미분 전체 문제

Different forces are acting on you (1) after you jump but before you land on the trampoline, and (2) while you are falling into the trampoline. In both instances, the acceleration is constant, so both height functions are quadratic, of the form $\frac{a}{2}t^2+vt+h$, where $a$ is the acceleration, $v$ is the velocity when $t=0$, and $h$ is the initial height.

• Let's consider (1) first, the time during your jump before your feet touch the trampoline. Let $t=0$ be the moment you jump, and let the rim of the trampoline be height $0$. Then, since your initial velocity was (positive) $1$ meter per second, your height is given by $$h_1(t)=\frac{-9.8}{2}t^2+t=t\left(\frac{-9.8}{2}t+1\right)$$ Notice that, because your acceleration is working against your positive velocity, it has a negative sign.
• We'll need to know your velocity when your feet first touch the trampoline on your fall. The time your feet first first touch the trampoline after your jump is precisely when $h_1(t)=0$ and $t>0$. That is, when $t=\frac{2}{9.8}$. Now, since $h'(t)=-9.8t+1$, $h'\left(\frac{2}{9.8}\right)=-9.8\left(\frac{2}{9.8}\right)+1=-1$. So, you are descending at a rate of 1 metre per second at the instant your feet touch the trampoline.

Remark: it is not only coincidence that this was your initial speed. Think about the symmetries of parabolas, and conservation of energy.

• Now we need to think about your height as the trampoline is slowing your fall. One thing to remember about our general equation $\frac{a}{2}t^2+vt+h$ is that $v$ is the velocity when $t=0$. But, you don't hit the trampoline at $t=0$, you hit it at $t=\frac{2}{9.8}$. In order to keep things simple, let's use a different time scale for this second part of your journey. Let's let $h_2(T)$ be your height at time $T$, from the moment your feet touch the trampoline skin ($T=0$) to the bottom of your fall. Now, we can use the fact that your initial velocity is $-1$ metres per second (negative, since your height is decreasing) and your acceleration is $4.9$ metres per second per second (positive, since your velocity is increasing from a negative number to zero): $$h_2(T)=\frac{4.9}{2}T^2-T$$ where still the height of the rim of the trampoline is taken to be zero.

Remark: if it seems very confusing that your free-falling acceleration is negative, while your acceleration in the trampoline is positive, remember that gravity is pushing you down, but the trampoline is pushing you up.

• How long were you falling in the trampoline? The equation $h_2(T)$ tells you your height only as long as the trampoline is slowing your fall. You reach the bottom of your fall when your velocity is zero. $$h_2'(T)=4.9T-1$$ so you reach the bottom of your fall at $T=\frac{1}{4.9}$. Be careful: this is $\frac{1}{4.9}$ seconds after you entered the trampoline, not after the peak of your fall, or after you jumped.
• The last piece of the puzzle is how long it took you to fall from the peak of your jump to the surface of the trampoline. We know the equation of your motion during that time: $h_1(t)=\frac{-9.8}{2}t^2+t$. You reached the peak when your velocity was zero: $$h_1'(t)=-9.8t+1 =0\qquad\Rightarrow\qquad t=\frac{1}{9.8}$$ So, you fell from your peak at $t=\frac{1}{9.8}$ and reached the level of the trampoline rim at $t=\frac{2}{1.98}$, which means the fall took $\frac{1}{9.8}$ seconds.

Remark: by the symmetry mentioned early, the time it took to fall from the peak of your jump to the surface of the trampoline is the same as one-half the time from the moment you jumped off the rim to the moment you're back on the surface of the trampoline.

• So, your time falling from the peak of your jump to its bottom was $\frac{1}{9.8}+\frac{1}{4.9}\approx 0.3$ seconds.

Let $v(t)$ be the velocity of the object. From the given information:

• $v(0)$ is some value, call it $v_0$,
• $v(1)=2v_0$ (since the speed doubled in the first second),
• $v(2)=2(2)v_0$ (since the speed doubled in the second second),
• $v(3)=2(2)(2)v_0$, and so on. So, for general $t$: $$v(t)=2^tv(0)$$ To find its acceleration, we simply differentiate. Recall $\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}\{2^x\}=2^x\log 2$, where $\log$ denotes logarithm base $e$. $$a(t)=2^tv_0\log 2$$ Remark: we can also write $a(t)=v(t)\log 2$. The acceleration doubles every second as well.

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.

For the first curve, the function's value at $x=2$ (that is, the $y$-value of the solid dot) is higher than anything around it. So, it's a local maximum.

For the second curve, the function's value at $x=2$ (that is, the $y$-value of the solid dot) is higher than everything to the left, but lower than values immediately to the right. (On the graph reproduced below, $f(x)$ is higher than everything in the red section, and lower than everything in the blue section.) So, it is neither a local max nor a local min.

Similarly, for the third curve, $f(2)$ is lower than the values to the right of it, and higher than values to the left of it, so it is neither a local minimum nor a local maximum.

In the final curve, $f(2)$ (remember–this is the $y$-value of the solid dot) is higher than everything immediately to the left or right of it (for instance, over the interval marked in red below), so it is a local 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.

Global extrema will occur at critical or singular points in the interval $(-5,5)$ or at the endpoints $x=5,\,x=-5$.

