롤의 정리와 평균값 정리

The Mean Value Theorem does not apply.

$c=4/\ln 5$

$c=\pm \sec^{-1} (2/\sqrt{\pi})$

$c=-2/3$

$c=\frac{5-7\sqrt{7}}{6}$

$c=\frac{\pm\sqrt{\pi^2-4}}{\pi}$

We know the top speed of the caribou, so we can use this to give the minimum possible number of hours the caribou spent travelling during its migration. If the caribou travels at 70kph, it will take $(5000 \mathrm{km})\left(\dfrac{1 \mathrm{hr}}{70 \mathrm{km}}\right) \approx 71.4 \mathrm{hrs}$ to travel 5000 kilometres. Probably the caribou wasn't sprinting the whole time, so probably it took it longer than that, but we can only say for sure that the caribou spent at least about 71.4 hours migrating.

If $f(x)$ is the position of the crane at time $x$, measured in hours, then (if we let $x=0$ be the beginning of the day) we know that $f(24)-f(0)=240$. Since $f(x)$ is the position of the bird, $f(x)$ is continuous and differentiable. So, the MVT says there is a $c$ in $(0,24)$ such that $f'(x)=\dfrac{f(24)-f(0)}{24-0}=\dfrac{240}{24}=10$. That is, at some point $c$ during the day, the speed of the crane was exactly 10 kph.

Since $f(x)$ is differentiable for all $x \in (0,10)$, then $f(x)$ is also continuous for all $x \in (0,10)$. If $f(x)$ were continuous on the closed interval $[0,10]$, then the MVT would guarantee $f'(x)=\dfrac{f(10)-f(0)}{10-0}=1$ for some $c \in (0,10)$; however, this is not the case. So, it must be that $f(x)$ is continuous for all $x \in (0,10)$, but not for all $x \in [0,10]$.

Since $f'(c)=0$ for $c \in (0,10)$, that means $f$ is constant on that interval. So, $f(x)$ is a function like this:

where the height of the constant function can be anything.

So, one possible answer is $f(x) = \left\{\begin{array}{lr} 0&x \neq 10\\ 10&x=10 \end{array}\right.$.

(a) No such function is possible: Rolle's Theorem guarantees $f'(c)=0$ for at least one point $c \in (1,2)$.

For the other functions, examples are below, but many answers are possible.

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 $$f'(c)=\frac{f(\pi/4)- f(0)}{\pi/4 - 0} = 0.$$

By inspection, we see $x=0$ is a root of $f(x)$. The question now is whether there could possibly be other roots. Since $f(x)$ is differentiable over all real numbers, if there is another root $a$, then by Rolle's Theorem, $f'(c)=0$ for some $c$ strictly between $0$ and $a$. However, $f'(x)=3-\cos x$ is never zero. So, there is no second root: $f(x)$ has precisely one root.

The function $f(x)$ is continuous and differentiable over all real numbers. If $a$ and $b$ are distinct roots of $f(x)$, then $f'(c)=0$ for some $c$ strictly between $a$ and $b$ (Rolle's Theorem). So, let's think about $f'(x)$. $$f'(x)=(4x+1)^3+1$$ This is simple enough that we can find its zero explicitly: $$(4x+1)^3 +1= 0 \quad \Leftrightarrow \quad (4x+1)^3 = -1 \quad \Leftrightarrow \quad 4x+1 = -1 \quad \Leftrightarrow \quad 4x=-2$$ Hence $f'(c)=0$ only when $c=\dfrac{-1}{2}$. That the derivative only has a single zero is very useful. It means (via Rolle's theorem) that if $f(x)$ has distinct roots $a$ and $b$ with $a<b$, then we must have $a < \dfrac{-1}{2} < b$. This also means that $f(x)$ cannot have 3 distinct roots $a,b$ and (say) $q$ with $a<b<q$, because then Rolle's theorem would imply that $f'(x)$ would have two zeros — one between $a$ and $b$ and another between $b$ and $q$.

We've learned that $f(x)$ has at most two roots, and we've learned something about where those roots can exist, if there are indeed two of them. But that means $f(x)$ could have 0, 1, or 2 roots.

It's not easy to find a root of $f(x)$ by inspection. But we can get a good enough picture of the graph of $y=f(x)$ to tell exactly how many roots there are, just by exploiting the following properties of $f(x)$ and $f'(x)$.

• As $x$ tends to $\pm \infty$, $f(x)$ tends to $+\infty$.
• The derivative $f'(x)=(4x+1)^3+1$ is negative for $x<-\frac{1}{2}$ and is positive for $x>-\frac{1}{2}$. That is, $f(x)$ is decreasing for $x<-\frac{1}{2}$ and increasing for $x>-\frac{1}{2}$.
• $f\left(-\frac{1}{2}\right)=\frac{1}{16}-\frac{1}{2}<0$.

This means the function must look something like the graph below:

except we are unsure of the locations of the $x$-intercepts. At present we only know that one is to the left of $x=-1/2$ and one is to the right.

Note what happens to $f(x)$ as $x$ increases from strongly negative values to strongly positive values.

• When $x$ is large and negative, $f(x)>0$.
• As $x$ increases, $f(x)$ decreases continuously until $x=-\frac{1}{2}$, where $f(x)<0$. In particular, since $f(-1)=\frac{65}{16}>0$ and $f\big(-\frac{1}{2}\big)<0$ and $f(x)$ is continuous, the intermediate value theorem guarantees that $f(x)$ takes the value zero for some $x$ between $-\frac{1}{2}$ and $-1$. More descriptively put, as $x$ increases from hugely negative numbers to $-\frac{1}{2}$, $f(x)$ passes through zero exactly once.
• As $x$ increases beyond $-\frac{1}{2}$, $f(x)$ increases continuously, starting negative and becoming very large and positive when $x$ becomes large and positive. In particular, since $f(0)=\frac{1}{16}>0$ and $f\big(-\frac{1}{2}\big)<0$ and $f(x)$ is continuous, the intermediate value theorem guarantees that $f(x)$ takes the value zero for some $x$ between $-\frac{1}{2}$ and $0$. So, as $x$ increases from $-\frac{1}{2}$ to near $+\infty$, $f(x)$ again passes through zero exactly once.

So $f(x)$ must have exactly two roots, one with $x<-\frac{1}{2}$ and one with $x>-\frac{1}{2}$.