AP Calculus AB Quiz: Estimating Derivatives Of A Function

What this quiz covers

This quiz focuses on Estimating Derivatives Of A Function, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus AB.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $40$ per unit (correct answer)
2. $80$ per unit
3. $2$ per unit
4. $30$ per unit

Explanation: The marginal cost when 50 units are produced is $C'(50)$. This can be estimated by the average rate of change of the cost function over the interval $[50, 52]$. The calculation is: $C'(50) \approx \frac{C(52) - C(50)}{52 - 50} = \frac{1580 - 1500}{2} = \frac{80}{2} = 40$ dollars per unit.

Question 2

1. The estimate is less than the actual value of $g'(3)$.
2. The estimate is greater than the actual value of $g'(3)$. (correct answer)
3. The estimate is equal to the actual value of $g'(3)$.
4. There is not enough information to compare the estimate to the actual value.

Explanation: The estimate given is the slope of the secant line between $(3, 10)$ and $(3.2, 10.8)$, which is $\frac{10.8-10}{3.2-3}=4$. For a function that is concave up, the tangent line at the left endpoint of an interval lies below the secant line over that interval. Thus, the slope of the tangent line at $x=3$, which is the actual value of $g'(3)$, is less than the slope of the secant line. Therefore, the estimate of 4 is greater than the actual value of $g'(3)$.

Question 3

1. $4$ m/s$^2$
2. $-2$ m/s$^2$
3. $-0.5$ m/s$^2$
4. $-4$ m/s$^2$ (correct answer)

Explanation: Acceleration is the derivative of velocity, $a(t) = v'(t)$. To estimate the acceleration at $t=10.2$, we find the average rate of change of velocity over the interval $[10, 10.5]$. The calculation is: $a(10.2) \approx \frac{v(10.5) - v(10)}{10.5 - 10} = \frac{10 - 12}{0.5} = \frac{-2}{0.5} = -4$ m/s$^2$.

Question 4

1. $-3$ (correct answer)
2. $-0.6$
3. $-0.3$
4. $3$

Explanation: The second derivative, $f''(x)$, is the derivative of the first derivative, $f'(x)$. We can estimate $f''(2.6)$ by calculating the average rate of change of $f'(x)$ over the interval $[2.5, 2.7]$. The calculation is: $f''(2.6) \approx \frac{f'(2.7) - f'(2.5)}{2.7 - 2.5} = \frac{5.8 - 6.4}{0.2} = \frac{-0.6}{0.2} = -3$.

Question 5

1. $8$ (correct answer)
2. $0.4$
3. $0.05$
4. $0.125$

Explanation: The derivative $f'(3)$ can be approximated by the slope of the secant line between the points $(3, 2)$ and $(3.05, 2.4)$. The slope is calculated as the average rate of change: $f'(3) \approx \frac{f(3.05) - f(3)}{3.05 - 3} = \frac{2.4 - 2}{0.05} = \frac{0.4}{0.05} = 8$.

Question 6

1. $30$ milligrams per hour
2. $-15$ milligrams per hour
3. $-30$ milligrams per hour (correct answer)
4. $15$ milligrams per hour

Explanation: The rate of change $C'(1)$ can be estimated by the average rate of change over the interval $[1, 1.5]$. The calculation is: $C'(1) \approx \frac{C(1.5) - C(1)}{1.5 - 1} = \frac{75 - 90}{0.5} = \frac{-15}{0.5} = -30$ milligrams per hour. The negative sign indicates the amount of caffeine is decreasing.

Question 7

1. $-4$ (correct answer)
2. $4$
3. $-0.8$
4. $0.25$

Explanation: The value $g'(5)$ can be approximated by the slope of the secant line over the symmetric interval $[4.9, 5.1]$ centered at $x=5$. The calculation is: $g'(5) \approx \frac{g(5.1) - g(4.9)}{5.1 - 4.9} = \frac{9.4 - 10.2}{0.2} = \frac{-0.8}{0.2} = -4$.

Question 8

1. $-2$ degrees Celsius per minute (correct answer)
2. $-4$ degrees Celsius per minute
3. $-8$ degrees Celsius per minute
4. $2$ degrees Celsius per minute

Explanation: The rate of change $T'(10)$ can be estimated by the average rate of change over the interval $[8, 12]$, since $t=10$ is the midpoint of this interval. The calculation is: $T'(10) \approx \frac{T(12) - T(8)}{12 - 8} = \frac{52 - 60}{4} = \frac{-8}{4} = -2$ degrees Celsius per minute.

Question 9

1. $2.2$
2. $3$ (correct answer)
3. $2.6$
4. $2.0$

Explanation: To get the best estimate for $h'(2)$, we should use the smallest interval available that starts at $x=2$. This is the interval $[2, 2.1]$. The average rate of change over this interval is: $h'(2) \approx \frac{h(2.1) - h(2)}{2.1 - 2} = \frac{5.3 - 5}{0.1} = \frac{0.3}{0.1} = 3$. The interval $[2, 2.5]$ would provide a less accurate estimate.

Question 10

1. $450$ (correct answer)
2. $500$
3. $525$
4. $1500$

Explanation: The rate of population growth in 2015 ($t=5$) is $P'(5)$. The best estimate is the average rate of change over the smallest interval containing $t=5$. Using the interval from $t=5$ to $t=6$: $P'(5) \approx \frac{P(6) - P(5)}{6 - 5} = \frac{15450 - 15000}{1} = 450$ people per year.

