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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,
(a) between 2001 and 2002?
(b) between 2001 and 2004?
Answer 1.
Answer 2.
\(-0.666667\)

5.

At the start of a trip, the odometer on a car reads 20801 miles. At the end of the trip, 12.5 hours later, the odometer reads 21391 miles.
What is the average speed the car traveled during this trip? miles/hour
Answer.

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 17 gallons.
What is the average rate of flow of the gasoline into the gas tank? gal/min
Answer.

9.

Find and simplify the difference quotient for \(\displaystyle{f(x)={-7x^{2}+1}}\text{.}\) Your answer must be simplified for it to be counted correct.
\(\displaystyle{ \frac{f(x+h)-f(x)}{h}=}\)
Answer.
\(-14x-7h\)

10.

Find and simplify the difference quotient for \(\displaystyle{f(x)={-5x^{2}-5x+1}}\text{.}\) Your answer must be simplified for it to be counted correct.
\(\displaystyle{ \frac{f(x+h)-f(x)}{h}=}\)
Answer.
\(-10x-5h-5\)

11.

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

12.

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

13.

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

14.

Let \(\ f(x) = 25 - x^2\text{.}\)
a) Compute each of the following expressions and interpret each as an average rate of change:
(i) \(\ \ \ \frac{f( 3 ) - f(0)}{ 3 - 0} = \ \)
(ii) \(\ \ \frac{f( 5 ) - f( 3 )}{5 - 3 } = \ \)
(iii) \(\ \frac{f( 5 ) - f(0)}{5 - 0} = \ \)
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.
The graph of a function with lines shown between various points on and off the graph.
Line A is horizontal from the point \((3,f(3))\) to the y axis.
Line B goes from the point \((0,f(0))\) to \((5,f(5))\text{.}\)
Line C goes from the point \((3,f(3))\) to the point \((5,f(5))\text{.}\)
Line D goes from the point \((0,f(0))\) to \((3,f(3))\text{.}\)
Line E goes from the point \((0,0)\) to \((3,f(3))\text{.}\)
Line F is vertical from the point \((3,f(3))\) to the x axis.
(click on image to enlarge)
\(\frac{f( 3 ) - f(0)}{ 3 - 0}\)
\(\frac{f( 5 ) - f( 3 )}{5 - 3 }\)
\(\frac{f( 5 ) - f(0)}{5 - 0}\)
Answer 1.
Answer 2.
Answer 3.
Answer 4.
Answer 5.
Answer 6.
Solution.
i) \(\frac{f( 3 ) - f(0)}{ 3 - 0} = \frac{ 16 - 25 }{3 - 0} = \frac{ -9 }{ 3} = -3\)
ii) \(\frac{f( 5 ) - f( 3 ) }{ 5 - 3 } = \frac{ 0 - 16}{5 - 3} = \frac{ -16 }{ 2} = -8\)
iii) \(\frac{f( 5 ) - f(0)}{ 5 - 0} = \frac{ 0 - 25 }{ 5 - 0} = \frac{ -16 }{ 5} = -5\)
b) The formulas in parts (i), (ii), and (iii) represent the average rate of change of \(f(x)\) over the intervals \(0 \leq x \leq 3\) , \(3 \leq x \leq 5\) , and \(0 \leq x \leq 5\) respectively. Thus line segment D connecting the points (0, 25 ) and ( 3, 16 ) illustrates formula (i), line segment C connecting the points ( 3, 16 ) and ( 5, 0 ) illustrates formula (ii), and line segment B connecting the points (0, 25 ) and ( 5, 0 ) illustrates formula (iii).

15.

Find and simplify the difference quotient for \(\displaystyle{f(x)={\frac{5}{x^{2}}}}\text{.}\) Your answer must be simplified for it to be counted correct.
\(\displaystyle{ \frac{f(x+h)-f(x)}{h}=}\)
Answer.
\(\frac{-10x-5h}{x^{2}\mathopen{}\left(x+h\right)^{2}}\)

16.

