naturally in applied settings.

Composition of Functions

Definition (Composition of Functions): Given two functions \(f\left(x\right)\) and \(g\left(x\right)\), as long as the range of \(g\) is a subset of the domain of \(f\), we define the composition of \(f\) and \(g\) as:

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

The input \(x\) is first transformed by \(g\), and then the output of \(g\) becomes the input for \(f\).

Think of it as a two-stage machine:

\[x \longrightarrow \boxed{~~g~~} \longrightarrow g\left(x\right) \longrightarrow \boxed{~~f~~} \longrightarrow f\left(g\left(x\right)\right)\]

Order of Composition Matters

Order matters. \(\left(f \circ g\right)\left(x\right)\) and \(\left(g \circ f\right)\left(x\right)\) are generally not equal – applying \(g\) first then \(f\) is different from applying \(f\) first then \(g\).

Completed Example 1

Problem: Consider \(f\left(x\right) = x + 4\) and \(g\left(x\right) = x^2\).

\(\left(a\right)\) Determine \(\left(f \circ g\right)\left(-7\right)\) by first computing \(g\left(-7\right)\), then using the result as the input to \(f\).

\(\left(b\right)\) Find the algebraic definition of \(\left(f \circ g\right)\left(x\right)\) by simplifying \(f\left(g\left(x\right)\right)\), then evaluate \(\left(f \circ g\right)\left(-7\right)\) directly.

Solution to \(\left(a\right)\). First, evaluate \(g\) at \(-7\):

\[\begin{align} g\left(-7\right) &= \left(-7\right)^2 = 49 \end{align}\]

Now use \(49\) as the input to \(f\):

\[\begin{align} f\left(49\right) &= 49 + 4 = \boxed{~53~} \end{align}\]

Completed Example 1 (Cont’d)

Problem: Consider \(f\left(x\right) = x + 4\) and \(g\left(x\right) = x^2\). Find \(\left(f \circ g\right)\left(x\right)\) algebraically, then evaluate at \(x = -7\).

Solution to \(\left(b\right)\). Substitute \(g\left(x\right)\) in place of every \(x\) in \(f\left(x\right)\):

\[\begin{align} \left(f \circ g\right)\left(x\right) &= f\left(g\left(x\right)\right) \end{align}\]

Completed Example 1 (Cont’d)

Problem: Consider \(f\left(x\right) = x + 4\) and \(g\left(x\right) = x^2\). Find \(\left(f \circ g\right)\left(x\right)\) algebraically, then evaluate at \(x = -7\).

Solution to \(\left(b\right)\). Substitute \(g\left(x\right)\) in place of every \(x\) in \(f\left(x\right)\):

\[\begin{align} \left(f \circ g\right)\left(x\right) &= f\left(g\left(x\right)\right)\\ &= f\left(x^2\right) \end{align}\]

Completed Example 1 (Cont’d)

Problem: Consider \(f\left(x\right) = x + 4\) and \(g\left(x\right) = x^2\). Find \(\left(f \circ g\right)\left(x\right)\) algebraically, then evaluate at \(x = -7\).

Solution to \(\left(b\right)\). Substitute \(g\left(x\right)\) in place of every \(x\) in \(f\left(x\right)\):

\[\begin{align} \left(f \circ g\right)\left(x\right) &= f\left(g\left(x\right)\right)\\ &= f\left(x^2\right)\\ &= x^2 + 4 \end{align}\]

Completed Example 1 (Cont’d)

Problem: Consider \(f\left(x\right) = x + 4\) and \(g\left(x\right) = x^2\). Find \(\left(f \circ g\right)\left(x\right)\) algebraically, then evaluate at \(x = -7\).

