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Reed College Department of Mathematics & Statistics

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Worksheet Average Rate of Change

4.

The table above gives the annual sales (in millions of dollars) of a product from 1998 to 2006. What was the average rate of change of annual sales,
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(a) between 2001 and 2002?
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(b) between 2001 and 2004?
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Answer 1.
\(9\)
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Answer 2.
\(2.33333\)
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5.

At the start of a trip, the odometer on a car reads 20764 miles. At the end of the trip, 15 hours later, the odometer reads 21360 miles.
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What is the average speed the car traveled during this trip? miles/hour
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Answer.
\(39.7333\)
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6.

A driver of a car stopped at a gas station to fill up his gas tank. He looked at his watch, and the time read exactly 3:40 p.m. At this time, he started pumping gas into the tank. At exactly 3:44, the tank was full and he noticed that he had pumped 13 gallons.
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What is the average rate of flow of the gasoline into the gas tank? gal/min
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Answer.
\(3.25\)
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9.

Find and simplify the difference quotient for \(\displaystyle{f(x)={6x^{2}-3}}\text{.}\) Your answer must be simplified for it to be counted correct.
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\(\displaystyle{ \frac{f(x+h)-f(x)}{h}=}\)
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Answer.
\(12x+6h\)
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10.

Find and simplify the difference quotient for \(\displaystyle{f(x)={8x^{2}+8x-8}}\text{.}\) Your answer must be simplified for it to be counted correct.
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\(\displaystyle{ \frac{f(x+h)-f(x)}{h}=}\)
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Answer.
\(16x+8h+8\)
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11.

Find and simplify the difference quotient for \(\displaystyle{f(x)={-7-x}}\text{.}\) Your answer must be simplified for it to be counted correct.
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\(\displaystyle{ \frac{f(x+h)-f(x)}{h}=}\)
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Answer.
\(-1\)
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12.

Find and simplify the difference quotient for \(\displaystyle{f(x)={-\frac{5}{x}}}\text{.}\) Your answer must be simplified for it to be counted correct.
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\(\displaystyle{ \frac{f(x+h)-f(x)}{h}=}\)
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Answer.
\(\frac{5}{x\mathopen{}\left(x+h\right)}\)
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13.

Find and simplify the difference quotient for \(\displaystyle{f(x)={-3x^{3}}}\text{.}\) Your answer must be simplified for it to be counted correct.
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\(\displaystyle{ \frac{f(x+h)-f(x)}{h}=}\)
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Answer.
\(-9x^{2}-9xh-3h^{2}\)
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14.

Let \(\ f(x) = 16 - x^2\text{.}\)
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a) Compute each of the following expressions and interpret each as an average rate of change:
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(i) \(\ \ \ \frac{f( 2 ) - f(0)}{ 2 - 0} = \ \)
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(ii) \(\ \ \frac{f( 4 ) - f( 2 )}{4 - 2 } = \ \)
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(iii) \(\ \frac{f( 4 ) - f(0)}{4 - 0} = \ \)
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b) Based on the graph sketched below, match each of your answers in (i) - (iii) with one of the lines labeled A - F. Type the corresponding letter of the line segment next to the appropriate formula. Clearly not all letters will be used.
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The graph of a function with lines shown between various points on and off the graph.
Line A is horizontal from the point \((2,f(2))\) to the y axis.
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Line B goes from the point \((0,f(0))\) to \((4,f(4))\text{.}\)
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Line C goes from the point \((2,f(2))\) to the point \((4,f(4))\text{.}\)
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Line D goes from the point \((0,f(0))\) to \((2,f(2))\text{.}\)
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Line E goes from the point \((0,0)\) to \((2,f(2))\text{.}\)
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Line F is vertical from the point \((2,f(2))\) to the x axis.
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(click on image to enlarge)
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\(\frac{f( 2 ) - f(0)}{ 2 - 0}\)
\(\frac{f( 4 ) - f( 2 )}{4 - 2 }\)
\(\frac{f( 4 ) - f(0)}{4 - 0}\)
Answer 1.
\(-2\)
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Answer 2.
\(-6\)
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Answer 3.
\(-4\)
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Answer 4.
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Answer 5.
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Answer 6.
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Solution.
i) \(\frac{f( 2 ) - f(0)}{ 2 - 0} = \frac{ 12 - 16 }{2 - 0} = \frac{ -4 }{ 2} = -2\)
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ii) \(\frac{f( 4 ) - f( 2 ) }{ 4 - 2 } = \frac{ 0 - 12}{4 - 2} = \frac{ -12 }{ 2} = -6\)
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iii) \(\frac{f( 4 ) - f(0)}{ 4 - 0} = \frac{ 0 - 16 }{ 4 - 0} = \frac{ -12 }{ 4} = -4\)
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b) The formulas in parts (i), (ii), and (iii) represent the average rate of change of \(f(x)\) over the intervals \(0 \leq x \leq 2\) , \(2 \leq x \leq 4\) , and \(0 \leq x \leq 4\) respectively. Thus line segment D connecting the points (0, 16 ) and ( 2, 12 ) illustrates formula (i), line segment C connecting the points ( 2, 12 ) and ( 4, 0 ) illustrates formula (ii), and line segment B connecting the points (0, 16 ) and ( 4, 0 ) illustrates formula (iii).
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15.

