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Section5.8Difference Quotients

Definition of the Difference Quotient.

The difference quotient is a new expression created from a template and a given function formula. The template for the difference quotient of the function $y=f(x)$ is

\begin{equation*} \frac{f(x+h)-f(x)}{h}\text{.} \end{equation*}

It will be discussed later why this is an important expression. For now, lets just get familiar the algebra associated with difference quotient.

Example5.8.1.

Determine the difference quotient for the function $f(x)=8x-7\text{.}$ Make sure that your expression is fully simplified.

Solution

The difference quotient for $f(x)=8x-7$ follows.

\begin{align*} \frac{\highlight{f(x+h)}-\highlightr{f(x)}}{x}\amp=\frac{(\highlight{8(x+h)-7})-(\highlightr{8x-7})}{h}\\ \amp=\frac{8x+8h-7-8x+7}{h}\\ \amp=\frac{8h}{h}\\ \amp=8,\,\,h \neq 0 \end{align*}
Example5.8.2.

Determine the difference quotient for the function $g(t)=6t-t^2\text{.}$ Make sure that your expression is fully simplified.

Solution

The difference quotient for $g(t)=6t-t^2$ follows.

\begin{align*} \frac{\highlight{g(t+h)}-\highlightr{g(t)}}{h}\amp=\frac{(\highlight{6(t+h)-(t+h)^2})-(\highlightr{6t-t^2})}{h}\\ \amp=\frac{6t+6h-(t+h)(t+h)-6t+t^2}{h}\\ \amp=\frac{6h-(t^2+2th+h^2)+t^2}{h}\\ \amp=\frac{6h-t^2-2th-h^2+t^2}{h}\\ \amp=\frac{6h-2th-h^2}{h}\\ \amp=\frac{h \cdot (6-2t-h)}{h}\\ \amp=-2t+6-h,\,\,h \neq 0 \end{align*}
Example5.8.3.

Determine the difference quotient for the function $m(x)=157\text{.}$ Make sure that your expression is completely simplified.

Solution

The difference quotient for the function $m(x)=157$ follows.

\begin{align*} \frac{\highlight{m(x+h)}-\highlightr{m(x)}}{h}\amp=\frac{\highlight{157}-\highlightr{157}}{h}\\ \amp=\frac{0}{h}\\ \amp=0,\,\,h \neq 0 \end{align*}
Example5.8.4.

Determine the difference quotient for the function $w(x)=\frac{5}{x-3}\text{.}$ Make sure that your expression is fully simplified.

Solution

The difference quotient for $w(x)=\frac{5}{x-3}$ follows.

\begin{align*} \frac{\highlight{w(x+h)}-\highlightr{w(x)}}{h}\amp=\frac{\highlight{\frac{5}{x+h-3}}-\highlightr{\frac{5}{x-3}}}{h}\\ \amp=\frac{\frac{5}{x+h-3} \cdot \highlightb{\frac{x-3}{x-3}}-\frac{5}{x-3} \cdot \highlightb{\frac{x+h-3}{x+h-3}}}{\frac{h}{1}}\\ \amp=\frac{5(x-3)-5(x+h-3)}{(x-3)(x+h-3)} \cdot \frac{1}{h}\\ \amp=\frac{5x-15-5x-5h+15}{(x-3)(x+h-3)h}\\ \amp=\frac{-5 \cdot h}{(x-3)(x+h-3) \cdot h}\\ \amp=-\frac{5}{(x-3)(x+h-3)},\,\, h \neq 0 \end{align*}
The Graphical Significance of the Difference Quotient.

A secant line for the graph of the function $y=f(x)$ is a line that connects two points on the curve. If you're thinking "gee, it's pretty easy to be a secant line" you're correct in that thought. For example, consider a function that graphs to a parabola. With one exception, any non-vertical line that intersects the parabola once intersects the parabola a second time as well. So with one exception, any non-vertical line that intersects a parabola is a secant line for the parabola. The only exception is the horizontal line that intersects the vertex.