Consider the function \(f(x) = 7 x - 2\) and find the following:
a) The average rate of change between the points \((-1, f(-1) )\) and \(( 1 , f( 1 ) )\) .
b) The average rate of change between the points \((a, f(a) )\) and \((b, f(b) )\) .
c) The average rate of change between the points \((x, f(x) )\) and \((x+h, f(x+h) )\) .
Answer 1.
Answer 2.
Answer 3.
Solution.
SOLUTION\((-1, f(-1))\)\(( 1 , f( 1 ))\)
\begin{equation*} \frac{ f( 1 ) - f(-1) }{1 - (-1)} = \frac{ 5 + 9 }{1 + 1} = \frac{ 14}{2 } = 7 \end{equation*}
\((a, f(a))\)\(( b , f(b))\)
\begin{equation*} \frac{ f(b) - f(a) }{b - a} = \frac{ (7 b - 2) - (7 a - 2) }{b - a} = \frac{ 7 b - 7 a}{b-a} = \frac{ 7 (b-a) }{b-a} = 7 \end{equation*}
\((x, f(x))\)\(( x+h , f(x+h))\)
\begin{equation*} \frac{ f(x+h) - f(x) }{(x+h) - x} = \frac{ (7 (x+h) - 2) - (7 x - 2) }{h} = \frac{ 7 x + 7 h -2 - 7 x + 2}{h} = \frac{ 7 h }{h} = 7 \end{equation*}

17.

Calculate the successive average rates of change for the function, \(H(x)\) in the table below.
\(x\) 12 14 16 18
\(H(x)\) 21.7 22.02 22.19 22.3
(a) The average rate of change over the interval \(12 \leq x \leq 14\) is
(Retain at least 3 decimal places in your answer.)
(b) The average rate of change over the interval \(14 \leq x \leq 16\) is
(Retain at least 3 decimal places in your answer.)
(c) The average rate of change over the interval \(16 \leq x \leq 18\) is
(Retain at least 3 decimal places in your answer.)
(d) Based your answers for the rates of change, the function \(H(x)\) is
Answer 1.
Answer 2.
Answer 3.
Solution.
SOLUTION\(x= 12\)\(x = 14\)
\begin{equation*} \frac{\triangle H}{\triangle x} = \frac{22.02 - 21.7}{14-12} = \frac{0.32}{2} = 0.16 \end{equation*}
\(x= 14\)\(x = 16\)
\begin{equation*} \frac{\triangle H}{\triangle x} = \frac{22.19 - 22.02}{16-14} = \frac{0.17}{2} = 0.085 \end{equation*}
\(x= 16\)\(x = 18\)
\begin{equation*} \frac{\triangle H}{\triangle x} = \frac{22.3 - 22.19}{18-16} = \frac{0.11}{2} = 0.055 \end{equation*}
\(x\)

18.

  1. What is the average rate of change of \(g(x) = {-2-7x}\) between the points \((-2, {12})\) and \((4,{-30})\text{?}\)
    The average rate of change is .
  2. The function \(g\) is
    on the interval \(-2 \leq x \leq 4\text{.}\)
Answer 1.
Answer 2.
\(\text{decreasing}\)
Solution.
  1. If \(y = g(x)\) then between \((-2 , {12})\) and \((4 , {-30} )\text{,}\)
    \(\displaystyle{\hbox{average rate of change of }g =\frac{\Delta y}{\Delta x}= \frac{{-30} - ({12})}{4 -(-2)}}\) \(\displaystyle{ = \frac{-30 - 12}{4 + 2}= \frac{{-42}}{{6}} = -7.}\)
  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 negative number. Hence \(g\) is decreasing on the given interval (and in fact over any interval).

21.

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

22.

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

23.

For the function \(\small{f(x) = x^{2}+3x}\text{,}\) simplify each expression as much as possible.
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.
Answer 2.
Solution.
\(\small{f(x) = x^{2}+3x, \;f(x + h) = \left(x+h\right)^{2}+3\mathopen{}\left(x+h\right)}\text{.}\)
\begin{align*} \small{\frac{f(x \;+\; h) \;-\; f(x)}{h}} \amp = \frac{\left(x+h\right)^{2}+3\mathopen{}\left(x+h\right) - (x^{2}+3x)}{h}\\ \amp = \small{\frac{2xh+h^{2}+3h}{h}} \\ \amp = \small{2x+3+h.} \end{align*}
\(\small{f(x) = x^{2}+3x, \;f(w) = w^{2}+3w}\text{.}\)
\begin{align*} \small{\frac{f(w) \;-\; f(x)}{w \;-\; x}} \amp = \frac{w^{2}+3w - (x^{2}+3x)}{w - x}\\ \amp = \small{\frac{w^{2}-x^{2}+3w-3x}{w-x}} \\ \amp = \small{\frac{\left(w+x\right)\mathopen{}\left(w-x\right)+3\mathopen{}\left(w-x\right)}{w-x}} \\ \amp = \small{\frac{\left(w-x\right)\mathopen{}\left(w+x+3\right)}{w-x}} \\ \amp = \small{w+x+3.} \end{align*}