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\centerline{Mathematics 133 -- Intensive Calculus for Science 1}
\centerline{Solutions -- Exam 3}
\centerline{\it December 3, 2001}
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\noindent
I.  
\item{A)} Using the derivative rule for $\ln$ and the 
chain rule:
$$f'(x) = {1\over x} + 2^{4x}\ln(2)\cdot 4 = {1\over x} + 4\ln(2)2^{4x}$$
\item{B)} Using the product and chain rules and the derivative
rules for $\sin$ and $\cos$:
$$g'(x) = \sin(x)\cdot -\sin(x^2)\cdot 2x + \cos(x^2)\cdot cos(x)= 
-2x\sin(x)\sin(x^2) + \cos(x)\cos(x^2)$$
\item{C)} Using the derivative rule for ${\rm arctan}$ and the chain
rule:
$$h'(x) = {1\over 1 + (x^2 + e^{2x} + 1)^2}\cdot (2x + 2e^{2x})$$
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\noindent
II. 
\item{A)}  Using the quotient rule,
$$\eqalign{f'(x) &= {(x^2+4)\cdot 1 - x\cdot 2x\over (x^2+4)^2}\cr
                 &= {-x^2+4\over (x^2+4)^2}\cr
                 &= {(2-x)(2+x)\over (x^2+4)^2}\cr}$$
(factoring using the pattern for a difference of squares).
\item{B)} From the factored form of the numerator, the 
critical points ($x$ where $f'(x) = 0$) are 
$x = \pm 2$.  The denominator of $f'(x)$ is always positive, so the 
sign of $f'$ is determined by the numerator.  $-x^2+4$ is $>0$ if 
$-2 < x < 2$ and positive when $x > 2$ or $x < -2$.  The
First Derivative Test shows that $x = -2$ is a {\it local minimum},
while $x = 2$ is a {\it local maximum}.
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\item{C)} Differentiating $f'$ with the quotient rule again,
we get
$$f''(x) = {(x^2+4)^2\cdot(-2x) - (-x^2+4)\cdot 2(x^2+4)\cdot 2x\over (x^2 +4)^4}$$
We want to simplify this.  {\it Note that there are several common factors
in the terms on the top of this expression.  WE DO NOT WANT TO MULTIPLY
THIS OUT!!}  Instead, we factor out $2x(x^2+4)$, cancel with the denominator,
and simplify to obtain:
$$\eqalign{f''(x) &= {2x (x^2+4)(-(x^2+4) - 2(-x^2+4))\over (x^2 +4)^4}\cr
                  &= {2x(x^2 - 12)\over (x^2+4)^2}\cr}$$
\item{D)} The inflection points of this function are at the roots
of the numerator of $f''(x)$ -- $x = 0, \pm\sqrt{12}$.  The second 
derivative changes sign at each of these.
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\noindent
III.  Let $x,y$ be the overall dimensions of the poster.  The printed region 
is $x - 4$ inches by $y - 3$ inches because of the margins.  Then we have
$xy = 400$, and we want to maximize the area of the printed region, which
is $A = (x-4)(y-3)$.  Solving for $y$ from the equation $xy = 400$,
$y = 400/x$, so 
$$\eqalign{A(x) &= (x-4)\left({400\over x} - 3\right)\cr
                &= 400 -{1600\over x} - 3x + 12\cr
                &= 412 -{1600\over x} - 3x\cr
           A'(x)&= {1600\over x^2} - 3\cr}$$
The critical points of the area function are at the solutions of $A'(x) = 0$, 
so 
$$\eqalign{{1600\over x^2} - 3 &= 0\cr
              {1600\over x^2} &=3\cr
              x^2 &= {1600\over 3}\cr
               x&=\sqrt{1600\over 3}\cr
                &\doteq 23.1\cr}$$
Then $y = {400\over 23.1} \doteq 17.3$.  This is a maximum for the area
because $A''(x) = -{3200\over x^3}$, so $A''\left(\sqrt{1600\over 3}\right) < 0$
(Second Derivative Test says that if $f''$ is negative at the critical point,
then $f$ has a local maximum there). 
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\noindent
IV.  
\item{A)} {\it True}:  Say $p_1 < p_2$.  
If we assume that $f'$ is continuous for all $x$, then $f'$ must
be zero somewhere between $p_1$ and $p_2$.  If not, $f'$ would
have to have the same sign everywhere between $p_1$ and $p_2$.
That sign would have to be negative because $p_1$ is a local max.
But $f'(x)$ cannot be negative to the left of $p_2$ if $p_2$
also a local max.
\item{B)} {\it True}: By implicit differentiation,
$2y{dy\over dx} + x{dy\over dx} + y = 0$, so 
$${dy\over dx} = {-y\over 2y + x}$$
In this case we can solve explicitly for $y$ as a function of
$x$ with the quadratic formula:
$$y = {-x\pm \sqrt{x^2 + 4}\over 2}$$
Each of these functions is defined for all $x$, 
and $2y + x = \sqrt{x^2 + 4} \ne 0$ for all $x$.  Hence ${dy\over dx}$
exists. 
\bye 
