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\begin{document}







To show there is only one root in the given interval, it suffices to
show that $p$ is strictly increasing or is strictly decreasing on this interval.
By the Mean Value Theorem, this will follow if we can show
$p^\prime(x)>0$ (strictly increasing) or $p^\prime(x)<0$ (strictly decreasing)
on the relevant interval, $(0,1)$.
Since $p$ is a polynomial, it is differentiable, and hence continuous, everywhere and hence
differentiable on $(0,1)$ and continuous on $[0,1]$ so the Mean Value Theorem
can be applied.

Compute $p^\prime(x) = 5x^4+6x^2+2$. 
This is obviously $\geq 2>0$ since $x^4\geq0$ and $x^2\geq0$.
Hence $p(x)$ is strictly increasing on $[0,1]$ and so can have at most one root there.
Begin with part b).
Intervals of increase/decrease are determined by the sign of the first derivative.
To determine these, locate the points where $f^\prime$ is $0$, does not exist, or
is not continuous. 
Since $f$ is a polynomial, the derivative is defined and continuous everywhere,
so we need to solve $f^\prime(x)= 12x^3 - 12x^2 = 0$.
The solutions are $x=0$ and $x=1$.
Since $f^\prime(-1)= -12 -12<0$; $f^\prime\bigl(\frac{1}{2}\bigr)=
12\cdot\Bigl(\frac{1}{8} - \frac{1}{4}\Bigr)<0$ and
$f^\prime(2)=12\cdot 8 - 12\cdot 4= 12\cdot 4>0$, the signs are
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Hence $f$ is decreasing on $(-\infty,1]$ and increasing on $[1,\infty)$.
Indeed both the increase and decrease are strict.

Now for part a).
There is a local minimum at $x=1$ and there are no local maxima.
Since the interval $(-\infty,\infty)$ has no endpoints, there can be no
extrema at the endpoints so there is no global maxima.
There is a global minimum at $x=1$ since $f$ decreases from $-\infty$ to $1$ and
then increases from $1$ to $\infty$.

For part c) we need to study $f^{\prime\prime}(x) = 36x^2 - 24x$.
Since $f^{\prime\prime}$ is defined and continuous everywhere, the only relevant points
are the solutions to $36x^2-24 x = 0$ or $x=0$ and $x = \frac{24}{36}=\frac{2}{3}$.
Since $f^{\prime\prime}(-1)=36+24>0$; $f^{\prime\prime}\bigl(\frac{1}{2}\bigr) = 
\frac{36}{4} - \frac{24}{2}=9-12<0$ and $f^{\prime\prime}(2)=36\cdot 4 -24\cdot 2>0$, the
signs are
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Hence $f$ is concave up on $(-\infty,0) \cup (\smfrac{2}{3},\infty)$ and concave
down on $(0,\smfrac{2}{3})$.
Inflection points occur at $x=0$ and at $x = \smfrac{2}{3}$.
\endp
\newp 13. %by LRT
The statement that the oil spill forms a circular region means that its area is $A = \pi r^2$,
where $r$ is its radius.
Since the area is increasing at a rate of $100$ square meters per hour,
we have $\frac{d A}{d t} = 100\ {\rm m}^2/{\rm hr}$.
We are asked to find $\frac{d r}{d t}$ when $r = 200\ {\rm m}$.
Well $\frac{d A}{d t} = 2\pi r \frac{d r}{d t}$ by the Chain Rule, so
$$\frac{d r}{d t} = \frac{\frac{d A}{d t}}{2\pi r} = \frac{100 {\rm m}^2/{\rm hr}}{2 \pi \ 200 {\rm m}} = \frac{1}{4\pi} {\rm m}/{\rm hr}
$$
\endp
\newp 14. %Barron
Answer: $12$

\noindent{\it Solution 1:} $\Delta n \approx dn = n'(t)dt$, and $n'(t) = 12t$, $t
= 5$, $dt = \Delta t = 5.2 - 5 = 0.2$.  Thus $\Delta n \approx dn =
60(0.2) = 12.$

\noindent{\it Solution 2:}  The linear approximation of $n(t)$ at $t = 5$ is $L(t)
= n(5) + n'(5)(t - 5) = 350 + 60(t-5)$.  Thus $n(5.2) \approx L(5.2) = 350
+ 60(0.2) = 350 + 12 = 362$.  Thus $\Delta n = n(5.2) - n(5) \approx
L(5.2) - L(5) = 362 - 350 = 12 = dn$.

\endp
\newp 15. %by LRT
The function is continuous everywhere, hence on $[-2,2]$.
Therefore the function has an absolute maximum and an absolute minimum.
The derivative is $f^{\prime}(x) = -\frac{2}{3} x^{-\smfrac{1}{3}}= \frac{-2}{3\root 3\of x}$
which is defined everywhere except $x=0$.
Hence $0$ is the only critical number.
The global extrema must occur at a critical point or at an end point and since we are looking
for the absolute extrema, we can proceed as follows.
Calculate $f(0)=2$; $f(-2)=2 - (-2)^{\smfrac{2}{3}} = 2 - \root 3 \of 4$ and
$f(2)=2 - (2)^{\smfrac{2}{3}} = 2 - \root 3 \of 4 = f(-2)$.
Since the cube root of $4$ is positive, $2 - \root 3 \of 4 < 2$ so
$2 = f(0)$ is the absolute maximum value and $2 - \root 3\of 4 = f(\pm2)$ is the
absolute minimum value.
\endp

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