Defining Average and Instantaneous Rates of Change Practice

Sample questions

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1. Sample 1difficulty 2/5

   The average rate of change of $f(x) = x^2$ on $[1, 4]$ is

   • A

     $15$

   • B

     $8$

   • C

     $5$

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   • D

     $3$

   Why

   $(f(4) - f(1))/(4-1) = (16-1)/3 = 5$.

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2. Sample 2difficulty 2/5

   Average velocity of $s(t) = t^2$ on $[1, 3]$ is

   • A

     $4$

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   • B

     $9$

   • C

     $2$

   • D

     $6$

   Why

   $(s(3) - s(1))/(3-1) = (9-1)/2 = 4$.

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3. Sample 3difficulty 2/5

   The secant slope between $(1,3)$ and $(5,9)$ on the graph of $f$ equals:

   • A

     $\tfrac{3}{2}$

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   • B

     $6$

   • C

     $\tfrac{2}{3}$

   • D

     $4$

   Why

   Slope $= \frac{9-3}{5-1} = \frac{6}{4} = \frac{3}{2}$.

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4. Sample 4difficulty 2/5

   $f'(a)$ represents the

   • A

     Average value of $f$ near $a$

   • B

     Instantaneous rate of change of $f$ at $a$

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   • C

     Area under $f$

   • D

     Maximum of $f$

   Why

   Definition: derivative is the instantaneous rate of change (slope of tangent).

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5. Sample 5difficulty 2/5

   Average rate of change of $f$ on $[1, 7]$ equals:

   • A

     $\tfrac{3}{4}$

   • B

     $\tfrac{4}{3}$

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   • C

     $6$

   • D

     $8$

   Why

   $\frac{f(7)-f(1)}{7-1} = \frac{10-2}{6} = \frac{8}{6} = \frac{4}{3}$.

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