미분 전체 문제

The Mean Value Theorem does not apply since the function is discontinuous at $x=\frac{1}{4},\frac{3}{4},\frac{5}{4},\frac{7}{4}.$

Yes

Yes

The Mean Value Theorem does not apply; not differentiable at $x=0.$

Yes

Let ${s}_{1}(t)$ represent the distance of Car 1 from Stoplight 1 at time $t$. Let ${s}_{2}(t)$ represent the distance of Car 2 from Stoplight 1 at time $t$. Then ${s}_{1}(0)=0={s}_{2}(0)$. Let $t=m$ represent the time that the cars arrive at Stoplight 2. Then ${s}_{1}(m)=d={s}_{2}(m)$ where $d$ is the distance between the stoplights.

Define a new function $f(t)={s}_{1}(t)-{s}_{2}(t)$. Then $f(0)=0=f(m)$. By the Mean Value Theorem, there is a time $0<c<m$ such that

$${f}^{'}(c)=\frac{f(m)-f(0)}{m-0}=\frac{0}{m}=0.$$

At time c,

$${f}^{'}(c)=0$$

$${s}_{1}^{'}(c)-{s}_{2}^{'}(c)=0$$

$${s}_{1}^{'}(c)={s}_{2}^{'}(c)$$

So, at time c, Car 1 and Car 2 are traveling at the same speed.

$h=\frac{62}{3}$ in.

$\frac{dy}{dx} = -x/y$, so the function is increasing in second and fourth quadrants, decreasing in the first and third quadrants.

$\frac{d^2y}{dx^2} = -1/y - x^2/y^3$, which is positive when $y<0$ and is negative when $y>0$. Hence the function is concave down in the first and second quadrants and concave up in the third and fourth quadrants.

By definition $$f'(0)=\lim_{h\rightarrow 0}\frac{f(h)-f(0)}{h} =\lim_{h\rightarrow 0}\frac{h|h|}{h} =\lim_{h\rightarrow 0}|h|=0$$ In particular, the limit exists, so the derivative exists (and is equal to zero).

When $x$ is not equal to 2, then the function is differentiable– the only place we have to worry about is when $x$ is exactly 2.

In order for $f$ to be differentiable at $x=2$, it must also be continuous at $x=2$. This forces $x^2\big|_{x=2}=\big[ax+b\big]_{x=2}$ or $$2a+b=4.$$

In order for a limit to exist, the left- and right-hand limits must exist and be equal to each other. Since a derivative is a limit, in order for $f$ to be differentiable at $x=2$, the left hand derivative of $ax+b$ at $x=2$ must be the same as the right hand derivative of $x^2$ at $x=2$. Since $ax+b$ is a line, its derivative is $a$ everywhere. We've already seen the derivative of $x^2$ is $2x$, so we need $$a=2x\big|_{x=2}=4.$$

So, the values of $a$ and $b$ that makes $f$ differentiable everywhere are $a=4$ and $b=-4$.

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 $0$).