미분 전체 문제

We are to find the number of positive solutions to the equation $e^x = 4\cos(2x)$. The figure below contains the graphs of $y=e^x$ and $y= 4\cos(2x)$. The solutions to $e^x = 4\cos(2x)$ are precisely the $x$'s where $y=e^x$ and $y= 4\cos(2x)$ cross.

It sure looks like there is exactly one crossing with $x\ge 0$ and that one crossing is somewhere between $x=0$ and $x=1$. Indeed since $\big[e^x- 4\cos(2x)\big]_{x=0} = -3<0$ and $\big[e^x- 4\cos(2x)\big]_{x=1} > e > 0$ and $f(x)=e^x- 4\cos(2x)$ is continuous, the intermediate value theorem guarantees that there is at least one root with $0<x<1$.

We still have to show that there is no second root — even if our graphs are not accurate.

Recall that the range of the cosine function is $[-1,1]$. If $e^x=4\cos(2x)$, then $e^x \leq 4$, so $x \leq \log(4)\approx 1.39$. So, we only need to search for roots of $f(x)$ on the interval $(0,1.4)$: we are guaranteed there are no roots elsewhere. Over this interval, $2x \in (0,2.8)$, so $\sin(2x)>0$, and thus $f'(x)=e^x+8\sin(2 x)>0$. Since $f'(x)$ has no roots in $(0,1.4)$, we conclude by Rolle's Theorem that $f(x)$ has at most one root in $(0,1.4)$ (and so at most one positive root total). Since we've already found that a root of $f(x)$ exists in $(0,1)$, we conclude $e^x=4\cos(2x)$ has precisely one positive-valued solution.

Write $f(x)=e^x$. Since $f(x)$ is continuous and differentiable, the Mean Value Theorem asserts that there exists some $c$ between $0$ and $T$ such that

$$\begin{aligned} f'(c)&=\frac{f(T)-f(0)}{T-0} \\ \text{The problem asks us to find this value of $c$. Solving:} \\ e^c&=\frac{e^T-e^0}{T}\\ e^c&=\frac{e^T-1}{T}\\ c&=\log\left(\frac{e^T-1}{T}\right) \end{aligned}$$

Let $s(t)$ be the distance your keys have travelled since they left your hand. The rate at which they are travelling, $s'(t)$, is decreasing by 0.25 metres per second. That is, $s''(t)=-0.25$. Therefore, $s'(t)=-0.25t+c$ for some constant $c$. Since $c=s'(0)=2$, we see $$s'(t)=2-0.25t$$ Then $s(t)$ has $2-0.25t$ as its derivative, so $s(t)=2t-\frac{1}{8}t^2+d$ for some constant $d$. At time $t=0$, the keys have not yet gone anywhere, so $0=s(0)=d$. Therefore, $$s(t)=2t-\frac{1}{8}t^2$$

The keys reach your friend when $s(t)=2$ and $t>0$. That is:

$$\begin{aligned} 2&=2t-\frac{1}{8}t^2\\ 0&=\frac{1}{8}t^2-2t+2\\ t&=8\pm4\sqrt{3} \end{aligned}$$

We need to figure out which of these values of $t$ is really the time when the keys reach your friend. The keys travel this way from $t=0$ to the time they reach your friend. (Then $s(t)$ no longer describes their motion.) So, we need to find the first value of $t$ that is positive with $s(t)=2$. Since $8-4\sqrt{3}>0$, this is the first time $s(t)=2$ and $t>0$. So, the keys take $$8-4\sqrt{3}\approx 1~\text{second}$$ to reach your friend.

Suppose the deceleration provided by the brakes is $d~\frac{\mathrm{km}}{\mathrm{hr}^2}$. Then if $v(t)$ is the velocity of the car, $v(t)=120-dt$ (at $t=0$, the velocity is 120, and it decreases by $d$ kph per hour). The car stops when $0=v(t)$, so $t=\frac{120}{d}$ hours.

Let $s(t)$ be the distance the car has travelled $t$ hours after applying the brakes. Then $s'(t)=v(t)$, so $s(t)=120t-\frac{d}{2}t^2+c$ for some constant $c$. Since $0=s(0)=c$, $$s(t)=120t-\frac{d}{2}t^2$$

The car needs to stop in 100 metres, which is $\frac{1}{10}$ kilometres. We already found that the stopping time is $t=\frac{120}{d}$. So:

$$\begin{aligned} \frac{1}{10}&=s\left(\frac{120}{d}\right)\\ \frac{1}{10}&=120\left(\frac{120}{d}\right)-\frac{d}{2}\left(\frac{120}{d}\right)^2\\ \\ \text{Multiplying both sides by $d$:} \\ \frac{d}{10}&=120^2-\frac{120^2}{2}\\ d&=5\cdot 120^2 \end{aligned}$$

So, the brakes need to apply 72 000 kph per hour of deceleration.

