미분 전체 문제

Many answers are possible; here is one.

The key is to realize that the few points you're given suggest a pattern, but don't guarantee it. You only know nine points; anything can happen in between.

The line $y=mx+b$ is tangent to $y=x^2$ at $x=\alpha$ if $$2\alpha=m\text{ and }\alpha^2=m\alpha+b \iff m=2\alpha\text{ and }b=-\alpha^2$$ The same line $y=mx+b$ is tangent to $y=-x^2+2x-5$ at $x=\beta$ if

$$\begin{aligned} -2\beta+2=m&\text{ and }-\beta^2+2\beta-5=m\beta+b\\ \iff m=2-2\beta&\text{ and }b=-\beta^2+2\beta-5-(2-2\beta)\beta=\beta^2-5 \end{aligned}$$

For the line to be simultaneously tangent to the two parabolas we need $$m=2\alpha=2-2\beta\text{ and }b=-\alpha^2=\beta^2-5$$ Substituting $\alpha=1-\beta$ into $-\alpha^2=\beta^2-5$ gives $-(1-\beta)^2=\beta^2-5$ or $2\beta^2-2\beta-4=0$ or $\beta=-1,2$. The corresponding values of the other parameters are $\alpha=2,-1$, $m=4,-2$ and $b=-4,-1$. The two lines are $y=4x-4$ and $y=-2x-1$.

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.

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.

(a) First, we differentiate. $$f(x) = x^5-x\qquad f'(x)=5x^4-1\qquad f''(x)=20x^3\qquad$$ The function and its first derivative tells us the following:

• $\displaystyle \lim_{x \to \infty} f(x)=\infty$, $\displaystyle \lim_{x \to -\infty} f(x)=-\infty$
• $f'(x)> 0$ (i.e. $f$ is increasing) for $|x|> \dfrac{1}{\sqrt[4]{5}}$
• $f'(x)=0$ (i.e. $f$ has critical points) for $x= \pm\dfrac{1}{\sqrt[4]{5}} \approx \pm 0.67$
• $f'(x)< 0$ (i.e. $f$ is decreasing) for $|x| < \dfrac{1}{\sqrt[4]{5}}$
• $f\left(\pm\dfrac{1}{\sqrt[4]{5}}\right)=\mp\dfrac{4}{5\sqrt[4]{5}}\approx\mp0.53$

This gives us a first idea of the shape of the graph.

We refine this skeleton using information from the second derivative.

• $f''(x)> 0$ (i.e. $f$ is concave up) for $x> 0$,
• $f''(x)=0$ (i.e. $f$ has an inflection point) for $x=0$, and
• $f''(x)< 0$ (i.e. $f$ is concave down) for $x<0$

Thus

• $f$ has no asymptotes
• $f$ has a local maximum at $x=-\dfrac{1}{\sqrt[4]{5}}$ and a local minimum at $x=\dfrac{1}{\sqrt[4]{5}}$
• $f$ has an inflection point at $x=0$
• $f$ is concave down for $x<0$ and concave up for $x>0$

(b) The function $x^5-x+k$ has a root at $x=x_0$ if and only if $x^5-x=-k$ at $x=x_0$. So the number of distinct real roots of $x^5-x+k$ is the number of times the curve $y=x^5-x$ crosses the horizontal line $y=-k$. The local maximum of $x^5-x$ (when $x=-\dfrac{1}{\sqrt[4]{5}}$) is $\dfrac{4}{5\sqrt[4]{5}}$, and the local minimum of $x^5-x$ (when $x=\dfrac{1}{\sqrt[4]{5}}$) is $-\dfrac{4}{5\sqrt[4]{5}}$. So, looking at the graph of $x^5-x$ above, we see that the number of distinct real roots of $x^5-x+k$ is

• 1 when $|k|>\dfrac{4}{5\sqrt[4]{5}}$
• 2 when $|k|=\dfrac{4}{5\sqrt[4]{5}}$
• 3 when $|k|<\dfrac{4}{5\sqrt[4]{5}}$