Solution to \(\left(b\right)\). Substitute \(g\left(x\right)\) in place of every \(x\) in \(f\left(x\right)\):

\[\begin{align} \left(f \circ g\right)\left(x\right) &= f\left(g\left(x\right)\right)\\ &= f\left(x^2\right)\\ &= x^2 + 4 \end{align}\]

Now evaluate at \(x = -7\):

\[\begin{align} \left(f \circ g\right)\left(-7\right) &= \left(-7\right)^2 + 4 = 49 + 4 = \boxed{~53~} \end{align}\]

Both methods give the same result – this will always be the case.

Composition of Functions Practice I

Try It! 1: Using the same functions \(f\left(x\right) = x + 4\) and \(g\left(x\right) = x^2\), evaluate \(\left(g \circ f\right)\left(-7\right)\) using both methods.

What do you notice? What does this tell us about the order in which we compose functions?

Reminder for method 1: Start by evaluating \(f\left(-7\right)\), then use that result as the input to \(g\).

Reminder for method 2: Find \(\left(g \circ f\right)\left(x\right) = g\left(f\left(x\right)\right)\) algebraically first, then plug in \(-7\).

Completed Example 2

Problem: Consider \(\displaystyle{f\left(x\right) = \frac{x}{x - 5}}\) and \(g\left(x\right) = x^2 - 5\).

Construct \(\left(f \circ g\right)\left(x\right)\) and evaluate it at \(x = 0\), \(x = 5\), and \(x = \sqrt{5}\).

Solution. We substitute \(g\left(x\right)\) into \(f\):

\[\begin{align} \left(f \circ g\right)\left(x\right) &= f\left(g\left(x\right)\right) \end{align}\]

Completed Example 2

Problem: Consider \(\displaystyle{f\left(x\right) = \frac{x}{x - 5}}\) and \(g\left(x\right) = x^2 - 5\).

Construct \(\left(f \circ g\right)\left(x\right)\) and evaluate it at \(x = 0\), \(x = 5\), and \(x = \sqrt{5}\).

Solution. We substitute \(g\left(x\right)\) into \(f\):

\[\begin{align} \left(f \circ g\right)\left(x\right) &= f\left(g\left(x\right)\right)\\ &= f\left(x^2 - 5\right) \end{align}\]

Completed Example 2

Problem: Consider \(\displaystyle{f\left(x\right) = \frac{x}{x - 5}}\) and \(g\left(x\right) = x^2 - 5\).

Construct \(\left(f \circ g\right)\left(x\right)\) and evaluate it at \(x = 0\), \(x = 5\), and \(x = \sqrt{5}\).

Solution. We substitute \(g\left(x\right)\) into \(f\):

\[\begin{align} \left(f \circ g\right)\left(x\right) &= f\left(g\left(x\right)\right)\\ &= f\left(x^2 - 5\right)\\ &= \frac{x^2 - 5}{\left(x^2 - 5\right) - 5} \end{align}\]

Completed Example 2

Problem: Consider \(\displaystyle{f\left(x\right) = \frac{x}{x - 5}}\) and \(g\left(x\right) = x^2 - 5\).

Construct \(\left(f \circ g\right)\left(x\right)\) and evaluate it at \(x = 0\), \(x = 5\), and \(x = \sqrt{5}\).

Solution. We substitute \(g\left(x\right)\) into \(f\):

\[\begin{align} \left(f \circ g\right)\left(x\right) &= f\left(g\left(x\right)\right)\\ &= f\left(x^2 - 5\right)\\ &= \frac{x^2 - 5}{\left(x^2 - 5\right) - 5}\\ &= \frac{x^2 - 5}{x^2 - 10} \end{align}\]

Completed Example 2

Problem: Consider \(\displaystyle{f\left(x\right) = \frac{x}{x - 5}}\) and \(g\left(x\right) = x^2 - 5\).

Construct \(\left(f \circ g\right)\left(x\right)\) and evaluate it at \(x = 0\), \(x = 5\), and \(x = \sqrt{5}\).

