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Section4.7Difference Quotients

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.

Example4.7.1

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}{8}\\ \amp=8,\,\,h \neq 0 \end{align*}
Example4.7.2

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*}
Example4.7.3

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*}
Example4.7.4

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*}

Let's turn our attention now to the geometric 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*}
Example4.7.5

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*}

Subsection4.7.1Exercises

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 Figureย 4.7.6 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.

A graph of the function \(f(x)=\frac{1}{2}x^2+2x-3\text{.}\)  A line is also graphed.  The line intersects the parabola at the points \((-2,-5)\) and \((2,3)\text{.}\)
Figure4.7.6\(f(x)=\frac{1}{2}x^2+2x-3\)

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.