AP Calculus AB Quiz: Estimating Derivatives Of A Function
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Estimating Derivatives Of A FunctionQuestion 1 of 20

The total cost to produce xx units of a certain product is given by a differentiable function C(x)C(x).

If the cost to produce 50 units is C(50)=$1500C(50) = \$1500 and the cost to produce 52 units is C(52)=$1580C(52) = \$1580, which of the following is the best estimate for the marginal cost when 50 units are produced?

4040 per unit
8080 per unit
22 per unit
3030 per unit
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AP Calculus AB Quiz

AP Calculus AB Quiz: Estimating Derivatives Of A Function

Practice Estimating Derivatives Of A Function in AP Calculus AB with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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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.

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Question 1

The total cost to produce xx units of a certain product is given by a differentiable function C(x)C(x).

If the cost to produce 50 units is C(50)=$1500C(50) = \$1500 and the cost to produce 52 units is C(52)=$1580C(52) = \$1580, which of the following is the best estimate for the marginal cost when 50 units are produced?

  1. 4040 per unit (correct answer)
  2. 8080 per unit
  3. 22 per unit
  4. 3030 per unit
Explanation: The marginal cost when 50 units are produced is C′(50)C'(50). This can be estimated by the average rate of change of the cost function over the interval [50,52][50, 52]. The calculation is: C′(50)≈C(52)−C(50)52−50=1580−15002=802=40C'(50) \approx \frac{C(52) - C(50)}{52 - 50} = \frac{1580 - 1500}{2} = \frac{80}{2} = 40 dollars per unit.

Question 2

The function gg is differentiable and its graph is known to be concave up on the interval [2,5][2, 5]. If g(3)=10g(3) = 10 and g(3.2)=10.8g(3.2) = 10.8, an estimate for g′(3)g'(3) is 4. How does this estimate relate to the actual value of g′(3)g'(3)?

  1. The estimate is less than the actual value of g′(3)g'(3).
  2. The estimate is greater than the actual value of g′(3)g'(3). (correct answer)
  3. The estimate is equal to the actual value of g′(3)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)(3, 10) and (3.2,10.8)(3.2, 10.8), which is 10.8−103.2−3=4\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=3x=3, which is the actual value of g′(3)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)g'(3).

Question 3

The velocity of a remote-controlled car, v(t)v(t), in meters per second, is a differentiable function of time tt in seconds.

At time t=10t=10 seconds, the velocity is 12 m/s. At time t=10.5t=10.5 seconds, the velocity is 10 m/s. Which of the following is the best estimate for the car's acceleration at t=10.2t=10.2 seconds?

  1. 44 m/s2^2
  2. −2-2 m/s2^2
  3. −0.5-0.5 m/s2^2
  4. −4-4 m/s2^2 (correct answer)
Explanation: Acceleration is the derivative of velocity, a(t)=v′(t)a(t) = v'(t). To estimate the acceleration at t=10.2t=10.2, we find the average rate of change of velocity over the interval [10,10.5][10, 10.5]. The calculation is: a(10.2)≈v(10.5)−v(10)10.5−10=10−120.5=−20.5=−4a(10.2) \approx \frac{v(10.5) - v(10)}{10.5 - 10} = \frac{10 - 12}{0.5} = \frac{-2}{0.5} = -4 m/s2^2.

Question 4

The function ff is twice differentiable. Values of the first derivative, f′(x)f'(x), are given by f′(2.5)=6.4f'(2.5) = 6.4 and f′(2.7)=5.8f'(2.7) = 5.8. Which of the following is the best estimate for f′′(2.6)f''(2.6)?

  1. −3-3 (correct answer)
  2. −0.6-0.6
  3. −0.3-0.3
  4. 33
Explanation: The second derivative, f′′(x)f''(x), is the derivative of the first derivative, f′(x)f'(x). We can estimate f′′(2.6)f''(2.6) by calculating the average rate of change of f′(x)f'(x) over the interval [2.5,2.7][2.5, 2.7]. The calculation is: f′′(2.6)≈f′(2.7)−f′(2.5)2.7−2.5=5.8−6.40.2=−0.60.2=−3f''(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

The amount of caffeine, C(t)C(t), in milligrams, in a person's bloodstream tt hours after drinking a coffee is a differentiable function.