It turns out that the difference quotient is the slope of a secant line to $f\text{.}$ Specifically, it is the slope of the line connecting the point $(x,f(x))$ with the point $(x+h,f(x+h))\text{.}$

To understand this, consider $y=f(x)\text{.}$ Let $x_1=x$ and $x_2=x+h\text{.}$ Let's now apply the classic slope-formula.

\begin{align*} m\amp=\frac{y_2-y_1}{x_2-x_1}\\ \amp=\frac{f(x_2)-f(x_1)}{x_2-x_1}\\ \amp=\frac{f(x+h)-f(x)}{(x+h)-x}\\ \amp=\frac{f(x+h)-f(x)}{h} \end{align*}

Note that this implies that for the two stated points the rise of the secant line is

\begin{equation*} \text{rise}=f(x+h)-f(x) \end{equation*}

and the run of the secant line is

\begin{equation*} \text{run}=h \end{equation*}
Example5.8.5.

Determine the difference quotient for the function $f(x)=2x^2-x-7\text{.}$ Then use the simplified difference quotient expression to determine the slope of the secant line for $f$ between the points where $x=-3$ and $x=5\text{.}$

Solution

We begin by determining the simplified difference quotient for $f\text{.}$

\begin{align*} \frac{\highlight{f(x+h)}-\highlightr{f(x)}}{h}\amp=\frac{(\highlight{2(x+h)^2-(x+h)-7})-(\highlightr{2x^2-x-7})}{h}\\ \amp=\frac{2(x+h)(x+h)-x-h-7-2x^2+x+7}{h}\\ \amp=\frac{2(x^2+2xh+h^2)-h-2x^2}{h}\\ \amp=\frac{2x^2+4xh+2h^2-h-2x^2}{h}\\ \amp=\frac{4xh+2h^2-h}{h}\\ \amp=\frac{h(4x+2h-1)}{h}\\ \amp=4x+2h-1,\,\,h \neq 0 \end{align*}

The run from the point where $x=-3$ to the point where $x=5$ is $8\text{.}$ So let's take our difference quotient formula and replace $x$ with $-3$ and $h$ with $8$ to determine the slope of the indicated secant line.

\begin{align*} m\amp=4x+2h-1\\ \amp=4(-3)+2 \cdot 8-1\\ \amp=3 \end{align*}

Let's confirm the slope using the traditional slope formula.

\begin{align*} m\amp=\frac{f(5)-f(-3)}{5-(-3)}\\ \amp=\frac{(2 \cdot 5^2-5-7)-(2 \cdot (-3)^2-(-3)-7)}{8}\\ \amp=\frac{38-14}{8}\\ \amp=\frac{24}{8}\\ \amp=3\,\,\checkmark \end{align*}

ExercisesExercises

Determine the difference quotient for each of the following functions. Make sure that you completely simplify each expression.

1.

$f(x)=7x-8$

Solution

The difference quotient for the function $f(x)=7x-8$ is derived below.

\begin{align*} \frac{\highlight{f(x+h)}-\highlightr{f(x)}}{h}\amp=\frac{\highlight{(7(x+h)-8)}-\highlightr{(7x-8)}}{h}\\ \amp=\frac{7x+7h-8-7x+8}{h}\\ \amp=\frac{7h}{h}\\ \amp=7,\,\,h \neq 0 \end{align*}
2.

$g(x)=x^2-3x+2$

Solution

The difference quotient for the function $g(x)=x^2-3x+2$ is derived below.

\begin{align*} \frac{\highlight{g(x+h)}-\highlightr{g(x)}}{h}\amp=\frac{\highlight{((x+h)^2-3(x+h)+2)}-\highlightr{(x^2-3x+2)}}{h}\\ \amp=\frac{x^2+2xh+h^2-3x-3h+2-x^2+3x-2}{h}\\ \amp=\frac{2xh-3h+h^2}{h}\\ \amp=\frac{h \cdot (2x-3+h)}{h}\\ \amp=2x-3+h,\,\,h \neq 0 \end{align*}
3.

$g(t)=5-t^2$

Solution

The difference quotient for the function $g(t)=5-t^2$ is derived below.

\begin{align*} \frac{\highlight{g(t+h)}-\highlightr{g(t)}}{h}\amp=\frac{\highlight{(5-(t+h)^2)}-\highlightr{(5-t^2)}}{h}\\ \amp=\frac{5-(t^2+2th+h^2)-5+t^2}{h}\\ \amp=\frac{5-t^2-2th-h^2-5+t^2}{h}\\ \amp=\frac{-2th-h^2}{h}\\ \amp=\frac{h \cdot (-2t-h)}{h}\\ \amp=-2t-h,\,\,h \neq 0 \end{align*}
4.