If the acceleration was constant, then it was $$\frac{17 500~\mathrm{mph}}{\frac{8.5}{60}~\mathrm{hr}} \approx 123 500 ~\frac{\mathrm{miles}}{\mathrm{hr}^2}$$

So, the velocity $t$ hours from liftoff is $$v(t)=123500t$$ Therefore, the position of the shuttle $t$ hours from liftoff (taking $s(0)=0$ to be its initial position) is $$s(t)=\frac{123500}{2}t^2=61750t^2$$ So, after $\frac{8.5}{60}$ hours, the shuttle has travelled $$s\left(\frac{8.5}{60}\right)=(61750)\left(\frac{8.5}{60}\right)^2\approx 1240 ~\mathrm{miles}$$ or a little less than 2000 kilometres.

The acceleration of a falling object due to gravity is $9.8$ metres per second squared. So, the object's velocity $t$ seconds after being dropped is $$v(t)=9.8t~\frac{\mathrm{m}}{\mathrm{s}}$$ We want $v(t)$ to be the speed of the peregrine's dive, so we should convert that to metres per second: $$325~\frac{\mathrm{km}}{\mathrm{hr}}\cdot \left(\frac{1000~\mathrm{m}}{1~\mathrm{km}}\right)\left(\frac{1~\mathrm{hr}}{3600~\mathrm{sec}}\right)=\frac{1625}{18}~\frac{\mathrm{m}}{\mathrm{s}}$$ The stone will reach this velocity when $$9.8t=\frac{1625}{18}\qquad\Rightarrow\qquad t=\frac{1625}{18(9.8)}$$

What is left to figure out is how far the stone will fall in this time. The position of the stone $s(t)$ has derivative $9.8t$, so $$s(t)=4.9t^2$$ if we take $s(0)=0$. So, if the stone falls for $\frac{1625}{18(9.8)}$ seconds, in that time it travels $$s\left(\frac{1625}{18(9.8)}\right)=4.9\left(\frac{1625}{18(9.8)}\right)^2\approx 416 ~\mathrm{m}$$ So, you would have to drop a stone from about 416 metres for it to fall as fast as the falcon.

Since gravity alone brings your cannon ball down, its acceleration is a constant $-9.8~\frac{\mathrm{m}}{\mathrm{s}^2}$. So, $v(t)=v_0-9.8t$ and thus its height is given by $s(t)=v_0t-4.9t^2$ (if we set $s(0)=0$).

We want to know what value of $v_0$ makes the maximum height 100 metres. The maximum height is reached when $v(t)=0$, which is at time $t=\frac{v_0}{9.8}$. So, we solve:

$$\begin{aligned} 100&=s\left(\frac{v_0}{9.8}\right)\\ 100&=v_0\left(\frac{v_0}{9.8}\right)-4.9\left(\frac{v_0}{9.8}\right)^2\\ 100&=\left(\frac{1}{9.8}-\frac{4.9}{9.8^2}\right)v_0^2\\ 100&=\frac{1}{2\cdot 9.8}v_0^2\\ v_0^2&=1960\\ v_0&=\sqrt{1960}\approx 44~\frac{\mathrm{m}}{\mathrm{s}} \end{aligned}$$

where we choose the positive square root because $v_0$ must be positive for the cannon ball to get off the ground.

The derivative of acceleration is constant, so the acceleration $a(t)$ has the form $mt+b$. We know $a(0)=-50~000$ and $a\left(\frac{3}{3600}\right)=-60~000$ (where we note that $3$ seconds is $\frac{3}{3600}$ hours). So, the slope of $a(t)$ is $\frac{-60~000+50~000}{\frac{3}{3600}}=-12~000~000$, which leads us to $$a(t)=-50~000-(12~000~000)t$$ where $t$ is measured in hours.

Since $v'(t)=a(t)=-50~000-(12~000~000)t$, we see $$v(t)=\frac{-12~ 000~000}{2}t^2-50~000t+c=-6~000~000t^2-50~000t+c$$ for some constant $c$. Since $120=v(0)=c$: $$v(t)=-6~000~000t^2-50~000t+120$$

Then after three seconds of braking,

$$\begin{aligned}v\left(\frac{3}{3600}\right) &=-6~000~000\left(\frac{3}{3600}\right)^2-50~000\left(\frac{3}{3600}\right)+120\\ &=-\frac{25}{6}-\frac{125}{3}+120\\ &\approx 74.2 \text{ kph} \end{aligned}$$

Remark: When acceleration is constant, the position function is a quadratic function, but we don't want you to get the idea that position functions are always quadratic functions–in the example you just did, it was the velocity function that was quadratic. Position, velocity, and acceleration functions don't have to be polynomial at all–it's only in this section, where we're dealing with the simplest cases, that they seem that way.

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 $x=0$.

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

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