Solution.

\[\left(f \circ g\right)\left(x\right) = \frac{x^2 - 5}{x^2 - 10}\]

Now evaluate at each input:

\[\begin{align} \left(f \circ g\right)\left(0\right) &= \frac{0 - 5}{0 - 10} = \frac{-5}{-10} = \boxed{~\frac{1}{2}~} \end{align}\]

\[\begin{align} \left(f \circ g\right)\left(5\right) &= \frac{25 - 5}{25 - 10} = \frac{20}{15} = \boxed{~\frac{4}{3}~} \end{align}\]

\[\begin{align} \left(f \circ g\right)\left(\sqrt{5}\right) &= \frac{5 - 5}{5 - 10} = \frac{0}{-5} = \boxed{~0~} \end{align}\]

Composition of Functions Practice II

Try It! 2: Consider the functions \(f\left(x\right) = 3x + 8\), \(\displaystyle{g\left(x\right) = \frac{x - 8}{3}}\), and \(\displaystyle{h\left(x\right) = x^2 - 5}\).

Construct \(\left(f\circ g\right)\left(x\right)\), and use it to evaluate:
- \(\left(f\circ g\right)\left(0\right)\)
- \(\left(f\circ g\right)\left(-8\right)\)
- \(\left(f\circ g\right)\left(\sqrt{2}\right)\)
Construct \(\left(g\circ f\right)\left(x\right)\), and use it to evaluate:
- \(\left(g\circ f\right)\left(-12\right)\)
- \(\left(g\circ f\right)\left(\frac{3}{4}\right)\)
- \(\left(g\circ f\right)\left(\pi\right)\)
What does this tell you about the functions \(f\) and \(g\)?

Construct \(\left(f\circ h\right)\left(x\right)\), and use it to evaluate:
- \(\left(f\circ h\right)\left(0\right)\)
- \(\left(f\circ h\right)\left(-8\right)\)
- \(\left(f\circ h\right)\left(\sqrt{2}\right)\)
Construct \(\left(h\circ f\right)\left(x\right)\), and use it to evaluate:
- \(\left(h\circ f\right)\left(-12\right)\)
- \(\left(h\circ f\right)\left(\frac{3}{4}\right)\)
- \(\left(h\circ f\right)\left(\pi\right)\)
Construct \(\left(g\circ g\right)\left(x\right)\), and use it to evaluate:
- \(\left(g\circ g\right)\left(0\right)\)
- \(\left(g\circ g\right)\left(5\right)\)
- \(\left(g\circ g\right)\left(\sqrt{5}\right)\)

Composition from Graphs

We can also evaluate compositions from graphs. The key is working from the inside out: evaluate the inner function first, then use the result as the input to the outer function.

Problem: The graphs of \(f\left(x\right)\) and \(g\left(x\right)\) appear below. Use them to evaluate each of the following.

\[\left(a\right)~\left(f \circ g\right)\left(2\right) \qquad \left(b\right)~f\left(g\left(-1\right)\right) \qquad \left(c\right)~\left(g \circ f\right)\left(0\right) \qquad \left(d\right)~\left(g \circ f\right)\left(-1\right)\] \[\left(e\right)~\left(g \circ g\right)\left(-3\right) \qquad \left(f\right)~\left(f \circ f\right)\left(-1\right) \qquad \left(g\right)~f\left(g\left(f\left(2\right)\right)\right) \qquad \left(h\right)~g\left(f\left(g\left(2\right)\right)\right)\]

Solution. Working from the inside out in each case:

\(\left(f \circ g\right)\left(2\right) = f\left(g\left(2\right)\right) = f\left(-1\right) = \boxed{~0~}\) (closed dot at \(x=-1\) on \(f\))

Composition from Graphs

We can also evaluate compositions from graphs. The key is working from the inside out: evaluate the inner function first, then use the result as the input to the outer function.

