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\centerline{MATH 133 -- Calculus with Fundamentals 1}
\centerline{The Derivative of a Function}
\centerline{October 16, 2017}
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{\it Background}
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We are now ready to begin Chapter 3 in our textbook.
In the video for today's class, we introduced the \emph{derivative} of a function 
$f$ at $x = a$ in the domain of $f$:
$$f'(a) = \lim_{h\to 0} \frac{f(a + h) - f(a)}{h} = \lim_{x \to a} \frac{f(x) - f(a)}{x - a},$$
provided the limit exists.  If the limit does exist, then by what we said in Section 2.1, 
$f'(a)$ will give the slope of the tangent line 
to the graph $y = f(x)$ at the point $(a,f(a))$.  If $x$ represents time, and $f(x)$ is a position,
then $f'(a)$ would be the instantaneous velocity.  

All the techniques we learned in Chapter 2 for 
computing indeterminate form limits were, in fact, \emph{set up to compute the limits giving}
$f'(a)$(!)  Let's practice (and review) some of those techniques!
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{\it Questions}
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\begin{enumerate}
\item[(1)]  Compute $f'(1)$ for $f(x) = x^3 + 2x + 1$ and use your result to 
find the equation of the tangent line to the graph $y = x^3 + 2x + 1$ at the point $(1,4)$.  
\item[(2)]  Compute $f'(3)$ for $f(x) = \sqrt{x + 1}$ and use your result to 
find the equation of the tangent line to the graph $y = \sqrt{x + 1}$ at the point $(3,2)$. 
\item[(3)]  Compute $f'(2)$ for $f(x) = \frac{1}{x}$ and use your result to 
find the equation of the tangent line to the graph $y = \frac{1}{x}$ at the point $(2,1/2)$. 
\end{enumerate}

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