I have been reading Calculus with Analytical Geometry by George F. Simmons. One of the practice problems is as follows:
A car starts from rest and travels 4mi along a straight road in 6 minutes. Use the Mean Value Theorem to show that at some moment during the trip its speed was exactly 40 mi/h.
Generally, to find a point $c$ that satisfies the MVT for some function $f(x)$, I solve the following:
$$f'(c) = \frac{f(b) – f(a)}{b-a}$$
However, I'm unsure what function $f(x)$ to construct that will fit the terms of the problem other than $f(x) = \frac{2}{3}x$, which is equivalent to a vehicle traveling at $\frac{2}{3}$ a mile per minute.
The derivative of that function, however, is a constant, which is not useful.
Thank you.
Best Answer
MVT state that for an interval $[a,b]$ there is c such that $f^{\prime}(c)=\dfrac{f(b)-f(a)}{b-a}$
Let us calculate the car's speed in mph $$\dfrac{4\mbox{ miles}}{6\mbox{ minutes}}=\dfrac{x\mbox{ miles}}{60\mbox{ minutes}}$$ $$x=40\mbox{ mph}$$ This is the Average speed.
We need to show that at some time, the car's Average speed is equaled to the Instantaneous speed. By MVT we can say that average rate of change is, at some time, is equal to the instantaneous rate of change. Hence proved.
Edit: Letting $f(t)$ be distance at time $t$ minutes,
$f(0)=0$
$f(6)=4$
$\dfrac{f(6)-f(0)}{6-0}=\dfrac23\dfrac{\mbox{ miles}}{\mbox{ minute}}$ which is $40$ mph
MVT states that at some $0\le c\le6,f^{\prime}(x)=\dfrac23$