Estimating Derivatives of a Function - AP Calculus BC

A function $f$ is differentiable for all real numbers. If $f(2.9) = 5.8$ and $f(3.1) = 6.2$ , which of the following is the best estimate for $f'(3)$ ?
Answer: $2$ The derivative at a point can be estimated by the slope of the secant line through two nearby points. The best estimate for $f'(3)$ using the given values is the average rate of change over the interval $[2.9, 3.1]$ . $f'(3) \approx \frac{f(3.1) - f(2.9)}{3.1 - 2.9} = \frac{6.2 - 5.8}{0.2} = \frac{0.4}{0.2} = 2$ .
Which of the following is the best estimate for the rate at which the population was growing at the beginning of 2019 ( $t=9$ ), in thousands of people per year?
Answer: $6$ The rate of population growth at $t=9$ is $P'(9)$ . We can estimate this value using the average rate of change over the interval $[8, 10]$ , for which $t=9$ is the midpoint. $P'(9) \approx \frac{P(10) - P(8)}{10 - 8} = \frac{162 - 150}{2} = \frac{12}{2} = 6$ . This is 6 thousand people per year.
The derivative of a function $f$ at $x=c$ , denoted $f'(c)$ , is best approximated by the slope of a secant line through two points on the graph of $f$ . For the most accurate approximation, the two points should be:
Answer: chosen symmetrically about $x=c$ and as close to $c$ as possible. The derivative is the limit of the slope of the secant line as the interval between the points approaches zero. Therefore, the best approximation is obtained when the points are very close to $x=c$ . A symmetric interval, using $x=c-h$ and $x=c+h$ , often provides a better approximation than a one-sided interval.
What is the best estimate for $D'(12)$ , the rate of change of the depth at $t=12$ hours, in feet per hour?
Answer: $-0.3$ We are given $D(10) = 50.8$ , $D(12) = 50$ , and $D(14) = 49.6$ . The best estimate for $D'(12)$ uses the symmetric interval around $t=12$ , which is $[10, 14]$ . $D'(12) \approx \frac{D(14) - D(10)}{14 - 10} = \frac{49.6 - 50.8}{4} = \frac{-1.2}{4} = -0.3$ feet per hour.
Which of the following is the best approximation for $f'(3.6)$ ?
Answer: $-1.5$ The point $x=3.6$ is the midpoint of the interval $[3.4, 3.8]$ . The best estimate for the derivative at the midpoint of an interval is the average rate of change over that interval. $f'(3.6) \approx \frac{f(3.8) - f(3.4)}{3.8 - 3.4} = \frac{6.5 - 7.1}{0.4} = \frac{-0.6}{0.4} = -1.5$ .
Which of the following is the best estimate for $h'(5.0)$ ?
Answer: $3.5$ The best estimate for $h'(5.0)$ can be found by using the symmetric difference quotient over the smallest symmetric interval $[4.8, 5.2]$ . This is equivalent to averaging the backward and forward difference quotients. $h'(5.0) \approx \frac{h(5.2) - h(4.8)}{5.2 - 4.8} = \frac{11.6 - 10.2}{0.4} = \frac{1.4}{0.4} = 3.5$ .
Which of the following is the best approximation for $f'(4)$ ?
Answer: $-7.7$ The derivative $f'(4)$ can be approximated by the slope of the secant line through the points at $x=3.99$ and $x=4.01$ . $f'(4) \approx \frac{f(4.01) - f(3.99)}{4.01 - 3.99} = \frac{6.277 - 6.431}{0.02} = \frac{-0.154}{0.02} = -7.7$ .
Based on this information, what is the best estimate for the rate at which the volume is changing at $t=2$ hours, in gallons per hour?
Answer: $40$ The rate of change of the volume at $t=2$ can be estimated by the average rate of change over the interval $[1.5, 2]$ . $V'(2) \approx \frac{V(2) - V(1.5)}{2 - 1.5} = \frac{300 - 280}{0.5} = \frac{20}{0.5} = 40$ gallons per hour.
Which of the following is the best estimate for $g'(2)$ ?
Answer: $3.5$ To find the best estimate for $g'(2)$ , we should use the smallest symmetric interval around $x=2$ . The values at $x=1.9$ and $x=2.1$ provide this interval. $g'(2) \approx \frac{g(2.1) - g(1.9)}{2.1 - 1.9} = \frac{4.9 - 4.2}{0.2} = \frac{0.7}{0.2} = 3.5$ .
Using this data, what is the approximate rate of change of the amount of the drug in the bloodstream at $t=2.25$ hours, in mg per hour?
Answer: $-6$ The rate of change at $t=2.25$ can be estimated by the average rate of change over the interval $[2, 2.5]$ , since $t=2.25$ is the midpoint of this interval. $A'(2.25) \approx \frac{A(2.5) - A(2)}{2.5 - 2} = \frac{12 - 15}{0.5} = \frac{-3}{0.5} = -6$ mg per hour.
Which of the following is the best estimate of the growth rate $H'(12)$ in centimeters per day?
Answer: $2.5$ To get the best estimate for $H'(12)$ , we should average the rates of change from the intervals on either side of $t=12$ . The backward difference (rate on $[10, 12]$ ) is $\frac{H(12)-H(10)}{12-10} = \frac{24-20}{2} = 2$ . The forward difference (rate on $[12, 15]$ ) is $\frac{H(15)-H(12)}{15-12} = \frac{33-24}{3} = 3$ . The best estimate is the average of these two rates: $\frac{2+3}{2} = 2.5$ cm/day.