Question 11

1. $3$
2. $0.6$
3. $1.8$
4. $9$ (correct answer)

Explanation: First, find the derivative of $h(x)$ with respect to $x$, which is $h'(x) = 3f'(x)$. Next, estimate $f'(1)$ using the given values: $f'(1) \approx \frac{f(1.2) - f(1)}{1.2 - 1} = \frac{10.6 - 10}{0.2} = \frac{0.6}{0.2} = 3$. Finally, substitute this estimate into the expression for $h'(1)$: $h'(1) = 3f'(1) \approx 3(3) = 9$.

Question 12

1. $4$ (correct answer)
2. $1.2$
3. $0.3$
4. $0.25$

Explanation: The best approximation for $f'(2)$ using the given points is the average rate of change over the interval $[1.8, 2.1]$, as $x=2$ lies within this interval. The calculation is: $f'(2) \approx \frac{f(2.1) - f(1.8)}{2.1 - 1.8} = \frac{8.2 - 7}{0.3} = \frac{1.2}{0.3} = 4$.

Question 13

1. $-0.4$ meters per day (correct answer)
2. $-2$ meters per day
3. $-2.5$ meters per day
4. $-0.5$ meters per day

Explanation: To estimate the instantaneous rate of change $W'(32)$, we use the average rate of change over the interval $[30, 35]$, which contains $t=32$. The calculation is: $W'(32) \approx \frac{W(35) - W(30)}{35 - 30} = \frac{48 - 50}{5} = \frac{-2}{5} = -0.4$ meters per day.

Question 14

1. $0.2$
2. $-5$
3. $5$ (correct answer)
4. $-0.2$

Explanation: To estimate $f'(4)$, we can use the average rate of change over the interval $[3.8, 4]$. This is often called a backward difference estimate. The calculation is: $f'(4) \approx \frac{f(4) - f(3.8)}{4 - 3.8} = \frac{12 - 11}{0.2} = \frac{1}{0.2} = 5$.

Question 15

1. The estimate is less than the actual value of $h'(1.5)$.
2. The estimate is greater than the actual value of $h'(1.5)$. (correct answer)
3. The estimate is equal to the actual value of $h'(1.5)$.
4. There is not enough information to compare the estimate to the actual value.

Explanation: The estimate is the slope of the secant line through $(1, 20)$ and $(1.5, 24)$, which is $\frac{24-20}{1.5-1}=8$. For a function that is concave down, the tangent line at the right endpoint of an interval lies below the secant line over that interval. Thus, the slope of the tangent line at $x=1.5$, which is the actual value of $h'(1.5)$, is less than the slope of the secant line. Therefore, the estimate of 8 is greater than the actual value of $h'(1.5)$.

Question 16

1. Alice's estimate
2. Bob's estimate
3. Charles's estimate (correct answer)
4. All three estimates are equally good.

Explanation: Charles computes the average rate of change over a symmetric interval $[3.9, 4.1]$ centered at $x=4$. This is called a centered difference approximation: $\frac{13.5 - 12.2}{4.1 - 3.9} = \frac{1.3}{0.2} = 6.5$. Alice's and Bob's estimates are one-sided (backward and forward differences, respectively). The centered difference approximation is typically a more accurate estimate of the instantaneous derivative than one-sided approximations.

Question 17

1. $1.2$ (correct answer)
2. $0.3$
3. $0.75$
4. $0.12$

Explanation: The instantaneous rate of growth at $t=2$ can be estimated by the average rate of growth over the interval $[2, 2.25]$. The calculation is: Rate $\approx \frac{B(2.25) - B(2)}{2.25 - 2} = \frac{5.4 - 5.1}{0.25} = \frac{0.3}{0.25} = 1.2$ millions of bacteria per hour.

Question 18

1. $-2$ (correct answer)
2. $-1.5$
3. $-1$
4. $2$

Explanation: To estimate $g''(1)$, we need to find the rate of change of $g'(x)$ near $x=1$. We can use the average rate of change on intervals containing $x=1$. On $[0,1]$, the rate is $\frac{3-5}{1-0}=-2$. On $[1,3]$, the rate is $\frac{-1-3}{3-1}=-2$. Since both intervals give the same rate, the best estimate for $g''(1)$ is -2. Using the larger interval $[0,3]$ also yields $\frac{-1-5}{3-0}=-2$.

Question 19

1. $0.1$ cm/s
2. $0.2$ cm/s
3. $2$ cm/s (correct answer)
4. $1$ cm/s

Explanation: The rate of change of the radius is $r'(t)$. We can estimate $r'(4)$ using the average rate of change over the interval $[4, 4.1]$. The calculation is: $r'(4) \approx \frac{r(4.1) - r(4)}{4.1 - 4} = \frac{10.2 - 10}{0.1} = \frac{0.2}{0.1} = 2$ cm/s.

Question 20

1. $V'(61) \approx -1.5$, meaning the stock's value is decreasing at a rate of approximately $1.50 per day. (correct answer)
2. $V'(61) \approx 1.5$, meaning the stock's value is increasing at a rate of approximately $1.50 per day.
3. $V'(61) \approx -3$, meaning the stock's value is decreasing by a total of $3.00 over two days.
4. $V'(61) \approx -1.5$, meaning the stock's value was $1.50 on day 61.

Explanation: First, estimate $V'(61)$ using the average rate of change over the interval $[60, 62]$: $V'(61) \approx \frac{V(62) - V(60)}{62 - 60} = \frac{247 - 250}{2} = \frac{-3}{2} = -1.5$. The units are dollars per day. A negative rate of change means the value is decreasing. Therefore, the stock's value is decreasing at an approximate rate of $1.50 per day.