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.

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.

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 $D(x)=k(x)$.

(a) Since $f(0)=2$, we solve

$$\begin{aligned} 2&=\sqrt{\log^2(0+p)}\\ &=\sqrt{\log^2 p}\\ &=\left|\log p\right|\\ \log p &= \pm 2\\ p&=e^{\pm 2}\\ p&=e^2 \text{ or } p=\frac{1}{e^2} \end{aligned}$$

We know that $p$ is $e^2$ or $\dfrac{1}{e^2}$, but we have to decide between the two. In both cases, $f(0)=2$. Let's consider the domain of $f(x)$. Since $\log^2(x+p)$ is never negative, the square root does not restrict our domain. However, we can only take the logarithm of positive numbers. Therefore, the domain is

$$\begin{aligned} x \text{ such that } &x+p>0\\ x \text{ such that } &x>-p\\ \end{aligned}$$

If $p=\dfrac{1}{e^2}$, then the domain of $f(x)$ is $\left(-\dfrac{1}{e^2},\infty\right)$. In particular, since $-\dfrac{1}{e^2}>-1$, the domain of $f(x)$ does not include $x=-1$. However, it is clear from the graph that $f(-1)$ exists. So, $p=e^2$.

(b) Now, we need to figure out what $b$ is. Notice that $b$ is the end of the domain of $f(x)$, which we already found to be $(-p,\infty)$. So, $b=-p=-e^2$.

(As a quick check, if we take $e\approx 2.7$, then $-e^2=-7.29$, and this looks about right on the graph.)

(c) The $x$-intercept is the value of $x$ for which $f(x)=0$:

$$\begin{aligned} 0&=\sqrt{\log^2(x+p)}\\ 0&=\log(x+p)\\ 1&=x+p\\ x&=1-p=1-e^2 \end{aligned}$$

The $x$-intercept is $1-e^2$.

(As another quick check, the $x$-intercept we found is a distance of 1 from the vertical asymptote, and this looks about right on the graph.)

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).