Problem: The graphs of \(f\left(x\right)\) and \(g\left(x\right)\) appear below. Use them to evaluate each of the following.

\[\left(a\right)~\left(f \circ g\right)\left(2\right) \qquad \left(b\right)~f\left(g\left(-1\right)\right) \qquad \left(c\right)~\left(g \circ f\right)\left(0\right) \qquad \left(d\right)~\left(g \circ f\right)\left(-1\right)\] \[\left(e\right)~\left(g \circ g\right)\left(-3\right) \qquad \left(f\right)~\left(f \circ f\right)\left(-1\right) \qquad \left(g\right)~f\left(g\left(f\left(2\right)\right)\right) \qquad \left(h\right)~g\left(f\left(g\left(2\right)\right)\right)\]

Solution. Working from the inside out in each case:

\(f\left(g\left(-1\right)\right) = f\left(-4\right) = \boxed{~7~}\)

Composition from Graphs

We can also evaluate compositions from graphs. The key is working from the inside out: evaluate the inner function first, then use the result as the input to the outer function.

Problem: The graphs of \(f\left(x\right)\) and \(g\left(x\right)\) appear below. Use them to evaluate each of the following.

\[\left(a\right)~\left(f \circ g\right)\left(2\right) \qquad \left(b\right)~f\left(g\left(-1\right)\right) \qquad \left(c\right)~\left(g \circ f\right)\left(0\right) \qquad \left(d\right)~\left(g \circ f\right)\left(-1\right)\] \[\left(e\right)~\left(g \circ g\right)\left(-3\right) \qquad \left(f\right)~\left(f \circ f\right)\left(-1\right) \qquad \left(g\right)~f\left(g\left(f\left(2\right)\right)\right) \qquad \left(h\right)~g\left(f\left(g\left(2\right)\right)\right)\]

Solution. Working from the inside out in each case:

\(\left(g \circ f\right)\left(0\right) = g\left(f\left(0\right)\right) = g\left(3\right) = \boxed{~4~}\)

Composition from Graphs

We can also evaluate compositions from graphs. The key is working from the inside out: evaluate the inner function first, then use the result as the input to the outer function.

Problem: The graphs of \(f\left(x\right)\) and \(g\left(x\right)\) appear below. Use them to evaluate each of the following.

\[\left(a\right)~\left(f \circ g\right)\left(2\right) \qquad \left(b\right)~f\left(g\left(-1\right)\right) \qquad \left(c\right)~\left(g \circ f\right)\left(0\right) \qquad \left(d\right)~\left(g \circ f\right)\left(-1\right)\] \[\left(e\right)~\left(g \circ g\right)\left(-3\right) \qquad \left(f\right)~\left(f \circ f\right)\left(-1\right) \qquad \left(g\right)~f\left(g\left(f\left(2\right)\right)\right) \qquad \left(h\right)~g\left(f\left(g\left(2\right)\right)\right)\]

Solution. Working from the inside out in each case:

\(\left(g \circ f\right)\left(-1\right) = g\left(f\left(-1\right)\right) = g\left(0\right) = \boxed{~-5~}\)

Composition from Graphs

We can also evaluate compositions from graphs. The key is working from the inside out: evaluate the inner function first, then use the result as the input to the outer function.

Problem: The graphs of \(f\left(x\right)\) and \(g\left(x\right)\) appear below. Use them to evaluate each of the following.

\[\left(a\right)~\left(f \circ g\right)\left(2\right) \qquad \left(b\right)~f\left(g\left(-1\right)\right) \qquad \left(c\right)~\left(g \circ f\right)\left(0\right) \qquad \left(d\right)~\left(g \circ f\right)\left(-1\right)\] \[\left(e\right)~\left(g \circ g\right)\left(-3\right) \qquad \left(f\right)~\left(f \circ f\right)\left(-1\right) \qquad \left(g\right)~f\left(g\left(f\left(2\right)\right)\right) \qquad \left(h\right)~g\left(f\left(g\left(2\right)\right)\right)\]

Solution. Working from the inside out in each case:

\(\left(g \circ g\right)\left(-3\right) = g\left(g\left(-3\right)\right) = g\left(4\right) = \boxed{~6~}\)

Composition from Graphs

We can also evaluate compositions from graphs. The key is working from the inside out: evaluate the inner function first, then use the result as the input to the outer function.

