평균변화율과 미분계수

(a) To find the average rate of change of $h$ on the interval $[40000,55000]$, we divide the change in $h$ values (in dollars) on this interval by the change in $m$ values (in miles) to obtain $$AV_{[40000,55000]} = \frac{h(55000)- h(40000)}{55000-40000} = \frac{13200-15500}{15000} = -\frac{2300}{15000} \approx -0.153$$ dollars per mile. This tells us that from the mileage of $40 000$ to $50 000$ miles, on average the car's value decreases by $0.153$ dollars per mile driven. (b) Since $40 000$ and $70 0000$ are equidistant from $55 000$, we use the central difference quotient to approximate the derivative of $h$ at $55 0000$ to obtain $$h'(55000) \approx \frac{h(70000) - h(40000)}{70000-40000} = \frac{11100-15500}{30000} = - \frac{4400}{30000} \approx -0.147$$ dollars per mile. This approximation tells us that at the instant the car has been driven $55 000$ miles, the car's value is decreasing at an instantaneous rate of $0.147$ dollars per mile. That is, during $55 0001$st mile, we'd expect the car's value to drop by $0.147$ dollars. (c) As we pile up miles on our car, we expect the value to go down. However, we should expect the depreciation to be more extreme at first, but less in magnitude as we drive more miles and our car gets older. Note, though, the the value of the car is always decreasing and $h'(m)$ is always negative. Thus, we should expect that the rate of change of $h$ is increasing (becoming less negative) as $m$ increases and that $h'(30000) < h'(80000)$. (d) The value of a car is never negative. So we know that $h(m) \geq 0$ for all $m$. A new car decreases in value very quickly at first, and as the car gets older with more mileage, its value will depreciate to near zero. So $h$ is a decreasing function. Since the value of our car always decreases, it follows that $h'(m) < 0$ for all $m > 0$. The fact that the depreciation slows as $m$ increases implies that $h'$ is an increasing (getting less negative) function. Since $h'$ is increasing, it follows that $h$ is bends upward; we will study this issue more in Section 1.6. So the graph of $h$ might have the general shape of the graph of $y = e^{-x}$ for positive values of $x$.