미분 전체 문제

We note that $f(0)=f(2\pi)=\sqrt{3} + \pi^2$. Then using the Mean Value Theorem (note that the function is differentiable for all real numbers since $3+\sin x>0$), we get that there exists $c\in (0,2\pi)$ such that $$f'(c)=\frac{f(2\pi)- f(0)}{2\pi - 0} = 0.$$

We note that $f(0)=0$ and $f(\pi/4)=0$. Then using the Mean Value Theorem (note that the function is differentiable for all real numbers), we get that there exists $c\in (0,\pi/4)$ such that $$f'(c)=\frac{f(\pi/4)- f(0)}{\pi/4 - 0} = 0.$$

By inspection, we see $x=0$ is a root of $f(x)$. The question now is whether there could possibly be other roots. Since $f(x)$ is differentiable over all real numbers, if there is another root $a$, then by Rolle's Theorem, $f'(c)=0$ for some $c$ strictly between $0$ and $a$. However, $f'(x)=3-\cos x$ is never zero. So, there is no second root: $f(x)$ has precisely one root.

The function $f(x)$ is continuous and differentiable over all real numbers. If $a$ and $b$ are distinct roots of $f(x)$, then $f'(c)=0$ for some $c$ strictly between $a$ and $b$ (Rolle's Theorem). So, let's think about $f'(x)$. $$f'(x)=(4x+1)^3+1$$ This is simple enough that we can find its zero explicitly: $$(4x+1)^3 +1= 0 \quad \Leftrightarrow \quad (4x+1)^3 = -1 \quad \Leftrightarrow \quad 4x+1 = -1 \quad \Leftrightarrow \quad 4x=-2$$ Hence $f'(c)=0$ only when $c=\dfrac{-1}{2}$. That the derivative only has a single zero is very useful. It means (via Rolle's theorem) that if $f(x)$ has distinct roots $a$ and $b$ with $a<b$, then we must have $a < \dfrac{-1}{2} < b$. This also means that $f(x)$ cannot have 3 distinct roots $a,b$ and (say) $q$ with $a<b<q$, because then Rolle's theorem would imply that $f'(x)$ would have two zeros — one between $a$ and $b$ and another between $b$ and $q$.

We've learned that $f(x)$ has at most two roots, and we've learned something about where those roots can exist, if there are indeed two of them. But that means $f(x)$ could have 0, 1, or 2 roots.

It's not easy to find a root of $f(x)$ by inspection. But we can get a good enough picture of the graph of $y=f(x)$ to tell exactly how many roots there are, just by exploiting the following properties of $f(x)$ and $f'(x)$.

• As $x$ tends to $\pm \infty$, $f(x)$ tends to $+\infty$.
• The derivative $f'(x)=(4x+1)^3+1$ is negative for $x<-\frac{1}{2}$ and is positive for $x>-\frac{1}{2}$. That is, $f(x)$ is decreasing for $x<-\frac{1}{2}$ and increasing for $x>-\frac{1}{2}$.
• $f\left(-\frac{1}{2}\right)=\frac{1}{16}-\frac{1}{2}<0$.

This means the function must look something like the graph below:

except we are unsure of the locations of the $x$-intercepts. At present we only know that one is to the left of $x=-1/2$ and one is to the right.

Note what happens to $f(x)$ as $x$ increases from strongly negative values to strongly positive values.

• When $x$ is large and negative, $f(x)>0$.
• As $x$ increases, $f(x)$ decreases continuously until $x=-\frac{1}{2}$, where $f(x)<0$. In particular, since $f(-1)=\frac{65}{16}>0$ and $f\big(-\frac{1}{2}\big)<0$ and $f(x)$ is continuous, the intermediate value theorem guarantees that $f(x)$ takes the value zero for some $x$ between $-\frac{1}{2}$ and $-1$. More descriptively put, as $x$ increases from hugely negative numbers to $-\frac{1}{2}$, $f(x)$ passes through zero exactly once.
• As $x$ increases beyond $-\frac{1}{2}$, $f(x)$ increases continuously, starting negative and becoming very large and positive when $x$ becomes large and positive. In particular, since $f(0)=\frac{1}{16}>0$ and $f\big(-\frac{1}{2}\big)<0$ and $f(x)$ is continuous, the intermediate value theorem guarantees that $f(x)$ takes the value zero for some $x$ between $-\frac{1}{2}$ and $0$. So, as $x$ increases from $-\frac{1}{2}$ to near $+\infty$, $f(x)$ again passes through zero exactly once.

So $f(x)$ must have exactly two roots, one with $x<-\frac{1}{2}$ and one with $x>-\frac{1}{2}$.

• We can see by inspection that $f(0)=0$, so there is at least one root. We have to determine how many any other roots there are, if any.
• The function $f(x)$ is the sum of the two terms $x^3$ and $\sin(x^5)$. A number $x$ is a root of $f(x)$ if and only if the two terms cancel each other exactly for that value of $x$. That is, $x$ is a root of $f$ if and only if $x^3=-\sin(x^5)$. To develope some intuition in our hunt for other roots, we sketch, in the same figure, the graphs $y=x^3$ and $y=-\sin(x^5)$. Then the roots of $f(x)$ are precisely the $x$'s where the two curves $y=x^3$ and $y= -\sin(x^5)$ intersect.

