Estimating Derivatives of a Function
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AP Calculus AB › Estimating Derivatives of a Function
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?
$$80$$ per unit
$$30$$ per unit
$$40$$ per unit
$$2$$ 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.
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?
$$4$$ m/s$$^2$$
$$-2$$ m/s$$^2$$
$$-0.5$$ m/s$$^2$$
$$-4$$ m/s$$^2$$
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$$^2$$.
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?
$$V'(61) \approx 1.5$$, meaning the stock's value is increasing at a rate of approximately $$\1.50$$ per day.
$$V'(61) \approx -1.5$$, meaning the stock's value was $$\1.50$$ on day 61.
$$V'(61) \approx -1.5$$, meaning the stock's value is decreasing at a rate of approximately $$\1.50$$ per day.
$$V'(61) \approx -3$$, meaning the stock's value is decreasing by a total of $$\3.00$$ over two days.
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.
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 $$f''(2.6)$$?
$$-3$$
$$3$$
$$-0.3$$
$$-0.6$$
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: $$f''(2.6) \approx \frac{f'(2.7) - f'(2.5)}{2.7 - 2.5} = \frac{5.8 - 6.4}{0.2} = \frac{-0.6}{0.2} = -3$$.
A function $$f$$ is differentiable. If $$f(3) = 2$$ and $$f(3.05) = 2.4$$, which of the following is the best approximation for $$f'(3)$$?
$$0.4$$
$$0.125$$
$$8$$
$$0.05$$
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: $$f'(3) \approx \frac{f(3.05) - f(3)}{3.05 - 3} = \frac{2.4 - 2}{0.05} = \frac{0.4}{0.05} = 8$$.
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?
$$30$$ milligrams per hour
$$15$$ milligrams per hour
$$-30$$ milligrams per hour
$$-15$$ 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.
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 $$g'(5)$$?
$$4$$
$$0.25$$
$$-4$$
$$-0.8$$
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: $$g'(5) \approx \frac{g(5.1) - g(4.9)}{5.1 - 4.9} = \frac{9.4 - 10.2}{0.2} = \frac{-0.8}{0.2} = -4$$.
If $$T(8) = 60$$ and $$T(12) = 52$$, which of the following is the best estimate for $$T'(10)$$?
$$-8$$ degrees Celsius per minute
$$-4$$ degrees Celsius per minute
$$-2$$ degrees Celsius per minute
$$2$$ 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.
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 $$h'(2)$$?
$$2.2$$
$$2.6$$
$$3$$
$$2.0$$
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.
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?
$$525$$
$$450$$
$$500$$
$$1500$$
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.