Find and simplify the difference quotient for \(\displaystyle{f(x)={\frac{1}{x^{2}}}}\text{.}\) Your answer must be simplified for it to be counted correct.
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\(\displaystyle{ \frac{f(x+h)-f(x)}{h}=}\)
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Answer.
\(\frac{-2x-h}{x^{2}\mathopen{}\left(x+h\right)^{2}}\)
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16.

Consider the function \(f(x) = 3 x - 8\) and find the following:
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a) The average rate of change between the points \((-1, f(-1) )\) and \(( 3 , f( 3 ) )\) .
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b) The average rate of change between the points \((a, f(a) )\) and \((b, f(b) )\) .
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c) The average rate of change between the points \((x, f(x) )\) and \((x+h, f(x+h) )\) .
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Answer 1.
\(3\)
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Answer 2.
\(3\)
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Answer 3.
\(3\)
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Solution.
SOLUTION\((-1, f(-1))\)\(( 3 , f( 3 ))\)
\begin{equation*} \frac{ f( 3 ) - f(-1) }{3 - (-1)} = \frac{ 1 + 11 }{3 + 1} = \frac{ 12}{4 } = 3 \end{equation*}
\((a, f(a))\)\(( b , f(b))\)
\begin{equation*} \frac{ f(b) - f(a) }{b - a} = \frac{ (3 b - 8) - (3 a - 8) }{b - a} = \frac{ 3 b - 3 a}{b-a} = \frac{ 3 (b-a) }{b-a} = 3 \end{equation*}
\((x, f(x))\)\(( x+h , f(x+h))\)
\begin{equation*} \frac{ f(x+h) - f(x) }{(x+h) - x} = \frac{ (3 (x+h) - 8) - (3 x - 8) }{h} = \frac{ 3 x + 3 h -8 - 3 x + 8}{h} = \frac{ 3 h }{h} = 3 \end{equation*}
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17.

Calculate the successive average rates of change for the function, \(H(x)\) in the table below.
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\(x\) 12 16 20 24
\(H(x)\) 21.5 21.66 21.87 22.19
(a) The average rate of change over the interval \(12 \leq x \leq 16\) is
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(Retain at least 3 decimal places in your answer.)
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(b) The average rate of change over the interval \(16 \leq x \leq 20\) is
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(Retain at least 3 decimal places in your answer.)
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(c) The average rate of change over the interval \(20 \leq x \leq 24\) is
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(Retain at least 3 decimal places in your answer.)
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(d) Based your answers for the rates of change, the function \(H(x)\) is
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Answer 1.
\(0.04\)
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Answer 2.
\(0.0525\)
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Answer 3.
\(0.08\)
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Solution.
SOLUTION\(x= 12\)\(x = 16\)
\begin{equation*} \frac{\triangle H}{\triangle x} = \frac{21.66 - 21.5}{16-12} = \frac{0.16}{4} = 0.04 \end{equation*}
\(x= 16\)\(x = 20\)
\begin{equation*} \frac{\triangle H}{\triangle x} = \frac{21.87 - 21.66}{20-16} = \frac{0.21}{4} = 0.0525 \end{equation*}
\(x= 20\)\(x = 24\)
\begin{equation*} \frac{\triangle H}{\triangle x} = \frac{22.19 - 21.87}{24-20} = \frac{0.32}{4} = 0.08 \end{equation*}
\(x\)
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18.