If C(1)=90C(1) = 90 and C(1.5)=75C(1.5) = 75, which of the following is the best estimate for C′(1)C'(1), the rate of change of the amount of caffeine in the bloodstream at t=1t=1 hour?

  1. 3030 milligrams per hour
  2. −15-15 milligrams per hour
  3. −30-30 milligrams per hour (correct answer)
  4. 1515 milligrams per hour
Explanation: The rate of change C′(1)C'(1) can be estimated by the average rate of change over the interval [1,1.5][1, 1.5]. The calculation is: C′(1)≈C(1.5)−C(1)1.5−1=75−900.5=−150.5=−30C'(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 6

The temperature of a cup of tea, T(t)T(t), in degrees Celsius, is a differentiable function of time tt in minutes.

If T(8)=60T(8) = 60 and T(12)=52T(12) = 52, which of the following is the best estimate for T′(10)T'(10)?

  1. −2-2 degrees Celsius per minute (correct answer)
  2. −4-4 degrees Celsius per minute
  3. −8-8 degrees Celsius per minute
  4. 22 degrees Celsius per minute
Explanation: The rate of change T′(10)T'(10) can be estimated by the average rate of change over the interval [8,12][8, 12], since t=10t=10 is the midpoint of this interval. The calculation is: T′(10)≈T(12)−T(8)12−8=52−604=−84=−2T'(10) \approx \frac{T(12) - T(8)}{12 - 8} = \frac{52 - 60}{4} = \frac{-8}{4} = -2 degrees Celsius per minute.

Question 7

The population of a town, P(t)P(t), is a differentiable function of time tt, where tt is measured in years since 2010.

The town's population was 15,000 in 2015 (t=5t=5), 15,450 in 2016 (t=6t=6), and 16,500 in 2018 (t=8t=8). Which of the following is the best estimate for the rate of population growth in 2015, in people per year?

  1. 450450 (correct answer)
  2. 500500
  3. 525525
  4. 15001500
Explanation: The rate of population growth in 2015 (t=5t=5) is P′(5)P'(5). The best estimate is the average rate of change over the smallest interval containing t=5t=5. Using the interval from t=5t=5 to t=6t=6: P′(5)≈P(6)−P(5)6−5=15450−150001=450P'(5) \approx \frac{P(6) - P(5)}{6 - 5} = \frac{15450 - 15000}{1} = 450 people per year.

Question 8

The depth of water in a reservoir, W(t)W(t), is a differentiable function of time tt in days.

At t=30t=30 days, the depth is 50 meters. At t=35t=35 days, the depth is 48 meters. Which of the following is the best estimate for W′(32)W'(32)?

  1. −0.4-0.4 meters per day (correct answer)
  2. −2-2 meters per day
  3. −2.5-2.5 meters per day
  4. −0.5-0.5 meters per day
Explanation: To estimate the instantaneous rate of change W′(32)W'(32), we use the average rate of change over the interval [30,35][30, 35], which contains t=32t=32. The calculation is: W′(32)≈W(35)−W(30)35−30=48−505=−25=−0.4W'(32) \approx \frac{W(35) - W(30)}{35 - 30} = \frac{48 - 50}{5} = \frac{-2}{5} = -0.4 meters per day.

Question 9

The function hh is differentiable and its graph is known to be concave down on the interval [0,4][0, 4]. If h(1)=20h(1) = 20 and h(1.5)=24h(1.5) = 24, an estimate for h′(1.5)h'(1.5) using the secant line over [1,1.5][1, 1.5] is 8. How does this estimate relate to the actual value of h′(1.5)h'(1.5)?

  1. The estimate is less than the actual value of h′(1.5)h'(1.5).
  2. The estimate is greater than the actual value of h′(1.5)h'(1.5). (correct answer)
  3. The estimate is equal to the actual value of h′(1.5)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)(1, 20) and (1.5,24)(1.5, 24), which is 24−201.5−1=8\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=1.5x=1.5, which is the actual value of h′(1.5)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)h'(1.5).