Problem: The graphs of \(f\left(x\right)\) and \(g\left(x\right)\) appear below. Use them to evaluate each of the following.

\[\left(a\right)~\left(f \circ g\right)\left(2\right) \qquad \left(b\right)~f\left(g\left(-1\right)\right) \qquad \left(c\right)~\left(g \circ f\right)\left(0\right) \qquad \left(d\right)~\left(g \circ f\right)\left(-1\right)\] \[\left(e\right)~\left(g \circ g\right)\left(-3\right) \qquad \left(f\right)~\left(f \circ f\right)\left(-1\right) \qquad \left(g\right)~f\left(g\left(f\left(2\right)\right)\right) \qquad \left(h\right)~g\left(f\left(g\left(2\right)\right)\right)\]

Solution. Working from the inside out in each case:

\(\left(f \circ f\right)\left(-1\right) = f\left(f\left(-1\right)\right) = f\left(0\right) = \boxed{~3~}\)

Composition from Graphs

We can also evaluate compositions from graphs. The key is working from the inside out: evaluate the inner function first, then use the result as the input to the outer function.

Problem: The graphs of \(f\left(x\right)\) and \(g\left(x\right)\) appear below. Use them to evaluate each of the following.

\[\left(a\right)~\left(f \circ g\right)\left(2\right) \qquad \left(b\right)~f\left(g\left(-1\right)\right) \qquad \left(c\right)~\left(g \circ f\right)\left(0\right) \qquad \left(d\right)~\left(g \circ f\right)\left(-1\right)\] \[\left(e\right)~\left(g \circ g\right)\left(-3\right) \qquad \left(f\right)~\left(f \circ f\right)\left(-1\right) \qquad \left(g\right)~f\left(g\left(f\left(2\right)\right)\right) \qquad \left(h\right)~g\left(f\left(g\left(2\right)\right)\right)\]

Solution. Working from the inside out in each case:

\(f\left(g\left(f\left(2\right)\right)\right) = f\left(g\left(3\right)\right) = f\left(4\right) = \boxed{~-5~}\)

Composition from Graphs

We can also evaluate compositions from graphs. The key is working from the inside out: evaluate the inner function first, then use the result as the input to the outer function.

Problem: The graphs of \(f\left(x\right)\) and \(g\left(x\right)\) appear below. Use them to evaluate each of the following.

\[\left(a\right)~\left(f \circ g\right)\left(2\right) \qquad \left(b\right)~f\left(g\left(-1\right)\right) \qquad \left(c\right)~\left(g \circ f\right)\left(0\right) \qquad \left(d\right)~\left(g \circ f\right)\left(-1\right)\] \[\left(e\right)~\left(g \circ g\right)\left(-3\right) \qquad \left(f\right)~\left(f \circ f\right)\left(-1\right) \qquad \left(g\right)~f\left(g\left(f\left(2\right)\right)\right) \qquad \left(h\right)~g\left(f\left(g\left(2\right)\right)\right)\]

Solution. Working from the inside out in each case:

\(g\left(f\left(g\left(2\right)\right)\right) = g\left(f\left(-1\right)\right) = g\left(0\right) = \boxed{~-5~}\)

Composition from Graphs Practice

Try It! 3: The graphs of \(p\left(x\right)\) and \(q\left(x\right)\) appear below. Use them to evaluate each of the following.

\[\left(a\right)~\left(p \circ q\right)\left(2\right) \qquad \left(b\right)~q\left(p\left(-2\right)\right) \qquad \left(c\right)~\left(q \circ p\right)\left(0\right) \qquad \left(d\right)~\left(p \circ p\right)\left(3\right) \qquad \left(e\right)~\left(q \circ q\right)\left(-1\right) \qquad \left(f\right)~p\left(q\left(p\left(2\right)\right)\right)\]

Applied Problems

Applied Problem 1: A forest fire leaves behind an expanding circular burn. The radius is increasing at rate \(r\left(t\right) = 2t + 1\) (feet, \(t\) in minutes).

