미분 전체 문제

Define $h(x)=f(x)-g(x)$, and notice $h(a)=f(a)-g(a)<0$ and $h(b)=f(b)-g(b)>0$. Since $h$ is the difference of two functions that are continuous over $[a,b]$ and differentiable over $(a,b)$, also $h$ is continuous over $[a,b]$ and differentiable over $(a,b)$. So, by the Mean Value Theorem, there exists some $c \in (a,b)$ with $$h'(c)=\frac{h(b)-h(a)}{b-a}$$ Since $(a,b)$ is an interval, $b>a$, so the denominator of the above expression is positive; since $h(b)>0>h(a)$, also the numerator of the above expression is positive. So, $h'(c)>0$ for some $c \in (a,b)$. Since $h'(c)=f'(c)-g'(c)$, we conclude $f'(c)>g'(c)$ for some $c \in (a,b)$.

Since $f(x)$ is differentiable over all real numbers, it is also continuous over all real numbers. We claim that $f(x)$ cannot have four or more distinct roots. For every two distinct roots $a<b$, Rolle's Theorem tells us there is a $c \in (a,b)$ such that $f'(c)=0$: that is, $c$ is a root of $f'$. Since $f'$ has only two distinct roots, $f$ can have at most three distinct roots.

We are asked to find the number of solutions to the equation $x^2+5x+1 = -\sin x$. The figure below contains the graphs of $y=x^2+5x+1$ and $y=-\sin x$. The solutions to $x^2+5x+1 = -\sin x$ are precisely the $x$'s where $y=x^2+5x+1$ and $y=-\sin x$ cross.

From the figure, it sure looks like there are two crossings. Since the function $-\sin x$ has range $[-1,1]$, if the two functions cross, then also $-1 \leq x^2+5x+1\leq 1$. This portion of the quadratic function is highlighted in blue in the figure. The $x$ coordinates of the end points of the blue arcs are found by solving $x^2+5x+1 = \pm1$, i.e. $x=0$, $-5$, and (using the quadratic equation) $x= \frac{-5\pm\sqrt{17}}{2}$.

We are now in a position to exploit the intuition that we have built using the above figure to write a concise argument showing that $f(x)$ has exactly two roots. Remember that, in general, if we want to show that a function has $n$ roots, we need to show that there exist $n$ distinct roots somewhere, and that there do not exist $n+1$ distinct roots. This argument is given below, in blue text.

• $f(x)$ is continuous over all real numbers
• $f(x)=0$ only when $-\sin x=x^2+5x+1$, which only happens when $|x^2+5x+1| \leq 1$. Thus, $f(x)$ only has roots in the intervals $\left[-5,\frac{-5-\sqrt{17}}{2}\right]$ and $\left[\frac{-5+\sqrt{17}}{2},0\right]$.
• $f(-5)=\sin(-5)+1 > 0$, and $f\left(\frac{-5-\sqrt{17}}{2}\right)=\sin\left(\frac{-5-\sqrt{17}}{2}\right)-1 <0$. So, by the IVT, $f(c)=0$ for some $c \in \left(-5,\frac{-5-\sqrt{17}}{2}\right)$.
• $f(0)=1>0$, and $f\left(\frac{-5+\sqrt{17}}{2}\right)=\sin\left(\frac{-5+\sqrt{17}}{2}\right)-1<0$. So, by the IVT, $f(c)=0$ for some $c \in \left(\frac{-5+\sqrt{17}}{2},0\right)$.
• $f'(x)=\cos x + 2x + 5$. If $f'(x)=0$, then $2x+5=-\cos x$, so $|2x+5| \leq 1$. So, the only interval that can contain roots of $f'(x)$ is $[-3,-2]$.
• Suppose $f(x)$ has more than two roots. Then it has two roots in the interval $\left[-5,\frac{-5-\sqrt{17}}{2}\right]$ OR it has two roots in the interval $\left[\frac{-5+\sqrt{17}}{2},0\right]$. Since $f(x)$ is differentiable for all real numbers, Rolle's Theorem tells us that $f'(x)$ has a root in $\left(-5,\frac{-5-\sqrt{17}}{2}\right)$ or in $\left(\frac{-5+\sqrt{17}}{2},0\right)$. However, since all roots of $f'(x)$ are in the interval $[-3,-2]$, and this interval shares no points with $\left(-5,\frac{-5-\sqrt{17}}{2}\right)$ or $\left(\frac{-5+\sqrt{17}}{2},0\right)$, this cannot be the case. Therefore $f(x)$ does not have more than two roots.

• Since $f(x)$ has at least two roots, and not more than two roots, $f(x)$ has exactly two roots.

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.

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.

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).

Let's consider $g(x)=\tan(3x)$ and $h(x)=\sin(4x)$ separately. Recall that $\pi$ is the period of tangent.

$$\begin{aligned} \tan X &= \tan(X+\pi)&&\text{for every $X$ in the domain of $\tan X$} \\ \text{Replacing $X$ with $3x$:} \\ \tan(3x)&=\tan(3x+\pi)&&\text{for every $x$ in the domain of $\tan 3x$}\\ \tan(3x)&=\tan\left(3\left(x+\frac{\pi}{3}\right)\right)&&\text{for every $x$ in the domain of $\tan 3x$}\\ g(x)&=g\left(x+\frac{\pi}{3}\right)&&\text{for every $x$ in the domain of $\tan 3x$} \\ \text{So, the period of $g(x)=\tan(3x)$ is $\dfrac{\pi}{3}$.} \\ \\ \text{Similarly, $2\pi$ is the period of sine.} \\ \sin(X)&=\sin(X+2\pi)&&\text{for every $X$ in the domain of $\sin(X)$} \\ \text{Replacing $X$ with $4x$:} \\ \sin(4x)&=\sin(4x+2\pi)&&\text{for every $x$ in the domain of $\sin(4x)$}\\ \sin(4x)&=\sin\left(4\left(x+\frac{\pi}{2}\right)\right)&&\text{for every $x$ in the domain of $\sin(4x)$}\\ h(x)&=h\left(x+\frac{\pi}{2}\right)&&\text{for every $x$ in the domain of $\sin(4x)$} \end{aligned}$$

So, the period of $h(x)=\sin(4x)$ is $\dfrac{\pi}{2}$.

All together, $f(x)=g(x)+h(x)$ will repeat when both $g(x)$ and $h(x)$ repeat. The least common integer multiple of $\dfrac{\pi}{3}$ and $\dfrac{\pi}{2}$ is $\pi$. Since $g(x)$ repeats every $\dfrac{\pi}{3}$ units, and $h(x)$ repeats every $\dfrac{\pi}{2}$ units, they will not both repeat until we move $\pi$ units. So, the period of $f(x)$ is $\pi$.