  1. What is the average rate of change of \(g(x) = {-6+7x}\) between the points \((-3, {-27})\) and \((4,{22})\text{?}\)
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    The average rate of change is .
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  2. The function \(g\) is
    on the interval \(-3 \leq x \leq 4\text{.}\)
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Answer 1.
\(7\)
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Answer 2.
\(\text{increasing}\)
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Solution.
  1. If \(y = g(x)\) then between \((-3 , {-27})\) and \((4 , {22} )\text{,}\)
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    \(\displaystyle{\hbox{average rate of change of }g =\frac{\Delta y}{\Delta x}= \frac{{22} - ({-27})}{4 -(-3)}}\) \(\displaystyle{ = \frac{22 + 27}{4 + 3}= \frac{{49}}{{7}} = 7.}\)
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  2. Since the graph of \(g\) is a straight line, the average rate of change of \(g\) is constant. By part (a), this constant average rate of change is equal to \(7\text{,}\) which is a positive number. Hence \(g\) is increasing on the given interval (and in fact over any interval).
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21.

Find the difference quotient for the function \(f(x) = -2 x^3\text{.}\) Simplify your answer as much as possible.
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\(\displaystyle\frac{f(x+h) - f(x)}{h} =\)
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Answer.
\(-2\mathopen{}\left(3x^{2}+3xh+h^{2}\right)\)
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22.

Find the difference quotient for the function \(\displaystyle f(x) = -\frac{5}{x}\text{.}\) Simplify your answer as much as possible.
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\(\displaystyle \frac{f(x+h)-f(x)}{h} =\)
Answer 1.
\(5\)
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Answer 2.
\(x\mathopen{}\left(x+h\right)\)
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23.

For the function \(\small{f(x) = x^{2}-x}\text{,}\) simplify each expression as much as possible.
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1. \(\large{\frac{f(x \;+\; h) \;-\; f(x)}{h}}, \small{h \ne 0}:\)
2. \(\large{\frac{f(w) \;-\; f(x)}{w \;-\; x}}, \small{x \ne w}:\)
Answer 1.
\(2x-1+h\)
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Answer 2.
\(w+x-1\)
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Solution.
\(\small{f(x) = x^{2}-x, \;f(x + h) = \left(x+h\right)^{2}-\left(x+h\right)}\text{.}\)
\begin{align*} \small{\frac{f(x \;+\; h) \;-\; f(x)}{h}} \amp = \frac{\left(x+h\right)^{2}-\left(x+h\right) - (x^{2}-x)}{h}\\ \amp = \small{\frac{2xh+h^{2}-h}{h}} \\ \amp = \small{2x-1+h.} \end{align*}
\(\small{f(x) = x^{2}-x, \;f(w) = w^{2}-w}\text{.}\)
\begin{align*} \small{\frac{f(w) \;-\; f(x)}{w \;-\; x}} \amp = \frac{w^{2}-w - (x^{2}-x)}{w - x}\\ \amp = \small{\frac{w^{2}-x^{2}-w+x}{w-x}} \\ \amp = \small{\frac{\left(w+x\right)\mathopen{}\left(w-x\right)-\left(w-x\right)}{w-x}} \\ \amp = \small{\frac{\left(w-x\right)\mathopen{}\left(w+x-1\right)}{w-x}} \\ \amp = \small{w+x-1.} \end{align*}
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