In previous activities we saw that if $p$ is a position function, then the difference quotient for $p$ can be used to calculate average velocities and the expression $\frac{\fe{p}{t_0+h}-\fe{p}{t_0}}{h}$ calculates the instantaneous velocity at time $t_0\text{.}$

Graphically, the difference quotient of a function $f$ can be used to calculate the slope of secant lines to $f\text{.}$ What happens when we take the run of the secant line to zero? Basically, we are connecting two points on the line that are really, really, (really), close to one another. As mentioned above, sending $h$ to zero turns an average velocity into an instantaneous velocity. Graphically, sending $h$ to zero turns a secant line into a tangent line.

##### Example3.2.1

The tangent line at $x=4$ to the function $f$ defined by $\fe{f}{x}=\sqrt{x}$ is shown in Figure 3.2.2. Here is a calculation of the slope of this line.

\begin{align*} m_{\text{tan}}&=\lim_{h\to0}\frac{\fe{f}{4+h}-\fe{f}{4}}{h}\\ &=\lim_{h\to0}\frac{\sqrt{4+h}-\sqrt{4}}{h}\\ &=\lim_{h\to0}\left(\frac{\sqrt{4+h}-2}{h}\cdot\frac{\sqrt{4+h}+2}{\sqrt{4+h}+2}\right)\\ &=\lim_{h\to0}\frac{4+h-4}{h\left(\sqrt{4+h}+2\right)}\\ &=\lim_{h\to0}\frac{h}{h\left(\sqrt{4+h}+2\right)}\\ &=\lim_{h\to0}\frac{1}{1\left(\sqrt{4+h}+2\right)}\\ &=\frac{1}{\sqrt{4+0}+2}\\ &=\frac{1}{4} \end{align*}

You should verify that the slope of the tangent line shown in Figure 3.2.2 is indeed $\frac{1}{4}\text{.}$ You should also verify that the equation of the tangent line is $y=\frac{1}{4}x+1\text{.}$

# Subsection3.2.1Exercises

Consider the function $g$ given by $\fe{g}{x}=5-\sqrt{4-x}$ graphed in Figure 3.2.3.

##### 1

Find the slope of the tangent line shown in Figure 3.2.3 using

\begin{equation*} m_{\text{tan}}=\lim_{h\to0}\frac{\fe{g}{3+h}-\fe{g}{3}}{h}\text{.} \end{equation*}

Show work consistent with that illustrated in Example 3.2.1.

##### 3

State the equation of the tangent line to $g$ at $x=3\text{.}$