Chapter 11.3, Problem 21E

To determine

To evaluate: the slope of the tangent line at the point $\left(0,-\frac{1}{4}\right)$ using the limit process and use the graph to confirm the slope of the tangent line.

Answer to Problem 21E

The slope of tangent line is $\frac{1}{6}$ .

Explanation of Solution

Given:

  $h\left(x\right)=\sqrt{x},\left(9,3\right)$

Using the formula,

  $f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$

So,

  $\begin{array}{c}f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{\sqrt{x+h}-\sqrt{x}}{h}\\ =\underset{h\to 0}{\lim }\left(\frac{\sqrt{x+h}-\sqrt{x}}{h}\times \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}\right)\\ =\underset{h\to 0}{\lim }\left(\frac{\left(x+h\right)-x}{h\left(\sqrt{x+h}+\sqrt{x}\right)}\right)\\ =\underset{h\to 0}{\lim }\left(\frac{h}{h\left(\sqrt{x+h}+\sqrt{x}\right)}\right)\\ =\underset{h\to 0}{\lim }\left(\frac{1}{\left(\sqrt{x+h}+\sqrt{x}\right)}\right)\\ =\frac{1}{2\sqrt{x}}\end{array}$

So,

  $\begin{array}{c}f\text{'}\left(9\right)=\frac{1}{2\sqrt{x}}\\ =\frac{1}{6}\end{array}$

The slope of tangent line is $\frac{1}{6}$ .

Using the graphing calculator to plot the graph,

  

So, it can be seen that the slope of tangent line is $\frac{1}{6}$ .