미분 전체 문제

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 $f'\left(\dfrac{169}{4}\right)=\dfrac{1}{13}$.

False. The acceleration of the ball is given by $h''(t)=-9.8$. This is constant throughout its trajectory (and is due to gravity).

Remark: the velocity of the ball at $t=2$ is zero, since $h'(2)=-9.8(2)+19.6=0$, but the velocity is only zero for an instant. Since the velocity is changing, the acceleration is nonzero.

The acceleration is constant, which means the rate of change of the velocity is constant. So, since it took 10 seconds for the velocity to increase by 1 metre per second (from $1~\frac{\mathrm{m}}{\mathrm{s}}$ to $2~\frac{\mathrm{m}}{\mathrm{s}}$), then it always takes 10 seconds for the velocity to increase by 1 metre per second.

So, it takes 10 seconds to accelerate from $2~\frac{\mathrm{m}}{\mathrm{s}}$ to $3~\frac{\mathrm{m}}{\mathrm{s}}$. To accelerate from $3~\frac{\mathrm{m}}{\mathrm{s}}$ to $13~\frac{\mathrm{m}}{\mathrm{s}}$ (that is, to change its velocity by 10 metres per second), it takes $10\times 10=100$ seconds.

When $x=0$, the curve $y=f(x)$ appears to have a flat tangent line, so the $x=0$ is a critical point. However, it is not a local extremum: it is not true that $f(0) \geq f(x)$ for all $x$ near 0, and it is not true that $f(0) \leq f(x)$ for all $x$ near 0.

To the right of the $x$-axis, there is a spike where the derivative of $f(x)$ does not exist. The $x$-value corresponding to this spike (call it $a$) is a singular point, and $f(x)$ has a local maximum at $x=a$.

The $x$-coordinate corresponding to the blue dot (let's call it $a$) is a critical point, because the tangent line to $f(x)$ at $x=a$ is horizontal. There is no lower point nearby, and actually no lower point on the whole interval shown, so $f(x)$ has both a local minimum and a global minimum at $x=a$.

If a function is not continuous at a point, then it is not differentiable at that point. So, the $x$-coordinate corresponding to the discontinuity (let's call it $b$) is a singular point. Values of $f(x)$ immediately to the right of $b$ are lower, and values immediately to the left of $b$ are higher, so $f(x)$ has no local (or global) extremum at $x=b$.

One possible answer is shown below.

For every $x$ in the red interval shown below, $f(2) \geq f(x)$, so $f(2)$ is a local maximum. However, the point marked with a blue dot shows that $f(x)>f(2)$ for some $x$, so $f(2)$ is not a global maximum.

Two examples are given below, but many are possible.

If $f(x)=-x^2$ or $f(x)=-\sqrt{|x|}$, then $f(x)$ has a global maximum at $x=0$. Since $f(x)$ keeps getting more and more strongly negative as $x$ gets farther and farther from 0, $f(x)$ has no global minimum.

Two examples are given below, but many are possible.

If $f(x)=e^x$, then $f(x) > 0$ for all $x$. As we move left along the $x$-axis, $f(x)$ gets smaller and smaller, approaching 0 but never reaching it. Since $f(x)$ gets smaller and smaller as we move left, there is no global minimum. Likewise, $f(x)$ increases more and more as we move right, so there is no maximum.

If $f(x)=\arctan(x)+2$, then $f(x) >\left(-\frac{\pi}{2}\right)+2>0$ for all $x$.

As we move left along the $x$-axis, $f(x)$ gets smaller and smaller, approaching $\left(-\frac{\pi}{2}+2\right)$ but never reaching it. Since $f(x)$ gets smaller and smaller as we move left, there is no global minimum.

Likewise, as we move right along the $x$-axis, $f(x)$ gets bigger and bigger, approaching $\left(\frac{\pi}{2}+2\right)$ but never reaching it. Since $f(x)$ gets bigger and bigger as we move right, there is no global maximum.

Since $f(5)$ is a global minimum, $f(5) \leq f(x)$ for all $x$, and so in particular $f(5) \leq f(-5)$.

Similarly, $f(-5) \leq f(x)$ for all $x$, so in particular $f(-5) \leq f(5)$.

Since $f(-5) \leq f(5)$ AND $f(5) \leq f(-5)$, it must be true that $f(-5)=f(5)$.

A sketch of one such graph is below.

In general, this is false. For example, the function $f(x) = \dfrac{x^2-9}{x^2-9}$ has no vertical asymptotes, because it is equal to 1 in every point in its domain (and is undefined when $x=\pm3$).

However, it is certainly possible that $f(x)$ has a vertical asymptote at $x=-3$. For example, $f(x)=\dfrac{1}{x^2-9}$ has a vertical asymptote at $x=-3$. More generally, if $g(x)$ is continuous and $g(-3)\ne 0$, then $f(x)$ has a vertical asympotote at $x=-3$.

Functions $A(x)$ and $B(x)$ share something in common that sets them apart from the others: they have a horizontal tangent line only once. In particular, $A'(-2) \neq 0$ and $B'(2) \neq 0$. The only listed functions that do not have two distinct roots are $l(x)$ and $p(x)$. Since $l(-2) \neq 0$ and $p(2) \neq 0$, we conclude $$A'(x)=l(x) \qquad B'(x)=p(x)$$

Function $C(x)$ is never decreasing. Its tangent line is horizontal when $x = \pm 2$, but the curve never decreases, so $C'(x) \geq 0$ for all $x$ and $C'(2)=C'(-2)=0$. The only function that matches this is $n(x)=(x-2)^2(x+2)^2$. Since its linear terms have even powers, it is never negative, and its roots are precisely $x=\pm 2$. $$C'(x)=n(x)$$

For the functions $D(x)$ and $E(x)$ we consider their behaviour near $x=0$. $D(x)$ is decreasing near $x=0$, so $D'(0)<0$, which matches with $o(0)<0$. Contrastingly, $E(x)$ is increasing near zero, so $E'(0)>0$, which matches with $m(0)>0$. $$D'(x)=o(x)\qquad E'(x)=m(x)$$

This function is symmetric across the $y$-axis, so it is even.

The function is not even, because it is not mirrored across the $y$-axis.

Assuming it continues as shown, the function is periodic, because the unit shown below is repeated:

Additionally, $f(x)$ is odd. In a function with odd symmetry, if we mirror the right-hand portion of the curve (the portion to the right of the $y$-axis) across both the $y$-axis and the $x$-axis, it lines up with the left-hand portion of the curve.

Since reflecting the right-hand portion of the graph across the $y$-axis, then the $x$-axis, gives us $f(x)$, we conclude $f(x)$ is odd.

Since the function is even, we simply reflect the portion shown across the $y$-axis to complete the sketch.

Since the function is odd, to complete the sketch, we reflect the portion shown across the $y$-axis (shown dashed), then the $x$-axis (shown in red).