The quantity $Q = \frac{g(a+0.01) - g(a)}{0.01}$ is calculated for a differentiable function $g$ . Which of the following statements about $Q$ is the most accurate?
Answer: $Q$ is the exact value of the average rate of change of $g$ on $[a, a+0.01]$ . The expression $\frac{g(b) - g(a)}{b - a}$ defines the exact average rate of change of the function $g$ over the interval $[a, b]$ . The given quantity $Q$ matches this form with $b=a+0.01$ . This value is often used to approximate the instantaneous rate of change $g'(a)$ , but the quantity itself is the exact average rate of change.
Let $k$ be a differentiable function. For a small positive value $h$ , it is known that $k(3) - k(3-h) \approx 2h$ . Which of the following is the best estimate for $k'(3)$ ?
Answer: $2$ The derivative $k'(3)$ can be estimated using a difference quotient. The given information $k(3) - k(3-h) \approx 2h$ can be rearranged by dividing by $h$ : $\frac{k(3) - k(3-h)}{h} \approx 2$ . The left side is the backward difference quotient $\frac{k(3) - k(3-h)}{3 - (3-h)}$ , which is an approximation for $k'(3)$ . Therefore, the best estimate for $k'(3)$ is 2.
Based on this information, which of the following is the best estimate for the rate of change of the temperature, in degrees Celsius per minute, at time $t=5$ minutes?
Answer: $16$ The rate of change of the temperature at $t=5$ can be estimated by the average rate of change over the interval $[5, 5.5]$ . $T'(5) \approx \frac{T(5.5) - T(5)}{5.5 - 5} = \frac{48 - 40}{0.5} = \frac{8}{0.5} = 16$ degrees Celsius per minute.
Based on this information, what is a possible estimate for the acceleration of the particle at $t=2$ seconds, in m/s $^2$ ?
Answer: $5$ Acceleration is the derivative of velocity, so we need to estimate $a(2) = v'(2)$ . We can use the average rate of change of velocity over the interval $[2, 2.1]$ . $a(2) \approx \frac{v(2.1) - v(2)}{2.1 - 2} = \frac{10.5 - 10}{0.1} = \frac{0.5}{0.1} = 5$ m/s $^2$ .
Let $f$ be a differentiable function. The average rate of change of $f$ on the interval $[c, c+h]$ is given by the expression $4 - 2h$ . Which of the following is the value of $f'(c)$ ?
Answer: $4$ The instantaneous rate of change, $f'(c)$ , is the limit of the average rate of change as the interval length $h$ approaches 0. We are given that the average rate of change on $[c, c+h]$ is $\frac{f(c+h)-f(c)}{h} = 4-2h$ . Taking the limit as $h \to 0$ gives: $f'(c) = \lim_{h \to 0} (4 - 2h) = 4 - 2(0) = 4$ .
Let $f$ be a differentiable function. Given $f(x_0 - \Delta x) = y_1$ and $f(x_0 + \Delta x) = y_2$ , where $\Delta x > 0$ . Which expression represents the best estimate for $f'(x_0)$ ?
Answer: $\frac{y_2 - y_1}{2 \Delta x}$ The best estimate for the derivative at a point $x_0$ is the slope of the secant line over a small symmetric interval around $x_0$ . The interval is $[x_0 - \Delta x, x_0 + \Delta x]$ . The change in $y$ is $y_2 - y_1$ . The change in $x$ is $(x_0 + \Delta x) - (x_0 - \Delta x) = 2 \Delta x$ . Therefore, the slope is $\frac{y_2 - y_1}{2 \Delta x}$ .
To estimate the value of $f'(2)$ for the function $f(x) = x^3 - x$ , an analyst uses the slope of the secant line connecting the points on the graph of $f$ at $x=1$ and $x=3$ . What is the value of this estimate?
Answer: $12$ First, find the function values at $x=1$ and $x=3$ . $f(1) = 1^3 - 1 = 0$ . $f(3) = 3^3 - 3 = 27 - 3 = 24$ . The slope of the secant line is the average rate of change between these points: $m_{sec} = \frac{f(3) - f(1)}{3 - 1} = \frac{24 - 0}{2} = 12$ . This value is an estimate for $f'(2)$ .
If $f(x) = x^3$ , which of the following is the value of $D(h)$ used to approximate $f'(c)$ ?
Answer: $3c^2 + h^2$ We substitute $f(x)=x^3$ into the expression for $D(h)$ : $D(h) = \frac{(c+h)^3 - (c-h)^3}{2h}$ $= \frac{(c^3+3c^2h+3ch^2+h^3) - (c^3-3c^2h+3ch^2-h^3)}{2h}$ $= \frac{6c^2h + 2h^3}{2h}$ $= 3c^2 + h^2$ . As $h \to 0$ , this expression approaches $3c^2$ , which is $f'(c)$ .
Using the secant line between $x=4.98$ and $x=5.02$ as an approximation to the tangent line at $x=5$ , what is the estimated value of $g'(5)$ ?
Answer: $2.50$ The slope of the tangent line at $x=5$ , which is $g'(5)$ , is approximated by the slope of the secant line through the nearby points at $x=4.98$ and $x=5.02$ . $g'(5) \approx \frac{g(5.02) - g(4.98)}{5.02 - 4.98} = \frac{1.35 - 1.25}{0.04} = \frac{0.10}{0.04} = 2.5$ .

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