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\centerline{Mathematics 133--Intensive Calculus for Science 1}
\centerline{Solutions for Exam 2}
\centerline{November 4, 2005}
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\noindent
I.  
\item{A)}  From the graph, $f(x)$ appears to be differentiable everywhere
except $x = 0$, where there is a sharp corner.
\item{B)}  The derivative $y = f'(x)$:
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Note:  $f'(x) > 0$ is positive on intervals where $f(x)$ is increasing
and negative on intervals where $f(x)$ is decreasing.  At $x=0$, there
is a jump discontinuity on the graph $y = f'(x)$ because of the sharp
corner on the graph $y = f(x)$. 
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\noindent
II.  
$$\eqalign{
f'(x) &= \lim_{h\to 0} {f(x+h) - f(x) \over h}\cr
&= \lim_{h\to 0} {(x+h)^2 - (x+h) + 4 - x^2 + x - 4 \over h}\cr
&= \lim_{h\to 0} {x^2 + 2xh + h^2 - x - h + 4 - x^2 + x - 4 \over h}\cr
&= \lim_{h\to 0} {2xh + h^2 - h \over h}\cr
&= \lim_{h\to 0} 2x + h - 1\cr
&= 2x - 1\cr}$$
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\noindent
III.
\item{A)}  (Product, exponential, and power rules) 
$$f'(x) = 2^x\ln(2)(x^4 + 3x^{-1/3}) + 2^x (4x^3 - x^{-4/3}) = 
2^x(\ln(2)x^4 + 3\ln(2)x^{-1/3} + 4x^3 - x^{-4/3})$$
\item{B)}  By the chain rule (twice):
$$g'(x) = 3(e^{\pi x} + 2)^2 \pi e^{\pi x}$$
\item{C)}  By the quotient rule:
$$h'(x) ={(2x^2+3x)(2x) - (x^2 - 3)(4x + 3)\over (2x^2 + 3)^2}
= {3x^2 + 12x  + 9\over (2x^2 + 3)^2}$$
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\noindent
IV.  The price in 2000 was $P(50) = 100(1.05)^50 \doteq \$ 1146.74$.
The price was changing at $P'(50) = 100(1.05)^50\ln(1.05) \doteq 55.95$
dollars per year.  It was increasing because $P'(50) > 0$.
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\noindent
V.  $C'(q) = {50\over 4} q^{-3/4}$, so $C''(q) = {-150\over 16} q^{-7/4}$.
The cost function graph is concave down for $q > 0$ since $C''(q) < 0$.
In everyday terms, this means that as the number of copies is 
increased, the {\it cost per copy decreases}.
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