Express the area burned as a function of time \(t\), and find the total area burned after 10 minutes.

Solution. The area of a circle is \(A = \pi r^2\), so \(A\left(r\right) = \pi r^2\). We substitute \(r\left(t\right)\) in place of \(r\):

\[\begin{align} A\left(r\left(t\right)\right) &= \pi\left(r\left(t\right)\right)^2 \end{align}\]

Applied Problems

Applied Problem 1: A forest fire leaves behind an expanding circular burn. The radius is increasing at rate \(r\left(t\right) = 2t + 1\) (feet, \(t\) in minutes).

Express the area burned as a function of time \(t\), and find the total area burned after 10 minutes.

Solution. The area of a circle is \(A = \pi r^2\), so \(A\left(r\right) = \pi r^2\). We substitute \(r\left(t\right)\) in place of \(r\):

\[\begin{align} A\left(r\left(t\right)\right) &= \pi\left(r\left(t\right)\right)^2\\ &= \pi\left(2t + 1\right)^2 \end{align}\]

Applied Problems

Applied Problem 1: A forest fire leaves behind an expanding circular burn. The radius is increasing at rate \(r\left(t\right) = 2t + 1\) (feet, \(t\) in minutes).

Express the area burned as a function of time \(t\), and find the total area burned after 10 minutes.

Solution. The area of a circle is \(A = \pi r^2\), so \(A\left(r\right) = \pi r^2\). We substitute \(r\left(t\right)\) in place of \(r\):

\[\begin{align} A\left(r\left(t\right)\right) &= \pi\left(r\left(t\right)\right)^2\\ &= \pi\left(2t + 1\right)^2\\ &= \pi\left(4t^2 + 4t + 1\right) \end{align}\]

Applied Problems

Applied Problem 1: A forest fire leaves behind an expanding circular burn. The radius is increasing at rate \(r\left(t\right) = 2t + 1\) (feet, \(t\) in minutes).

Express the area burned as a function of time \(t\), and find the total area burned after 10 minutes.

Solution. The area of a circle is \(A = \pi r^2\), so \(A\left(r\right) = \pi r^2\). We substitute \(r\left(t\right)\) in place of \(r\):

\[\begin{align} A\left(r\left(t\right)\right) &= \pi\left(r\left(t\right)\right)^2\\ &= \pi\left(2t + 1\right)^2\\ &= \pi\left(4t^2 + 4t + 1\right) \end{align}\]

After \(10\) minutes: \(A\left(r\left(10\right)\right) = \pi\left(4\left(100\right) + 4\left(10\right) + 1\right) = \boxed{~441\pi \approx 1385.4 \text{ sq. ft.}~}\)

Applied Problems (Cont’d)

Applied Problem 2: The number of bacteria in a refrigerated food product is given by

\[N\left(T\right) = 23T^2 - 56T + 1 \qquad 3 < T < 33\]

where \(T\) is the temperature (°F). After the food is removed from the refrigerator, the temperature is given by \(T\left(t\right) = 5t + 1.5\), where \(t\) is time in hours.

\(\left(a\right)\) Find the composite function \(N\left(T\left(t\right)\right)\).

\(\left(b\right)\) Find the time (to two decimal places) when the bacteria count reaches \(6{,}752\).

Solution to \(\left(a\right)\). Substitute \(T\left(t\right) = 5t + 1.5\) into \(N\left(T\right)\):

\[\begin{align} N\left(T\left(t\right)\right) &= 23\left(5t + 1.5\right)^2 - 56\left(5t + 1.5\right) + 1 \end{align}\]

Applied Problems (Cont’d)

Applied Problem 2: The number of bacteria in a refrigerated food product is given by

\[N\left(T\right) = 23T^2 - 56T + 1 \qquad 3 < T < 33\]

where \(T\) is the temperature (°F). After the food is removed from the refrigerator, the temperature is given by \(T\left(t\right) = 5t + 1.5\), where \(t\) is time in hours.

