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\centerline{\bf Solutions to Math 126, Test 2}
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\noindent
{\bf1a.}  We have
$$\frac{f(x+h)-(f(x)}{h}=\frac{\sqrt{(x+h)^2 +1} - \sqrt{x^2+1}}{h}=
\frac{(x+h)^2 +1 - x^2 - 1}{h( \sqrt{(x+h)^2 + 1} + \sqrt{x^2 + 1)}}=$$
$$\frac{2 x h + h^2}{h(\sqrt{(x+h)^2 + 1} + \sqrt{x^2 + 1}}=
\frac{2x + h}{\sqrt {(x+h)^2+1} + \sqrt {x^2 + 1}}
\underset{h\rightarrow0}\to\longrightarrow \frac{2x}{2 \sqrt{x^2 + 1}}$$


Therefore $f'(x) = \frac{x}{\sqrt{x^2 + 1}}.$ \newline
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\noindent
{\bf 1b.}We have that
$$f(x) = \cases \text{sin } x, & x \in \left[2k \Pi, (2k+1) \Pi\right]\\
-\text{sin } x, & x \in \left((2k-1)\Pi\right) \endcases$$

At the points $x=k \pi, k \in Z,f(x)$ is not differentiable since the
derivative from the right is equal to $1$ and the derivative from the left is
$-1$.  If $x \ne k\pi,$ then
$$f'(x) = \cases \text{cos }x, x \in \left( 2k \pi, (2k+1)\pi \right)\\
- \text{cos }x, x \in \left((2k-1)\pi, 2k\pi \right).\endcases$$ 
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\noindent
{\bf 2a.}
We have $$\frac{1}{x \sin x} - \frac{1}{x^2} = \frac{x - \sin x}{x^2 \sin x}
\underset {x \to 0} \to \longrightarrow \frac{0}{0}.$$


By applying L'Hopital's rule
three times we have \newline
$\lim_{x \to 0}(\frac{1}{x \sin x} - \frac{1}{x^2})=
\lim_{x \to 0} \frac{1 - \cos x}{2x \sin x + x^2 \cos x}=$ 
$$\lim_{x \to 0} \frac{\sin x}{2 \sin x + 4x \cos x - x^2 \sin x}= \lim_{x \to 0}
\frac{\cos x}{6 \cos x - 6x \sin x - x^2 \cos x} = \frac 16.$$

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\noindent
{\bf 2b.} We have that $W(x) \cong W(0) + W'(0)x.$ \newline
Since $$W'(x) = \frac{-2mgR^2}{(R + x)^3}$$
We have $W(x) \cong mg - \frac{2mg}{R}x = mg(1 - 2 \frac{x}{R}).$
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\noindent
{\bf 3a.}  We have $f'(x) = 20x^4 - 20x^3 = 20x^3(x-1) = 0 \ \text{if} \  x = 0,1.$
\newline
$f''(x) = 80x^3 - 60x^2 = 20x^2(4x-3) = 0 \ \text{if} \ x =0, 3.$
\newline Thus, $f'(x)< 0$ for $x \in (0,1)$ and \newline
$f'(x) > 0$ for $x \in (-\infty, 0) \cup (1, \infty).$ \newline
$f''(x) > 0 \  \text{for} \  x > \frac 34.$ \newline
$f''(x) <0 \ \text{for} \ x < \frac 34, x \ne 0.$ \newline
$f(0) = 2 \ \text{and} \ f(1) = 1.$
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\noindent
{\bf 3b.}
Since $f$ is decreasing on $(0,1),$ increasing on $(1, \infty)$ and $f(1) = 1 > 0$
we have that $f(x) \ge 1 > 0$ on $(0, \infty).$  Thus, there is no real zero of
$f(x)$ in $0, \infty).$ \newline
Since $f(0) = 2$ and $f(-1) = -7$ by the intermediate value theorem $f$ has a zero
in the interval $(-1,0).$  \newline  Now we will show that there is no other real
zero of $f(x)$ in $(- \infty, 0).$  In fact, if there were two zeros $x_1, x_2 \in
(- \infty, 0), x_1 < x_2$ then, by the mean value theorem, we would have $0 =
f(x_2) - f(x_1) = f'(\xi)(x_2=x_1)$ for some $\xi \in (x_1 , x_2).$ \newline

This would imply $f'(\xi) = 0, \xi < 0$ which is impossible since $f'$ vanishes
only for $x=0 \ \text{or} \  1.$
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\noindent
{\bf 4.} Let $r$ be the radius of the base and $h$ the height of the cylinder.  We
have Volume $V = \pi r^2 h.$  Surface area $S = 2 \pi rh + 2 \pi r^2.$  
\newline
If $V$ is fixed, then $h = \frac{V}{\pi r^2}$ and $S = S(r) = 2 \pi r \frac{V}{\pi
r^2} + 2\pi r^2 = \frac{2V}{V} + 2 \pi r^2.$  \newline

Want to minimize $S(r)$ for $r \in (0, \infty).$  We have
$$S'(r) = - \frac{2V}{r^2} + 4 \pi r \ \text{and} \ S'(r) = 0 \ \text{if} \ r
= \root 3 \of {\frac{V}{2 \pi}}.$$

Since $S'(r) > 0$ for $r > \root 3 \of {\frac{V}{2 \pi}}$ and $S'(r) < 0$ for 
$r < \root 3 \of {\frac{V}{2\pi}}$ 
we have that $S(r)$ is minimum for 
$\boxed{r=\root 3 \of {\frac{V}{2 \pi}}}.$

Thus the surface area takes its minimum for $r=\root 3 \of { \frac{V}{2 \pi} }$ 
and
$h= \frac{V} {(\frac{V}{2 \pi})^\frac23}.$ 
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\noindent


{\bf 5a.}  Let $f(x) = \frac{1}{1+x^2}.$  We have $f'(x) = \frac{-2x}{(1 +
x^2)^2} = 0$ \ \  Iff $x = 0.$

Since $f'(x) > 0$ for $x < 0$ and $f'(x) < 0$ for $x > 0,$ $f$ takes its global
maximum at $x = 0.$  Thus $\boxed{SupA = f(0) = 1.}$ \newline 
Since $f(x) > 0$ for all $x \in \Bbb R$ and $\displaystyle \lim_{x\to\pm\infty} f(x)
= 0$ we have $\boxed {inf A = 0}.$
\newline

{\bf 5b.}  By the chain rule, we have $f'(x) = \cos \left( g(x) \right) \cdot 
g'(x).$ \newline
Thus, $f'(1) = \cos \left(g(1) \right)g'(1) = \cos(0) \cdot g'(1) = 1 \cdot 5 = 5.$
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