$k(x)=12$

Solution

The difference quotient for the function $k(x)=12$ is derived below.

\begin{align*} \frac{\highlight{k(x+h)}-\highlightr{k(x)}}{h}\amp=\frac{\highlight{12}-\highlightr{12}}{h}\\ \amp=\frac{0}{h}\\ \amp=0,\,\,h \neq 0 \end{align*}
5.

$r(t)=\frac{3}{t+4}$

Solution

The difference quotient for the function $r(t)=\frac{3}{t+4}$ is derived below.

\begin{align*} \frac{\highlight{r(t+h)}-\highlightr{r(t)}}{h}\amp=\frac{\highlight{\frac{3}{t+h+4}}-\highlightr{\frac{3}{t+4}}}{h}\\ \amp=\frac{\frac{3}{t+h+4} \cdot \highlightg{\frac{t+4}{t+4}}-\frac{3}{t+4} \cdot \highlightb{\frac{t+h+4}{t+h+4}}}{h}\\ \amp=\frac{\frac{3t+12-3t-3h-12}{(t+4)(t+h+4)}}{\frac{h}{1}}\\ \amp=\frac{-3h}{(t+4)(t+h+4)} \cdot \frac{1}{h}\\ \amp=\frac{-3 \cdot h}{(t+4)(t+h+4) \cdot h}\\ \amp=-\frac{3}{(t+4)(t+4+h)},\,\, h \neq 0 \end{align*}
6.

$y(x)=\frac{1}{7-2x}$

Solution

The difference quotient for the function $y(x)=\frac{1}{7-2x}$ is derived below.

\begin{align*} \frac{\highlight{y(x+h)}-\highlightr{y(x)}}{h}\amp=\frac{\highlight{\frac{1}{7-2(x+h)}}-\highlightr{\frac{1}{7-2x}}}{h}\\ \amp=\frac{\frac{1}{7-2x-2h}-\frac{1}{7-2x}}{h}\\ \amp=\frac{\frac{1}{7-2x-2h} \cdot \highlightg{\frac{7-2x}{7-2x}}-\frac{1}{7-2x} \cdot \highlightb{\frac{7-2x-2h}{7-2x-2h}}}{h}\\ \amp=\frac{\frac{7-2x-7+2x+2h}{(7-2x-2h)(7-2x)}}{h}\\ \amp=\frac{\frac{2h}{(7-2x-2h)(7-2x)}}{\frac{h}{1}}\\ \amp=\frac{2h}{(7-2x-2h)(7-2x)} \cdot \frac{1}{h}\\ \amp=\frac{2 \cdot h}{(7-2x-2h)(7-2x) \cdot h}\\ \amp=\frac{2}{(7-2x-2h)(7-2x)},\,h\neq 0 \end{align*}

Interpret a difference quotient value as a slope.

7.

The function $f(x)=\frac{1}{2}x^2+2x-3$ is shown in Figure 5.8.6 along with the secant line connecting the points $(-2,-5)$ and $(2,3)\text{.}$ Determine the difference quotient of $f$ and then use that to determine the slope of the indicated secant line.

Solution

We begin by deriving the difference quotient for $f(x)=\frac{1}{2}x^2+2x-3\text{.}$

\begin{align*} \frac{\highlight{f(x+h)}-\highlightr{f(x)}}{h}\amp=\frac{\highlight{\left(\frac{1}{2}(x+h\right)^2+2(x+h)-3)}-\highlightr{\left(\frac{1}{2}x^2+2x-3\right)}}{h}\\ \amp=\frac{\frac{1}{2}(x^2+2xh+h^2)+2x+2h-3-\frac{1}{2}x^2-2x+3}{h}\\ \amp=\frac{\frac{1}{2}x^2+xh+\frac{1}{2}h^2+2h-\frac{1}{2}x^2}{h}\\ \amp=\frac{xh+\frac{1}{2}h^2+2h}{h}\\ \amp=\frac{h \cdot \left(x+\frac{1}{2}h+2\right)}{h}\\ \amp=x+2+\frac{1}{2}h,\,\,h \neq 0 \end{align*}

We now find the slope of the indicated secant line by letting $x=-2$ and $h=4$ (the left-most of the two $x$-coordinates and the run between the two indicated points). The calculation follows.

\begin{align*} m\amp=-2+2+\frac{1}{2} \cdot 4\\ \amp=2 \end{align*}

A slope of $2$ is confirmed by looking at the $\frac{\text{rise}}{\text{run}}$ between the two points.