Question 10

For a differentiable function ff, it is known that f(3.9)=12.2f(3.9) = 12.2, f(4)=12.8f(4) = 12.8, and f(4.1)=13.5f(4.1) = 13.5. Three students compute estimates for f′(4)f'(4). Alice uses the interval [3.9,4][3.9, 4]. Bob uses the interval [4,4.1][4, 4.1]. Charles uses the interval [3.9,4.1][3.9, 4.1]. Which student's computation is generally considered the best estimate?

  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][3.9, 4.1] centered at x=4x=4. This is called a centered difference approximation: 13.5−12.24.1−3.9=1.30.2=6.5\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 11

The number of bacteria in a petri dish, B(t)B(t), is a differentiable function of time tt in hours.

At t=2t=2 hours, there are approximately 5.1 million bacteria. At t=2.25t=2.25 hours, there are approximately 5.4 million bacteria. Which of the following is the best estimate for the instantaneous rate of growth of the bacteria population at t=2t=2 hours, in millions of bacteria per hour?

  1. 1.21.2 (correct answer)
  2. 0.30.3
  3. 0.750.75
  4. 0.120.12
Explanation: The instantaneous rate of growth at t=2t=2 can be estimated by the average rate of growth over the interval [2,2.25][2, 2.25]. The calculation is: Rate ≈B(2.25)−B(2)2.25−2=5.4−5.10.25=0.30.25=1.2 \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 12

A function g(x)g(x) is twice-differentiable. Selected values of its derivative, g′(x)g'(x), are given by g′(0)=5g'(0)=5, g′(1)=3g'(1)=3, and g′(3)=−1g'(3)=-1. Which of the following is the best estimate for g′′(1)g''(1)?

  1. −2-2 (correct answer)
  2. −1.5-1.5
  3. −1-1
  4. 22
Explanation: To estimate g′′(1)g''(1), we need to find the rate of change of g′(x)g'(x) near x=1x=1. We can use the average rate of change on intervals containing x=1x=1. On [0,1][0,1], the rate is 3−51−0=−2\frac{3-5}{1-0}=-2. On [1,3][1,3], the rate is −1−33−1=−2\frac{-1-3}{3-1}=-2. Since both intervals give the same rate, the best estimate for g′′(1)g''(1) is -2. Using the larger interval [0,3][0,3] also yields −1−53−0=−2\frac{-1-5}{3-0}=-2.

Question 13

A balloon is being inflated. The radius rr, in centimeters, is a differentiable function of time tt in seconds.

At time t=4t=4 seconds, the radius is 10 cm. At time t=4.1t=4.1 seconds, the radius is 10.2 cm. What is the best estimate for the rate of change of the radius with respect to time at t=4t=4 seconds?

  1. 0.10.1 cm/s
  2. 0.20.2 cm/s
  3. 22 cm/s (correct answer)
  4. 11 cm/s
Explanation: The rate of change of the radius is r′(t)r'(t). We can estimate r′(4)r'(4) using the average rate of change over the interval [4,4.1][4, 4.1]. The calculation is: r′(4)≈r(4.1)−r(4)4.1−4=10.2−100.1=0.20.1=2r'(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 14

The value of a stock, V(t)V(t) in dollars, is a differentiable function of time tt, where tt is the number of days after purchase.

Suppose V(60)=250V(60) = 250 and V(62)=247V(62) = 247. Based on this information, which of the following provides the best estimate for V′(61)V'(61) and its interpretation?

  1. V′(61)≈−1.5V'(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)≈1.5V'(61) \approx 1.5, meaning the stock's value is increasing at a rate of approximately $1.50 per day.
  3. V′(61)≈−3V'(61) \approx -3, meaning the stock's value is decreasing by a total of $3.00 over two days.
  4. V′(61)≈−1.5V'(61) \approx -1.5, meaning the stock's value was $1.50 on day 61.
Explanation: First, estimate V′(61)V'(61) using the average rate of change over the interval [60,62][60, 62]: V′(61)≈V(62)−V(60)62−60=247−2502=−32=−1.5V'(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.

Question 15

The function ff is differentiable. If f(1.8)=7f(1.8) = 7 and f(2.1)=8.2f(2.1) = 8.2, which of the following is the best approximation for f′(2)f'(2)?

