Chapter 12.3, Problem 22E

(a)

To determine

To find: The slope of the graph of the function at the specified point.

Answer to Problem 22E

The slope of the graph of the function at the specified point is $\frac{1}{2}$ .

Explanation of Solution

Given information:

The given function is,

  $f\left(x\right)=\sqrt{x-4}$

The given point is $\left(5,1\right)$ .

Concept used:

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

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

The given function is,

  $f\left(x\right)=\sqrt{x-4}$

The slope of the graph of the function at the specified point is calculated as,

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

Simplify further,

  $\begin{array}{c}m=\underset{h\to 0}{\lim }\frac{1}{\left[\sqrt{x+h-4}\right]+\left[\sqrt{x-4}\right]}\\ =\frac{1}{2\sqrt{x-4}}\end{array}$

The slope at the point $\left(5,1\right)$ is calculated as,

  $\begin{array}{c}m=\frac{1}{2\sqrt{x-4}}\\ =\frac{1}{2\sqrt{5-4}}\\ =\frac{1}{2}\end{array}$

Therefore, the slope of the graph of the function at the specified point is $\frac{1}{2}$ .

(b)

To determine

To find: The slope of the graph of the function at the specified point.

Answer to Problem 22E

The slope of the graph of the function at the specified point is $\frac{1}{4}$ .

Explanation of Solution

Given Information:

The given function is,

  $f\left(x\right)=\sqrt{x-4}$

The given point is $\left(8,2\right)$ .

Concept used:

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

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

The given function is,

  $f\left(x\right)=\sqrt{x-4}$

The slope of the graph of the function at the specified point is calculated as,

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

Simplify further,

  $\begin{array}{c}m=\underset{h\to 0}{\lim }\frac{1}{\left[\sqrt{x+h-4}\right]+\left[\sqrt{x-4}\right]}\\ =\frac{1}{2\sqrt{x-4}}\end{array}$

The slope at the point $\left(8,2\right)$ is calculated as,

  $\begin{array}{c}m=\frac{1}{2\sqrt{x-4}}\\ =\frac{1}{2\sqrt{8-4}}\\ =\frac{1}{4}\end{array}$

Therefore, the slope of the graph of the function at the specified point is $\frac{1}{4}$ .