Chapter 11.3, Problem 28E

a.

To determine

ToCalculate:The formula for the slope of the graph of $f\left(x\right)=\sqrt{x-4}$ at the point $\left(x,f\left(x\right)\right)$ . Then use it to find the slope at the point $\left(5,1\right)$

Answer to Problem 28E

The formula for the slope of the graph of $f\left(x\right)=\sqrt{x-4}$ at the point $\left(x,f\left(x\right)\right)$ is $m=\frac{1}{2\sqrt{x-4}}$

At $\left(5,1\right)$ , the slope is $m=\frac{1}{2}$

Explanation of Solution

Given:

The function $f\left(x\right)=\sqrt{x-4}$ and the point $\left(5,1\right)$

Concept Used:

The slope $m$ of the graph of $f$ at the point $\left(x,f\left(x\right)\right)$ is equal to the slope of its tangent line at $\left(x,f\left(x\right)\right)$ , and is given by

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

Calculation:

Given the function $f\left(x\right)=\sqrt{x-4}$

The slope $m$ of the graph of $f$ at the point $\left(x,f\left(x\right)\right)$ is given by

  $\begin{array}{c}m=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}\\ =\underset{h\to 0}{\lim }\left(\frac{\sqrt{x+h-4}-\sqrt{x-4}}{h}\right)\\ =\underset{h\to 0}{\lim }\left(\frac{\sqrt{x+h-4}-\sqrt{x-4}}{h}\right)\left(\frac{\sqrt{x+h-4}+\sqrt{x-4}}{\sqrt{x+h-4}+\sqrt{x-4}}\right)\\ =\underset{h\to 0}{\lim }\left(\frac{{\left(\sqrt{x+h-4}\right)}^2-{\left(\sqrt{x-4}\right)}^2}{h\left(\sqrt{x+h-4}+\sqrt{x-4}\right)}\right)\end{array}$

  $\begin{array}{c}=\underset{h\to 0}{\lim }\left(\frac{x+h-4-x+4}{h\left(\sqrt{x+h-4}+\sqrt{x-4}\right)}\right)\\ =\underset{h\to 0}{\lim }\left(\frac{h}{h\left(\sqrt{x+h-4}+\sqrt{x-4}\right)}\right)\\ =\underset{h\to 0}{\lim }\left(\frac{1}{\sqrt{x+h-4}+\sqrt{x-4}}\right)\\ =\frac{1}{\sqrt{x+0-4}+\sqrt{x-4}}\text{ Substituting limit}\end{array}$

  $=\frac{1}{2\sqrt{x-4}}$

Therefore, the formula for the slope of the graph of $f\left(x\right)=\sqrt{x-4}$ at the point $\left(x,f\left(x\right)\right)$ is $m=\frac{1}{2\sqrt{x-4}}$

At $\left(5,1\right)$ , the slope is

  $\begin{array}{c}m=\frac{1}{2\sqrt{5-4}}\\ =\frac{1}{2\sqrt{1}}\\ =\frac{1}{2\left(1\right)}\\ =\frac{1}{2}\end{array}$

Conclusion:

The formula for the slope of the graph of $f\left(x\right)=\sqrt{x-4}$ at the point $\left(x,f\left(x\right)\right)$ is $m=\frac{1}{2\sqrt{x-4}}$

At $\left(5,1\right)$ , the slope is $m=\frac{1}{2}$

b.

To determine

ToCalculate: The formula for the slope of the graph of $f\left(x\right)=\sqrt{x-4}$ at the point $\left(x,f\left(x\right)\right)$ . Then use it to find the slope at the point $\left(8,2\right)$

Answer to Problem 28E

The formula for the slope of the graph of $f\left(x\right)=\sqrt{x-4}$ at the point $\left(x,f\left(x\right)\right)$ is $m=\frac{1}{2\sqrt{x-4}}$

At $\left(8,2\right)$ , the slope is $m=\frac{1}{4}$

Explanation of Solution

Given:

The function $f\left(x\right)=\sqrt{x-4}$ and the point $\left(8,2\right)$

Concept Used:

The slope $m$ of the graph of $f$ at the point $\left(x,f\left(x\right)\right)$ is equal to the slope of its tangent line at $\left(x,f\left(x\right)\right)$ , and is given by

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

Calculation:

Given the function $f\left(x\right)=\sqrt{x-4}$

The slope $m$ of the graph of $f$ at the point $\left(x,f\left(x\right)\right)$ is given by

  $\begin{array}{c}m=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}\\ =\underset{h\to 0}{\lim }\left(\frac{\sqrt{x+h-4}-\sqrt{x-4}}{h}\right)\\ =\underset{h\to 0}{\lim }\left(\frac{\sqrt{x+h-4}-\sqrt{x-4}}{h}\right)\left(\frac{\sqrt{x+h-4}+\sqrt{x-4}}{\sqrt{x+h-4}+\sqrt{x-4}}\right)\\ =\underset{h\to 0}{\lim }\left(\frac{{\left(\sqrt{x+h-4}\right)}^2-{\left(\sqrt{x-4}\right)}^2}{h\left(\sqrt{x+h-4}+\sqrt{x-4}\right)}\right)\end{array}$

  $\begin{array}{c}=\underset{h\to 0}{\lim }\left(\frac{x+h-4-x+4}{h\left(\sqrt{x+h-4}+\sqrt{x-4}\right)}\right)\\ =\underset{h\to 0}{\lim }\left(\frac{h}{h\left(\sqrt{x+h-4}+\sqrt{x-4}\right)}\right)\\ =\underset{h\to 0}{\lim }\left(\frac{1}{\sqrt{x+h-4}+\sqrt{x-4}}\right)\\ =\frac{1}{\sqrt{x+0-4}+\sqrt{x-4}}\text{ Substituting limit}\end{array}$

  $=\frac{1}{2\sqrt{x-4}}$

Therefore, the formula for the slope of the graph of $f\left(x\right)=\sqrt{x-4}$ at the point $\left(x,f\left(x\right)\right)$ is $m=\frac{1}{2\sqrt{x-4}}$

At $\left(8,2\right)$ , the slope is

  $\begin{array}{c}m=\frac{1}{2\sqrt{8-4}}\\ =\frac{1}{2\sqrt{4}}\\ =\frac{1}{2\left(2\right)}\\ =\frac{1}{4}\end{array}$

Conclusion:

The formula for the slope of the graph of $f\left(x\right)=\sqrt{x-4}$ at the point $\left(x,f\left(x\right)\right)$ is $m=\frac{1}{2\sqrt{x-4}}$

At $\left(8,2\right)$ , the slope is $m=\frac{1}{4}$