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\centerline{MATH 136 -- Advanced Placement Calculus}
\centerline{Solutions for Resubmitted Problems -- Problem Set 4}
\centerline{October 26, 2009}
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\noindent
4.3/12.  (a) We have $f(x) = \cos^2(x) - 2\sin(x)$, so 
$f'(x) = -2\cos(x)\sin(x) - 2\cos(x) = -2 \cos(x)(\sin(x) + 1)$.
So $f'(x) = 0$ at $x = {\pi\over 2}$ and ${3\pi\over 2}$ in
the interval $[0,2\pi]$.  Since $\sin(x) + 1 \ge 0$,
the sign of $-\cos(x)$ gives the sign of $f'$ and this
is negative on $\left[0,{\pi\over 2}\right)$ and $\left({3\pi\over 2},2\pi\right]$, 
positive on $\left({\pi\over 2},{3\pi\over 2}\right)$.

(b) By the first derivative test, $f(x)$ has a local maximum
at $x = {3\pi\over 2}$ and a local minimum at $x = {\pi\over 2}$.  

(c)  The second derivative is 
$$\eqalign{
f''(x) &= 2\sin^2(x) - 2\cos^2(x) + 2\sin(x)\cr
       &= 2\sin^2(x) - 2(1 - \sin^2(x)) + 2\sin(x)\cr
       &= 4\sin^2(x) + 2\sin(x) - 2\cr
       &= 2(\sin(x) + 1)(2\sin(x) - 1)\cr
}$$
Now, $\sin(x) = -1$ at $x = {3\pi\over 2}$, while $\sin(x) = {1\over 2}$
at $x = {\pi\over 6}$ and $x = {5\pi\over 6}$.  Of these, only  
$x = {\pi\over 6}$ and $x = {5\pi\over 6}$ are inflection points
because $f''(x)$ does not change sign at $x = {3\pi\over 2}$.
(Note that $f'(x)$ has a double root at $x = {3\pi\over 2}$.
This means that the graph of $f(x)$ is very ``flat'' around
the maximum, but the concavity does not change there.)
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\noindent
4.4/14.  The limit $\lim_{x\to\infty} {(\ln(x))^2\over x}$ is indeterminate
of the form $\infty/\infty$.  Applying L'Hopital a first time, we
examine 
$$\lim_{x\to\infty} {2\ln(x){1\over x}\over 1} = \lim_{x\to\infty} {2\ln(x)\over x}.$$
Since this is still indeterminate of the form $\infty/\infty$, we apply L'Hopital's
Rule a second time:
$$\lim_{x\to\infty} {{2\over x}\over 1} = 0.$$
Hence the original limit is $0$ also.
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\noindent
4.6/12.  Let $x$ be the side of the base of the box and $y$ be the height.
We have $32000 = x^2y$, so $y = {32000\over x^2}$.  The surface area of 
the bottom and the sides (no top) is 
$$S = x^2 + 4xy = x^2 + 4x{32000\over x^2} = x^2 + {168000\over x}.$$
Now we differentiate with respect to $x$ and set the derivative equal
to zero to find the critical numbers:
$$0 = S' = 2x - {168000\over x^2} = 0 \Rightarrow x^3 = 64000$$
so $x = 40$.  Substituting back, $y = {32000\over 1600}= 20$.  
This is a {\it minimum} of the $S$ function (and the material
needed for the box), since $S'' = 2 + {336000\over x^3} > 0$
for all $x > 0$.   The second derivative test shows that
$S$ has a local minimum at $x = 40$, and this is a global
minimum since it is the only critical number for $x > 0$ and 
$S$ is differentiable for all $x > 0$.  
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\noindent
4.8/30.  Given $f''(x) = 8x^3 + 5$, and $f'(1) = 8$, $f(1) = 0$.
Integrating once, $f'(x) = 2x^4 + 5x + c$.  Then substituting
$x = 1$ gives $8 = f'(1) = 2 + 5 + c$.  Hence $c = 1$.  
Integrating again, $f(x) = {2\over 5} x^5 + {5\over 2} x^2 + x + d$.
Substituting $x = 1$ again,
${2\over 5} + {5\over 2} + 1 + d = 0$.  Hence $d = {-(4 + 25 + 10)\over 10} = {-39\over 10}$.
Then 
$$f(x) = {2\over 5} x^5 + {5\over 2} x^2 + x - {39\over 10}.$$
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