미분 전체 문제

Using the product rule seems faster than expanding.

$$\begin{aligned} f'(x)&=\frac{\mathrm{d}}{\mathrm{d}x}\left\{3\sqrt[5]{x}+15\sqrt[3]{x}+8\right\}\left(3x^2+8x-5\right)+ \left(3\sqrt[5]{x}+15\sqrt[3]{x}+8\right)\frac{\mathrm{d}}{\mathrm{d}x}\left\{3x^2+8x-5\right\}\\ &=\frac{\mathrm{d}}{\mathrm{d}x}\left\{3{x}^{\frac{1}{5}}+15{x}^{\frac{1}{3}}+8\right\}\left(3x^2+8x-5\right)+ \left(3\sqrt[5]{x}+15\sqrt[3]{x}+8\right)\frac{\mathrm{d}}{\mathrm{d}x}\left\{3x^2+8x-5\right\}\\ &=\left(\frac{3}{5}{x}^{\frac{-4}{5}}+5{x}^{\frac{-2}{3}}\right)\left(3x^2+8x-5\right)+ \left(3\sqrt[5]{x}+15\sqrt[3]{x}+8\right)\left(6x+8\right) \end{aligned}$$

To avoid the quotient rule, we can divide through the denominator:

$$\begin{aligned} f(x)&=\dfrac{(x^2+5x+1)(\sqrt{x}+\sqrt[3]{x})}{x} =(x^2+5x+1)\dfrac{(\sqrt{x}+\sqrt[3]{x})}{x}\\ &=(x^2+5x+1)(x^{-1/2}+x^{-2/3}) \\ \text{Now, product rule:} \\ f'(x)&=(2x+5)(x^{-1/2}+x^{-2/3})+(x^2+5x+1)\left(\frac{-1}{2}x^{-3/2}-\frac{2}{3}x^{-5/3}\right) \end{aligned}$$

(If you simplified differently, or used the quotient rule, you probably came up with a different-looking answer. There is only one derivative, though, so all correct answers will look the same after sufficient algebraic manipulation.)

Denote by $m$ the slope of the common tangent, by $(x_1,y_1)$ the point of tangency with $y=x^2$, and by $(x_2,y_2)$ the point of tangency with $y=x^2-2x+2$. Then we must have $$y_1=x_1^2\qquad y_2=x_2^2-2x_2+2\qquad m=2x_1=2x_2-2=\frac{y_2-y_1}{x_2-x_1}$$ From the “$m$” equations we get $x_1=\frac{m}{2}$, $x_2=\frac{m}{2}+1$ and

$$\begin{aligned} m&=\frac{y_2-y_1}{x_2-x_1}\\ &=y_2-y_1\\ &=x_2^2-2x_2+2-x_1^2\\ \\& =(x_2-x_1)(x_2+x_1)-2(x_2-1)\\ &=\left(\frac{m}{2}+1-\frac{m}{2}\right)\left(\frac{m}{2}+1+\frac{m}{2}\right) -2\left(\frac{m}{2}+1-1\right) \\ &=(1)(m+1)-2\frac{m}{2}\\ &=1\\ \text{So,}~~~m&=1,\qquad x_1=\tfrac{1}{2},\qquad y_1=\frac{1}{4},\qquad x_2=\frac{3}{2},\qquad y_2=\frac{9}{4}-3+2=\frac{5}{4} \end{aligned}$$

An equation of the common tangent is $y=x-\frac{1}{4}$.

This limit represents the derivative computed at $x=2$ of the function $f(x)=x^{2015}$. To see this, simply use the definition of the derivative at $a=2$ with $f(x)=x^{2015}$:

$$\begin{aligned} \left.\frac{\mathrm{d}}{\mathrm{d}x}\{f(x)\}\right|_{a} &= \lim_{x \to a}\frac{f(x)-f(a)}{x-a}\\ \left.\frac{\mathrm{d}}{\mathrm{d}x}\{x^{2015}\}\right|_{2} &=\lim_{x\to2} \frac{x^{2015}-2^{2015}}{x-2} \end{aligned}$$

Since the derivative of $f(x)$ is $2015\cdot x^{2014}$, then its value at $x=2$ is exactly $2015\cdot 2^{2014}$.

Since $f(x)$ is differentiable for all $x \in (0,10)$, then $f(x)$ is also continuous for all $x \in (0,10)$. If $f(x)$ were continuous on the closed interval $[0,10]$, then the MVT would guarantee $f'(x)=\dfrac{f(10)-f(0)}{10-0}=1$ for some $c \in (0,10)$; however, this is not the case. So, it must be that $f(x)$ is continuous for all $x \in (0,10)$, but not for all $x \in [0,10]$.

Since $f'(c)=0$ for $c \in (0,10)$, that means $f$ is constant on that interval. So, $f(x)$ is a function like this:

where the height of the constant function can be anything.

So, one possible answer is $f(x) = \left\{\begin{array}{lr} 0&x \neq 10\\ 10&x=10 \end{array}\right.$.

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.