• Looking at the sketch, we see that the two curves cannot possibly intersect at any point having $|x|>1$ — if $|x|>1$, then $|x^3|>1$ but $|-\sin (x^5)|\le 1$ and we cannot possibly have $x^3=-\sin(x^5)$. That is, all roots of $f(x)$ are in the interval $[-1,1]$.
• From the sketch, we would probably guess that $y=x^3$ and $y=-\sin(x^5)$ cross only at $x=0$. We can use Rolle's Theorem to verify that that is indeed the case.
• If there is a root $a \neq 0$, then by Rolle's Theorem (since $f(x)$ is continuous and differentiable for all real numbers $x$) $f'(c)=0$ for some $c$ strictly between $0$ and $a$. In particular, since we already know any roots $a$ will be between $-1$ and $1$, if $f(x)$ has two roots then $f'(c)=0$ for some $c \in (-1,0) \cup (0,1)$.
• The derivative of $f$ is

$$f'(x)=3x^2+5x^4\cos\big(x^5\big)=x^2\left(3+5x^2\cos\big(x^5\big)\right)$$

So, if $f'(x)=0$, then $x=0$ or $\left(3+5x^2\cos\left(x^5\right)\right)=0$. If $x \in (-1,0) \cup (0,1)$, then

$$\begin{aligned} |x|<1 &\implies |x^5|<1<\tfrac{\pi}{2} \implies \cos\big(x^5\big)>0 \\ &\implies 3+5x^2\cos\big(x^5\big) > 3 \\ &\implies f'(x) \neq 0 \end{aligned}$$

That is, there is no $c \in (-1,0) \cup (0,1)$ with $f'(c)=0$. Therefore, following our last bullet point, $f(x)$ has only one root.

Note here that $f'(x)$ has many zeroes — infinitely many, in fact. However, $x=0$ is the only root of $f'(x)$ in the interval $(-1,1)$.

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

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

If $2x+\sin x$ is one–to–one over an interval, it never takes the same value for two distinct numbers in that interval. By Rolle's Theorem, if $f(a)=f(b)$ for distinct $a$ and $b$, then $f'(c)=0$ for some $c$ between $a$ and $b$. However, $f'(x)=2+\cos x$, which is never zero. In fact, $f'(x)\ge 1$ for all $x$, so $f(x)$ is strictly increasing over its entire domain. Therefore, our function $f$ never takes the same value twice, so it is one–to–one over all the real numbers, $(-\infty,\infty)$.

When we define the inverse function $f^{-1}(x)$, the domain of $f$ is the range of $f^{-1}$, and vice-versa. In general, we might have to restrict the domain of $f$ (and hence the range of $f^{-1}$) to an interval where $f$ is one–to–one, but in our case, this isn't necessary. So, the range of $f^{-1}$ is $(-\infty,\infty)$ and the domain of $f^{-1}$ is the range of $f$: $(-\infty,\infty)$.

If $f(x)=\dfrac{x}{2}+\sin x$ is one–to–one over an interval, it never takes the same value for two distinct numbers in that interval. By Rolle's Theorem, if $f(a)=f(b)$ for distinct $a$ and $b$, then $f'(c)=0$ for some $c$ between $a$ and $b$. Since $f'(x)=\frac{1}{2}+\cos x$, $f'(x)=0$ when $x=2n\pi\pm\frac{2\pi}{3}$ for some integer $n$. So, in particular, if $a$ and $b$ are distinct numbers in the interval $\left[-\frac{2\pi}{3},\frac{2\pi}{3}\right]$, then for every $c$ strictly between $a$ and $b$, $f'(c) \neq 0$, so by Rolle's Theorem $f(a) \neq f(b)$. Therefore $f(x)$ is one–to–one on the interval $\left[-\frac{2\pi}{3},\frac{2\pi}{3}\right]$.

We should also show that the interval $\left[-\frac{2\pi}{3},\frac{2\pi}{3}\right]$ cannot be extended to a larger interval over which $f(x)$ is still one–to–one. Consider the derivative $f'(x) = \frac{1}{2}+\cos x$. For all $-\frac{2\pi}{3} < x < \frac{2\pi}{3}$, we have $\cos x > -\frac{1}{2}$ (sketch the graph of $\cos x$ yourself) so that $f'(x)>0$ and $f(x)$ is increasing. But at $x=\frac{2\pi}{3}$, $f'(x)=0$, and then for $x$ a bit bigger than $\frac{2\pi}{3}$ we have $\cos x < -\frac{1}{2}$ so that $f'(x)<0$ and $f(x)$ is decreasing. So the graph “reverses direction", and $f(x)$ repeats values. (See the graph of $y=f(x)$ below.) The same is true for $x$ a little smaller than $-\frac{2\pi}{3}.$

When we define the inverse function $f^{-1}(x)$, first we restrict $f$ to $\left[-\frac{2\pi}{3},\frac{2\pi}{3}\right]$. Then the range of $f^{-1}$ is also $\left[-\frac{2\pi}{3},\frac{2\pi}{3}\right]$. The domain of $f^{-1}$ is the range of $f$ over this interval, so $\left[-\left(\frac{\pi}{3}+\frac{\sqrt{3}}{2}\right),\left(\frac{\pi}{3}+\frac{\sqrt{3}}{2}\right)\right]$.