롤의 정리와 평균값 정리

Let ${s}_{1}(t)$ represent the distance of Car 1 from Stoplight 1 at time $t$. Let ${s}_{2}(t)$ represent the distance of Car 2 from Stoplight 1 at time $t$. Then ${s}_{1}(0)=0={s}_{2}(0)$. Let $t=m$ represent the time that the cars arrive at Stoplight 2. Then ${s}_{1}(m)=d={s}_{2}(m)$ where $d$ is the distance between the stoplights.

Define a new function $f(t)={s}_{1}(t)-{s}_{2}(t)$. Then $f(0)=0=f(m)$. By the Mean Value Theorem, there is a time $0<c<m$ such that

$${f}^{'}(c)=\frac{f(m)-f(0)}{m-0}=\frac{0}{m}=0.$$

At time c,

$${f}^{'}(c)=0$$

$${s}_{1}^{'}(c)-{s}_{2}^{'}(c)=0$$

$${s}_{1}^{'}(c)={s}_{2}^{'}(c)$$

So, at time c, Car 1 and Car 2 are traveling at the same speed.

The function $f(x)$ is continuous over all real numbers, but it is only differentiable when $x \neq 0$. So, if we want to apply the MVT, our interval must consist of only positive numbers or only negative numbers: the interval $(-4,13)$ is not valid.

It is possible to use the mean value theorem to prove what we want: if $a=1$ and $b=144$, then $f(x)$ is differentiable over the interval $(1,144)$ (since 0 is not contained in that interval), and $f(x)$ is continuous everywhere, so by the mean value theorem there exists some point $c$ where $f'(x)=\dfrac{\sqrt{|144|}-\sqrt{|1|}}{144-1}=\dfrac{11}{143}=\dfrac{1}{13}$.

That being said, an easier way to prove that a point exists is to simply find it–without using the MVT. When $x>0$, $f(x)=\sqrt{x}$, so $f'(x)=\dfrac{1}{2\sqrt {x}}$. Then $f'\left(\dfrac{169}{4}\right)=\dfrac{1}{13}$.

We note that $f(0)=f(2\pi)=0$. Then using the Mean Value Theorem (note that the function is differentiable for all real numbers), we conclude that there exists $c$ in $(0,2\pi)$ such that $$f'(c)=\frac{f(2\pi)- f(0)}{2\pi - 0} = 0.$$

We note that $f(0)=f(1)=0$. Then using the Mean Value Theorem (note that the function is differentiable for all real numbers), we get that there exists $c\in (0,1)$ such that $$f'(c)=\frac{f(1)- f(0)}{1 - 0} = 0.$$

We note that $f(0)=f(2\pi)=\sqrt{3} + \pi^2$. Then using the Mean Value Theorem (note that the function is differentiable for all real numbers since $3+\sin x>0$), we get that there exists $c\in (0,2\pi)$ such that $$f'(c)=\frac{f(2\pi)- f(0)}{2\pi - 0} = 0.$$

We note that $f(0)=0$ and $f(\pi/4)=0$. Then using the Mean Value Theorem (note that the function is differentiable for all real numbers), we get that there exists $c\in (0,\pi/4)$ such that $$f'(c)=\frac{f(\pi/4)- f(0)}{\pi/4 - 0} = 0.$$

Define $h(x)=f(x)-g(x)$, and notice $h(a)=f(a)-g(a)<0$ and $h(b)=f(b)-g(b)>0$. Since $h$ is the difference of two functions that are continuous over $[a,b]$ and differentiable over $(a,b)$, also $h$ is continuous over $[a,b]$ and differentiable over $(a,b)$. So, by the Mean Value Theorem, there exists some $c \in (a,b)$ with $$h'(c)=\frac{h(b)-h(a)}{b-a}$$ Since $(a,b)$ is an interval, $b>a$, so the denominator of the above expression is positive; since $h(b)>0>h(a)$, also the numerator of the above expression is positive. So, $h'(c)>0$ for some $c \in (a,b)$. Since $h'(c)=f'(c)-g'(c)$, we conclude $f'(c)>g'(c)$ for some $c \in (a,b)$.