  1. 44 (correct answer)
  2. 1.21.2
  3. 0.30.3
  4. 0.250.25
Explanation: The best approximation for f′(2)f'(2) using the given points is the average rate of change over the interval [1.8,2.1][1.8, 2.1], as x=2x=2 lies within this interval. The calculation is: f′(2)≈f(2.1)−f(1.8)2.1−1.8=8.2−70.3=1.20.3=4f'(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 16

A function ff is differentiable. If f(3)=2f(3) = 2 and f(3.05)=2.4f(3.05) = 2.4, which of the following is the best approximation for f′(3)f'(3)?

  1. 88 (correct answer)
  2. 0.40.4
  3. 0.050.05
  4. 0.1250.125
Explanation: The derivative f′(3)f'(3) can be approximated by the slope of the secant line between the points (3,2)(3, 2) and (3.05,2.4)(3.05, 2.4). The slope is calculated as the average rate of change: f′(3)≈f(3.05)−f(3)3.05−3=2.4−20.05=0.40.05=8f'(3) \approx \frac{f(3.05) - f(3)}{3.05 - 3} = \frac{2.4 - 2}{0.05} = \frac{0.4}{0.05} = 8.

Question 17

A differentiable function gg has values g(4.9)=10.2g(4.9) = 10.2 and g(5.1)=9.4g(5.1) = 9.4. Which of the following is the best approximation for g′(5)g'(5)?

  1. −4-4 (correct answer)
  2. 44
  3. −0.8-0.8
  4. 0.250.25
Explanation: The value g′(5)g'(5) can be approximated by the slope of the secant line over the symmetric interval [4.9,5.1][4.9, 5.1] centered at x=5x=5. The calculation is: g′(5)≈g(5.1)−g(4.9)5.1−4.9=9.4−10.20.2=−0.80.2=−4g'(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 18

The function hh is differentiable. The following values for h(x)h(x) are known: h(2)=5h(2) = 5, h(2.1)=5.3h(2.1) = 5.3, and h(2.5)=6.1h(2.5) = 6.1. Which of the following is the best estimate for h′(2)h'(2)?

  1. 2.22.2
  2. 33 (correct answer)
  3. 2.62.6
  4. 2.02.0
Explanation: To get the best estimate for h′(2)h'(2), we should use the smallest interval available that starts at x=2x=2. This is the interval [2,2.1][2, 2.1]. The average rate of change over this interval is: h′(2)≈h(2.1)−h(2)2.1−2=5.3−50.1=0.30.1=3h'(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][2, 2.5] would provide a less accurate estimate.

Question 19

A function ff is differentiable. Let h(x)=3f(x)−4h(x) = 3f(x) - 4. If f(1)=10f(1) = 10 and f(1.2)=10.6f(1.2) = 10.6, which of the following is the best estimate for h′(1)h'(1)?

  1. 33
  2. 0.60.6
  3. 1.81.8
  4. 99 (correct answer)
Explanation: First, find the derivative of h(x)h(x) with respect to xx, which is h′(x)=3f′(x)h'(x) = 3f'(x). Next, estimate f′(1)f'(1) using the given values: f′(1)≈f(1.2)−f(1)1.2−1=10.6−100.2=0.60.2=3f'(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): h′(1)=3f′(1)≈3(3)=9h'(1) = 3f'(1) \approx 3(3) = 9.

Question 20

The function ff is differentiable. If f(4)=12f(4) = 12 and f(3.8)=11f(3.8) = 11, which of the following is the best estimate for f′(4)f'(4)?

  1. 0.20.2
  2. −5-5
  3. 55 (correct answer)
  4. −0.2-0.2
Explanation: To estimate f′(4)f'(4), we can use the average rate of change over the interval [3.8,4][3.8, 4]. This is often called a backward difference estimate. The calculation is: f′(4)≈f(4)−f(3.8)4−3.8=12−110.2=10.2=5f'(4) \approx \frac{f(4) - f(3.8)}{4 - 3.8} = \frac{12 - 11}{0.2} = \frac{1}{0.2} = 5.