\(\left(a\right)\) Find the composite function \(N\left(T\left(t\right)\right)\).

\(\left(b\right)\) Find the time (to two decimal places) when the bacteria count reaches \(6{,}752\).

Solution to \(\left(a\right)\). Substitute \(T\left(t\right) = 5t + 1.5\) into \(N\left(T\right)\):

\[\begin{align} N\left(T\left(t\right)\right) &= 23\left(5t + 1.5\right)^2 - 56\left(5t + 1.5\right) + 1\\ &= 23\left(25t^2 + 15t + 2.25\right) - 280t - 84 + 1 \end{align}\]

Applied Problems (Cont’d)

Applied Problem 2: The number of bacteria in a refrigerated food product is given by

\[N\left(T\right) = 23T^2 - 56T + 1 \qquad 3 < T < 33\]

where \(T\) is the temperature (°F). After the food is removed from the refrigerator, the temperature is given by \(T\left(t\right) = 5t + 1.5\), where \(t\) is time in hours.

\(\left(a\right)\) Find the composite function \(N\left(T\left(t\right)\right)\).

\(\left(b\right)\) Find the time (to two decimal places) when the bacteria count reaches \(6{,}752\).

Solution to \(\left(a\right)\). Substitute \(T\left(t\right) = 5t + 1.5\) into \(N\left(T\right)\):

\[\begin{align} N\left(T\left(t\right)\right) &= 23\left(5t + 1.5\right)^2 - 56\left(5t + 1.5\right) + 1\\ &= 23\left(25t^2 + 15t + 2.25\right) - 280t - 84 + 1\\ &= 575t^2 + 345t + 51.75 - 280t - 83 \end{align}\]

Applied Problems (Cont’d)

Applied Problem 2: The number of bacteria in a refrigerated food product is given by

\[N\left(T\right) = 23T^2 - 56T + 1 \qquad 3 < T < 33\]

where \(T\) is the temperature (°F). After the food is removed from the refrigerator, the temperature is given by \(T\left(t\right) = 5t + 1.5\), where \(t\) is time in hours.

\(\left(a\right)\) Find the composite function \(N\left(T\left(t\right)\right)\).

\(\left(b\right)\) Find the time (to two decimal places) when the bacteria count reaches \(6{,}752\).

Solution to \(\left(a\right)\). Substitute \(T\left(t\right) = 5t + 1.5\) into \(N\left(T\right)\):

\[\begin{align} N\left(T\left(t\right)\right) &= 23\left(5t + 1.5\right)^2 - 56\left(5t + 1.5\right) + 1\\ &= 23\left(25t^2 + 15t + 2.25\right) - 280t - 84 + 1\\ &= 575t^2 + 345t + 51.75 - 280t - 83\\ &= \boxed{~575t^2 + 65t - 31.25~} \end{align}\]

Applied Problems (Cont’d)

Applied Problem 2 (Cont’d): Find the time when \(N\left(T\left(t\right)\right) = 6{,}752\).

Solution to \(\left(b\right)\). Set \(575t^2 + 65t - 31.25 = 6{,}752\) and solve.

Quadratic Formula

As a reminder, one way to solve a quadratic equation of the form \(ax^2 + bx + c = 0\) is via the quadratic formula, which yields \(\displaystyle{x = \frac{-b \pm \sqrt{\left(b^2 - 4ac\right)}}{2a}}\).

\[\begin{align} 575t^2 + 65t - 31.25 &= 6752 \end{align}\]

Applied Problems (Cont’d)

Applied Problem 2 (Cont’d): Find the time when \(N\left(T\left(t\right)\right) = 6{,}752\).

Solution to \(\left(b\right)\). Set \(575t^2 + 65t - 31.25 = 6{,}752\) and solve.

