Section 6.10 Difference Quotients
A linear function changes at a constant rate of change determined by its slope. For example, the value of function \(f(x) = -2.3x +7\) decreases \(2.3\) units per unit increase in \(x\text{.}\) Non-linear functions have a non-constant rate of change, but it is still useful to attempt to study the way they change. In fact, a good portion of calculus addresses this particular question.
The best we can do right now is measure the
average rate of change of a function over some interval. To do this, we measure the slope of a line joining two points on the graph called a
secant line. In the graph below, I’ve sketched the secant line joining between
\(x=a\) and
\(x=b\) on the graph of
\(y=f(x)\text{.}\)
Definition 6.50. Average Rate of Change.
Suppose \(f\) is a function defined on an interval \([a,b]\text{.}\) the average rate of change of \(f\) on \([a,b]\) is
\begin{equation}
\frac{f(b)-f(a)}{b-a}.\tag{6.1}
\end{equation}
The expression in
(6.1) is called a
difference quotient.
Example 6.51. Finding Average Rate of Change.
Let’s find the average rate of change of \(f(x) = x^3 - 2x + 1\) on the interval \([-1,2]\text{.}\)
To do so, we evaluate the difference quotient of \(f\) on \([-1,2]\text{.}\)
\begin{align*}
\frac{f(2)-f(-1)}{2-(-1)}
&= \frac{[2^3-4+1]-[-1+2+1]}{3} \\
&= \frac{5-2}{3} \\
&= \frac{3}{3} \\
&= 1
\end{align*}
Observe the slope of the secant on the graph below.
Example 6.52. Data Transfer.
The graph below shows the amount of data transferred (in gigabytes) over a secure network during a
\(6\) second period of time.
Find the average rate of data transfer during the time interval starting at \(t = 2\) seconds and ending at \(t = 5\) seconds. What units does this have? Draw the corresponding secant line on the graph.
Find the average rate of data transfer during the time interval starting at \(t = 2\) seconds and ending at \(t = 3\) seconds.
Solution.
Let \(D(t)\) denote the amount of data transferred after \(t\) seconds. Then the average rate on \([2,5]\) is
\begin{equation*}
\frac{D(5)-D(2)}{5-2} = \frac{5.5-1.5}{3} = \frac{4}{3} \approx 1.33
\end{equation*}
with units of Gigabytes per second.
Repeating for
\([2,3]\text{,}\)
\begin{equation*}
\frac{D(3)-D(2)}{3-2} = \frac{2-1.5}{1} = 0.5 \, {\rm Gb/sec}.
\end{equation*}
which appears to very accurately indicate the rate of data transfer.
Difference Quotients in Calculus.
In calculus, we will want to consider the difference quotient over small intervals
\([x,x+h]\text{,}\) where
\(h\neq 0\) is a small non-zero amount. This puts the two points on the graph close together.
The slope of this secant line does a good job of measuring the rate of change of the function at
\(x\text{.}\) In this case, the difference quotient (
(6.1)) takes the form
\begin{equation}
\frac{f(x+h)-f(x)}{h}.\tag{6.2}
\end{equation}
Example 6.53. Evaluating a Difference Quotient.
Let’s find and simplify the difference quotient
(6.2) for
\({\color{blue} f(x) = -x^2 + 2x + 1}\text{.}\)
First, note that \({\color{green} f(x+h) = -(x+h)^2 +2(x+h)+1}.\) So,
\begin{align*}
\frac{{\color{green}f(x+h)}-{\color{blue}f(x)}}{h}
&=
\frac{{\color{green}-(x+h)^2 +2(x+h)+1} -\left[{\color{blue}-x^2 + 2x + 1}\right]}{h}\\
&= \frac{-(x^2+2xh+h^2) +2x+2h+1 +x^2 - 2x - 1}{h}\\
&= \frac{-x^2-2xh-h^2 +2x+2h+1 +x^2 - 2x - 1}{h}\\
&= \frac{-2xh-h^2 +2h}{h}\\
&= \frac{h(-2x-h +2)}{h}\\
&= -2x-h +2
\end{align*}
Example 6.54. Additional Difference Quotient Examples.
Find and simplify the difference quotient
(6.2) for each function below. Video solutions follow.
\(\displaystyle f(x) = 3x-4\)
\(\displaystyle f(x) = x-2x^2\)
\(\displaystyle f(x) = \frac{1}{3x}\)
\(\displaystyle f(x) = \frac{1}{x+3}\)
