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\begin{document}
\noindent
\centerline{MATH 131 section 1 -- Calculus for Physical and Life Sciences}
\centerline{Discussion 2 -- October 3, 2007}
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\noindent
{\it Background}
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Recall that last time we introduced the {\it derivative} at $a$ of a function
$f(x)$:
\[
f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}
\]
(if the limit exists).  If $f'(a)$ exists for all $a$ in an interval, we can 
also use this to define the values of a {\it new function}, also called $f'$.
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\noindent
{\it Discussion Questions}  
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\noindent
Below is the graph $y = f(x)$.  
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\noindent
A.  In terms of the graph, what do each of the following quantities represent?
\[
f(a) - f(b),  \quad \frac{f(b) - f(a)}{b - a}, \quad f'(a), \quad f'(b)
\]
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\noindent
B.  Based on the information in the graph, put the following quantities
in numerical order from smallest to largest:  $f'(a)$, $f'(b)$, the 
slope of the secant line containing $P$ and $Q$, the slope of the secant
line containing $Q$ and $R$, $\frac{f(c) - f(a)}{c - a}$.   (Don't forget
that for negative numbers we have $-10 < -4$, for instance!)
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\noindent
C.  Redraw the same graph $y = f(x)$ as accurately as you can, 
and indicate {\it new points} $S = (s,f(s))$, $T = (t,f(t))$, and $U = (u,f(u))$ 
on the graph such that $s < t < u$ and
\[
f'(u) < m_{TU} < f'(t) < f'(s)
\]
where $m_{TU}$ is the slope of the secant line containing your $T$ and $U$.
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\noindent
D.  Draw the graph of the {\it function} $f'(x)$, where as above $f'(x)$ is the 
derivative function -- its value at $x$ is the slope
of the tangent to $y = f(x)$ at $x$.  
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