Quadratic Formula

As a reminder, one way to solve a quadratic equation of the form \(ax^2 + bx + c = 0\) is via the quadratic formula, which yields \(\displaystyle{x = \frac{-b \pm \sqrt{\left(b^2 - 4ac\right)}}{2a}}\).

\[\begin{align} 575t^2 + 65t - 31.25 &= 6752\\ \implies 575t^2 + 65t - 6783.25 &= 0 \end{align}\]

Applied Problems (Cont’d)

Applied Problem 2 (Cont’d): Find the time when \(N\left(T\left(t\right)\right) = 6{,}752\).

Solution to \(\left(b\right)\). Set \(575t^2 + 65t - 31.25 = 6{,}752\) and solve.

Quadratic Formula

As a reminder, one way to solve a quadratic equation of the form \(ax^2 + bx + c = 0\) is via the quadratic formula, which yields \(\displaystyle{x = \frac{-b \pm \sqrt{\left(b^2 - 4ac\right)}}{2a}}\).

\[\begin{align} 575t^2 + 65t - 31.25 &= 6752\\ \implies 575t^2 + 65t - 6783.25 &= 0\\ \implies t &= \frac{-65 \pm \sqrt{65^2 + 4\left(575\right)\left(6783.25\right)}}{2\left(575\right)} \end{align}\]

Applied Problems (Cont’d)

Applied Problem 2 (Cont’d): Find the time when \(N\left(T\left(t\right)\right) = 6{,}752\).

Solution to \(\left(b\right)\). Set \(575t^2 + 65t - 31.25 = 6{,}752\) and solve.

Quadratic Formula

As a reminder, one way to solve a quadratic equation of the form \(ax^2 + bx + c = 0\) is via the quadratic formula, which yields \(\displaystyle{x = \frac{-b \pm \sqrt{\left(b^2 - 4ac\right)}}{2a}}\).

\[\begin{align} 575t^2 + 65t - 31.25 &= 6752\\ \implies 575t^2 + 65t - 6783.25 &= 0\\ \implies t &= \frac{-65 \pm \sqrt{65^2 + 4\left(575\right)\left(6783.25\right)}}{2\left(575\right)}\\ &= \frac{-65 \pm \sqrt{4225 + 15{,}601{,}475}}{1150}\\ &\approx \frac{-65 \pm 3950.5}{1150} \end{align}\]

Taking the positive root (since \(t \geq 0\)): \(\displaystyle{t \approx \frac{3885.5}{1150} \approx \boxed{~3.38 \text{ hours}~}}\)

Exit Ticket Task

Navigate to our MAT142 Exit Ticket Form, answer the questions, and complete the task below.

Note. Today’s discussion is listed as 5. Composition of Functions

Task: Consider \(f\left(x\right) = 2x - 1\) and \(g\left(x\right) = x^2 + 3\).

\(\left(a\right)\) Find \(\left(f \circ g\right)\left(x\right)\) and evaluate at \(x = 2\).

\(\left(b\right)\) Find \(\left(g \circ f\right)\left(x\right)\) and evaluate at \(x = 2\).

Summary and Next Time…

Ideas From Today

The composition \(\left(f \circ g\right)\left(x\right) = f\left(g\left(x\right)\right)\) chains two functions together – \(g\) acts first, then \(f\) acts on the result.
Order matters: \(\left(f \circ g\right)\left(x\right)\) and \(\left(g \circ f\right)\left(x\right)\) are generally different functions.
We can evaluate compositions numerically (step by step) or algebraically (substitute \(g\left(x\right)\) into \(f\)) – both give the same result.
We can evaluate compositions from graphs by reading output values and using them as the next input.
Composition arises naturally in applied settings whenever one quantity depends on another that is itself changing.

Looking Ahead

Try It! 2 previewed something important. When \(\left(f \circ g\right)\left(x\right) = x\) and \(\left(g \circ f\right)\left(x\right) = x\), we say \(f\) and \(g\) are inverses of each other.
That idea – invertibility – is something we’ll encounter later in our course.

Next Time:
Exam I

Homework:
Complete Homework 4 on MyOpenMath