미분 전체 문제

True, examine derivative

The function is differentiable at $x=0$ if the following limit: $$\lim_{x\to 0}\frac{f(x)-f(0)}{x-0} = \lim_{x\to 0}\frac{f(x)-0}{x}=\lim_{x\to 0} \frac{f(x)}{x}$$ exists (note that we used the fact that $f(0)=0$ as per the definition of the first branch which includes the point $x=0$). We start by computing the left limit. For this computation, recall that if $x<0$ then $\sqrt{x^2}=|x|=-x$. $$\lim_{x\to 0^-}\frac{f(x)}{x}=\lim_{x\to 0^-}\frac{\sqrt{x^2+x^4}}{x}=\lim_{x\to 0^-} \frac{\sqrt{x^2}\sqrt{1+x^2}}{x}=\lim_{x \rightarrow 0}\frac{-x\sqrt{1+x^2}}{x}=-1$$ Now, from the right: $$\lim_{x\to 0^+}\frac{x\cos x}{x}=\lim_{x\to 0^+}\cos x = 1.$$ Since the limit from the left does not equal the limit from the right, the derivative does not exist at $x=0$.

The function is differentiable at $x=0$ if the following limit: $$\lim_{x\to 0}\frac{f(x)-f(0)}{x-0} = \lim_{x\to 0}\frac{f(x)-0}{x}=\lim_{x\to 0} \frac{f(x)}{x}$$ exists (note that we used the fact that $f(0)=0$ as per the definition of the first branch which includes the point $x=0$).

We start by computing the left limit.

$$\begin{aligned} \lim_{x\to 0^-}\frac{f(x)}{x}=\lim_{x\to 0^-} \frac{x\cos x}{x} =\lim_{x\to 0^-} \cos x = 1. \end{aligned}$$

Now, from the right:

$$\begin{aligned} \lim_{x\to 0^+}\frac{\sqrt{1+x}-1}{x} &= \lim_{x\to 0^+}\frac{\sqrt{1+x}-1}{x} \cdot \frac{\sqrt{1+x}+1}{\sqrt{1+x}+1} \\ &= \lim_{x\to 0^+} \frac{1+x-1}{x(\sqrt{1+x}+1)} = \lim_{x\to 0^+} \frac{1}{\sqrt{1+x}+1} = \frac{1}{2} \end{aligned}$$

Since the limit from the left does not equal the limit from the right, the derivative does not exist at $x=0$.

The function is differentiable at $x=0$ if the following limit:

$$\begin{aligned} \lim_{x\to 0}\frac{f(x)-f(0)}{x-0} = \lim_{x\to 0}\frac{f(x)-0}{x}=\lim_{x\to 0} \frac{f(x)}{x} \end{aligned}$$

exists (note that we used the fact that $f(0)=0$ as per the definition of the first branch which includes the point $x=0$). We compute left and right limits; so

$$\begin{aligned} \lim_{x\to 0^-}\frac{f(x)}{x}=\lim_{x\to 0^-}\frac{x^3-7x^2}{x}=\lim_{x\to 0^-} x^2-7x=0 \end{aligned}$$

and

$$\begin{aligned} \lim_{x\to 0^+}\frac{x^3\cos\left(\frac{1}{x}\right)}{x}=\lim_{x\to 0^+}x^2\cdot \cos\left(\frac{1}{x}\right). \end{aligned}$$

This last limit equals $0$ by the Squeeze Theorem since

$$\begin{aligned} -1\le \cos\left(\frac{1}{x}\right)\le 1 \end{aligned}$$

and so,

$$\begin{aligned} -x^2\le x^2\cdot \cos\left(\frac{1}{x}\right)\le x^2, \end{aligned}$$

where in these inequalities we used the fact that $x^2\ge 0$. Finally, since $\lim_{x\to 0^+}-x^2=\lim_{x\to 0^+}x^2=0$, the Squeeze Theorem yields that also $\lim_{x\to 0^+}x^2\cos\left(\frac{1}{x}\right) =0$, as claimed.

Since the left and right limits match (they're both equal to $0$), we conclude that indeed $f(x)$ is differentiable at $x=0$ (and its derivative at $x=0$ is actually equal to $0$).

The function is differentiable at $x=1$ if the following limit: $$\lim_{x\to 1}\frac{f(x)-f(1)}{x-1} = \lim_{x\to 1}\frac{f(x)-0}{x-1}=\lim_{x\to 1} \frac{f(x)}{x-1}$$ exists (note that we used the fact that $f(1)=0$ as per the definition of the first branch which includes the point $x=0$). We compute left and right limits; so $$\lim_{x\to 1^-}\frac{f(x)}{x-1}=\lim_{x\to 1^-}\frac{4x^2-8x+4}{x-1}=\lim_{x\rightarrow 1^-}\frac{4(x-1)^2}{x-1} =\lim_{x\to 1^-} 4(x-1)=0$$ and $$\lim_{x\to 1^+}\frac{(x-1)^2\sin\left(\frac{1}{x-1}\right)}{x-1}=\lim_{x\to 1^+}(x-1)\cdot \sin\left(\frac{1}{x-1}\right).$$ This last limit equals $0$ by the Squeeze Theorem since $$-1\le \sin\left(\frac{1}{x-1}\right)\le 1$$ and so, $$-(x-1)\le (x-1)\cdot \sin\left(\frac{1}{x-1}\right)\le x-1,$$ where in these inequalities we used the fact that $x\to 1^+$ yields positive values for $x-1$. Finally, since $\lim_{x\to 1^+}-x+1=\lim_{x\to 1^+}x-1=0$, the Squeeze Theorem yields that also $\lim_{x\to 1^+}(x-1)\sin\left(\frac{1}{x-1}\right) =0$, as claimed.

Since the left and right limits match (they're both equal to $0$), we conclude that indeed $f(x)$ is differentiable at $x=1$ (and its derivative at $x=1$ is actually equal to $0$).

$$\begin{aligned} \\ \text{First expression, $f(x)=\dfrac{g(x)}{h(x)}$:} \\ f'(x)&=\frac{h(x)g'(x)-g(x)h'(x)}{h^2(x)} \\ \text{Second expresson, $f(x)=\dfrac{g(x)}{k(x)}\cdot\dfrac{k(x)}{h(x)}$:} \\ f'(x)&=\left(\frac{k(x)g'(x)-g(x)k'(x)}{k^2(x)}\right)\left(\frac{k(x)}{h(x)}\right)+\left(\frac{g(x)}{k(x)}\right)\left(\frac{h(x)k'(x)-k(x)h'(x)}{h^2(x)}\right)\\ &=\frac{k(x)g'(x)-g(x)k'(x)}{k(x)h(x)}+ \frac{g(x)h(x)k'(x)-g(x)k(x)h'(x)}{k(x)h^2(x)}\\ &=\frac{h(x)k(x)g'(x)-h(x)g(x)k'(x)}{k(x)h^2(x)}+ \frac{g(x)h(x)k'(x)-g(x)k(x)h'(x)}{k(x)h^2(x)}\\ &=\frac{h(x)k(x)g'(x)-h(x)g(x)k'(x)+g(x)h(x)k'(x)-g(x)k(x)h'(x)}{k(x)h^2(x)}\\ &=\frac{h(x)k(x)g'(x)-g(x)k(x)h'(x)}{k(x)h^2(x)}\\ &=\frac{h(x)g'(x)-g(x)h'(x)}{h^2(x)} \\ \text{and this is exactly what we got from differentiating the first expression.} \\ \end{aligned}$$

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 $c \in (a,b)$.

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).