Estimating Derivatives of a Function - AP Calculus AB

For a differentiable function $f$ , it is known that $f(3.9) = 12.2$ , $f(4) = 12.8$ , and $f(4.1) = 13.5$ . Three students compute estimates for $f'(4)$ . Alice uses the interval $[3.9, 4]$ . Bob uses the interval $[4, 4.1]$ . Charles uses the interval $[3.9, 4.1]$ . Which student's computation is generally considered the best estimate?
Answer: Charles's estimate 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.
The function $g$ is differentiable and its graph is known to be concave up on the interval $[2, 5]$ . If $g(3) = 10$ and $g(3.2) = 10.8$ , an estimate for $g'(3)$ is 4. How does this estimate relate to the actual value of $g'(3)$ ?
Answer: The estimate is greater than the actual value of $g'(3)$ . 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 $g'(3)$ .
If the cost to produce 50 units is $C(50) = \$1500$ and the cost to produce 52 units is $C(52) = \$1580$ , which of the following is the best estimate for the marginal cost when 50 units are produced?
Answer: $40$ per unit 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.
The function $f$ is differentiable. If $f(4) = 12$ and $f(3.8) = 11$ , which of the following is the best estimate for $f'(4)$ ?
Answer: $5$ 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: $f'(4) \approx \frac{f(4) - f(3.8)}{4 - 3.8} = \frac{12 - 11}{0.2} = \frac{1}{0.2} = 5$ .
Suppose $V(60) = 250$ and $V(62) = 247$ . Based on this information, which of the following provides the best estimate for $V'(61)$ and its interpretation?
Answer: $V'(61) \approx -1.5$ , meaning the stock's value is decreasing at a rate of approximately $1.50 per day. 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.
The town's population was 15,000 in 2015 ( $t=5$ ), 15,450 in 2016 ( $t=6$ ), and 16,500 in 2018 ( $t=8$ ). Which of the following is the best estimate for the rate of population growth in 2015, in people per year?
Answer: $450$ 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.
At time $t=4$ seconds, the radius is 10 cm. At time $t=4.1$ 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 $t=4$ seconds?
Answer: $2$ cm/s 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.
If $C(1) = 90$ and $C(1.5) = 75$ , which of the following is the best estimate for $C'(1)$ , the rate of change of the amount of caffeine in the bloodstream at $t=1$ hour?
Answer: $-30$ milligrams per hour 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.
The function $h$ is differentiable. The following values for $h(x)$ are known: $h(2) = 5$ , $h(2.1) = 5.3$ , and $h(2.5) = 6.1$ . Which of the following is the best estimate for $h'(2)$ ?
Answer: $3$ 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.
The function $f$ is differentiable. If $f(1.8) = 7$ and $f(2.1) = 8.2$ , which of the following is the best approximation for $f'(2)$ ?
Answer: $4$ 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: $f'(2) \approx \frac{f(2.1) - f(1.8)}{2.1 - 1.8} = \frac{8.2 - 7}{0.3} = \frac{1.2}{0.3} = 4$ .
The function $f$ is twice differentiable. Values of the first derivative, $f'(x)$ , are given by $f'(2.5) = 6.4$ and $f'(2.7) = 5.8$ . Which of the following is the best estimate for $f''(2.6)$ ?
Answer: $-3$ 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: $f''(2.6) \approx \frac{f'(2.7) - f'(2.5)}{2.7 - 2.5} = \frac{5.8 - 6.4}{0.2} = \frac{-0.6}{0.2} = -3$ .
The function $h$ is differentiable and its graph is known to be concave down on the interval $[0, 4]$ . If $h(1) = 20$ and $h(1.5) = 24$ , an estimate for $h'(1.5)$ using the secant line over $[1, 1.5]$ is 8. How does this estimate relate to the actual value of $h'(1.5)$ ?
Answer: The estimate is greater than the actual value of $h'(1.5)$ . 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 $h'(1.5)$ .
At $t=30$ days, the depth is 50 meters. At $t=35$ days, the depth is 48 meters. Which of the following is the best estimate for $W'(32)$ ?
Answer: $-0.4$ meters per day 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.
A function $g(x)$ is twice-differentiable. Selected values of its derivative, $g'(x)$ , are given by $g'(0)=5$ , $g'(1)=3$ , and $g'(3)=-1$ . Which of the following is the best estimate for $g''(1)$ ?
Answer: $-2$ 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 $\frac{-1-5}{3-0}=-2$ .
A function $f$ is differentiable. Let $h(x) = 3f(x) - 4$ . If $f(1) = 10$ and $f(1.2) = 10.6$ , which of the following is the best estimate for $h'(1)$ ?
Answer: $9$ 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 $h'(1)$ : $h'(1) = 3f'(1) \approx 3(3) = 9$ .
At time $t=10$ seconds, the velocity is 12 m/s. At time $t=10.5$ seconds, the velocity is 10 m/s. Which of the following is the best estimate for the car's acceleration at $t=10.2$ seconds?
Answer: $-4$ m/s $^2$ 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 $^2$ .
A differentiable function $g$ has values $g(4.9) = 10.2$ and $g(5.1) = 9.4$ . Which of the following is the best approximation for $g'(5)$ ?
Answer: $-4$ 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: $g'(5) \approx \frac{g(5.1) - g(4.9)}{5.1 - 4.9} = \frac{9.4 - 10.2}{0.2} = \frac{-0.8}{0.2} = -4$ .
At $t=2$ hours, there are approximately 5.1 million bacteria. At $t=2.25$ 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 $t=2$ hours, in millions of bacteria per hour?
Answer: $1.2$ 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.
If $T(8) = 60$ and $T(12) = 52$ , which of the following is the best estimate for $T'(10)$ ?
Answer: $-2$ degrees Celsius per minute 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.
A function $f$ is differentiable. If $f(3) = 2$ and $f(3.05) = 2.4$ , which of the following is the best approximation for $f'(3)$ ?
Answer: $8$ 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: $f'(3) \approx \frac{f(3.05) - f(3)}{3.05 - 3} = \frac{2.4 - 2}{0.05} = \frac{0.4}{0.05} = 8$ .

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