Difference Quotients

Section 7.10 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.

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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 \(y=f(x)\text{.}\)

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Definition 7.10.1. 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{7.10.1} \end{equation}

The expression in (7.10.1) is called a difference quotient.

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Example 7.10.2. Finding Average Rate of Change.

Let’s find the average rate of change of \(f(x) = x^3 - 2x + 1\) on the interval \([-1,2]\text{.}\)

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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.

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Example 7.10.3. 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.
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Find the average rate of data transfer during the time interval starting at \(t = 2\) seconds and ending at \(t = 3\) seconds.
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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.

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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 ((7.10.1)) takes the form

\begin{equation} \frac{f(x+h)-f(x)}{h}.\tag{7.10.2} \end{equation}

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Example 7.10.4. Evaluating a Difference Quotient.

Let’s find and simplify the difference quotient (7.10.2) for \({\color{blue} f(x) = -x^2 + 2x + 1}\text{.}\)

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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*}

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Example 7.10.5. Additional Difference Quotient Examples.

Find and simplify the difference quotient (7.10.2) for each function below. Video solutions follow.

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\(\displaystyle f(x) = 3x-4\)

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\(\displaystyle f(x) = x-2x^2\)

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\(\displaystyle f(x) = \frac{1}{3x}\)

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\(\displaystyle f(x) = \frac{1}{x+3}\)

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Solution.

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