$f'(x)=2x+6$. Since this is defined for all real numbers, there are no singular points. The only time $f'(x)=0$ is when $x=-3$. This is inside the interval $[-5,5]$. So, our points to check are $x=-3,\,x=-5,$ and $x=5$.

$$\begin{array}{|c||c|c|c|}\hline c & -3 & -5 & 5 \\ \hline \text{type} & \text{critical point} & \text{endpoint} & \text{endpoint} \\ \hline f(c) & -19 & -15 & 45 \\ \hline \end{array}$$

The global maximum is 45 at $x=5$ and the global minimum is $-19$ at $x=-3$.

Global extrema will occur at the endpoints of the interval, $x=-4$ and $x=0$, or at singular or critical points inside the interval. Since $f(x)$ is a polynomial, it is differentiable everywhere, so there are no singular points. To find the critical points, we set the derivative equal to zero.

$$\begin{aligned} f'(x)&=2x^2-4x-30\\ 0&=2x^2-4x-30 &=(2x-10)(x+3)\\ x&=5,\,-3 \end{aligned}$$

The only critical point inside the interval is $x=-3$.

$$\begin{array}{|c||c|c|c|}\hline c & -3 & -4 & 0 \\ \hline \text{type} & \text{critical point} & \text{endpoint} & \text{endpoint} \\ \hline f(c) & 61 & \frac{157}{3}=52+\frac{1}{3} & 7 \\ \hline \end{array}$$

The global maximum over the interval is $61$ at $x=-3$, and the global minimum is $7$ at $x=0$.

Call the length of the wire $L$ units and suppose that it is cut $\ell$ units from one end. Make the square from the piece of length $\ell$, and make the circle from the remaining piece of length $L-\ell$.

The square has perimeter $\ell$, so its side length is $\ell/4$ and its area is $\left(\dfrac{\ell}{4}\right)^2$. The circle has circumference $L-\ell$, so its radius is $\dfrac{L-\ell}{2\pi}$ and its area is $\pi\left(\dfrac{L-\ell}{2\pi}\right)^2=\dfrac{(L-\ell)^2}{4\pi}$.

The area enclosed by the shapes, when the square is made from a length of size $\ell$, is

$$\begin{aligned} A(\ell)&=\dfrac{\ell^2}{16}+\dfrac{(L-\ell)^2}{4\pi} \\ \text{We want to find the global max and min for this function, given the constraint $0 \leq \ell \leq L$, so we find its derivative:} \\ A'(\ell)&=\dfrac{\ell}{8} -\dfrac{L-\ell}{2\pi} =\dfrac{\pi+4}{8\pi}\ell-\dfrac{L}{2\pi} \\ \text{Now, we find the critical point.} \\ A'(\ell)&=0\\ \frac{\pi+4}{8\pi}\ell&=\frac{L}{2\pi}\\ \ell&=\frac{4L}{\pi+4} \end{aligned}$$

$$\begin{array}{|c||c|c|c|}\hline \ell & 0 & L & \dfrac{.^.}{.}\frac{4L}{\pi+4} \\ \hline \text{type} & \text{endpoint} & \text{endpoint} & \text{critical point} \\ \hline A(\ell) & \dfrac{.^.}{.}\frac{L^2}{4\pi} & \frac{L^2}{16}_{} & A\left(\frac{4L}{\pi+4}\right) \\ \hline \end{array}$$

It seems obnoxious to evaluate $A\left(\frac{4L}{\pi+4}\right)$, and the problem doesn't ask for it–but we still have to figure out whether it is a global max or min.

When $\ell<\frac{4L}{\pi+4}$, $A'(\ell)<0$, and when $\ell>\frac{4L}{\pi+4}$, $A'(\ell)>0$. So, $A(\ell)$ is decreasing until $\ell=\frac{4L}{\pi+4}$, then increasing. That means our critical point $\ell=\frac{4L}{\pi+4}$ is a local minimum.

So, the minimum occurs at the only critical point, which is $\ell=\left(\dfrac{4}{4+\pi}\right)L$. This corresponds to $\dfrac{\ell}{L}=\dfrac{4}{4+\pi}$: the proportion of the wire that is cut is $\dfrac{4}{4+\pi}$.

The maximum has to be either at $\ell=0$ or at $\ell=L$. As $A(0)=\dfrac{L^2}{4\pi} > A(L)=\dfrac{L^2}{16}$, the maximum has $\ell=0$ (that is, no square).

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 $x=-3$.

Since $x^2+1$ and $x^2+4$ are always positive, $f(x)$ and $h(x)$ are defined over all real numbers. So, $f(x)$ and $h(x)$ correspond to $A(x)$ and $B(x)$. Which is which? $A(0)=1=f(0)$ while $B(0)=2=h(0)$, so $A(x)=f(x)$ and $B(x)=h(x)$.

That leaves $g(x)$ and $k(x)$ matching to $C(x)$ and $D(x)$. The domain of $g(x)$ is all $x$ such that $x^2-1\ge0$. That is, $|x|\ge1$, like $C(x)$. The domain of $k(x)$ is all $x$ such that $x^2-4\ge0$. That is, $|x|\ge2$, like $D(x)$. So, $C(x)=g(x)$ and $D(x)=k(x)$.