Chapter 12.3, Problem 19E

(a)

To determine

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

Answer to Problem 19E

The slope of the graph of the function at the specified point is $\frac{-1}{16}$ .

Explanation of Solution

Given information:

The given function is,

  $f\left(x\right)=\frac{1}{x+4}$

The given point is $\left(0,\frac{1}{4}\right)$ .

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,

  $f\left(x\right)=\frac{1}{x+4}$

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

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

Simplify further,

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

The slope at the point $\left(0,\frac{1}{4}\right)$ is calculated as,

  $\begin{array}{c}m=\frac{-1}{{\left(x+4\right)}^2}\\ =\frac{-1}{{\left(0+4\right)}^2}\\ =\frac{-1}{16}\end{array}$

Therefore, the slope of the graph of the function at the specified point is $\frac{-1}{16}$ .

(b)

To determine

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

Answer to Problem 19E

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,

  $f\left(x\right)=\frac{1}{x+4}$

The given point is $\left(-2,\frac{1}{2}\right)$ .

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,

  $f\left(x\right)=\frac{1}{x+4}$

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

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

Simplify further,

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

The slope at the point $\left(-2,\frac{1}{2}\right)$ is calculated as,

  $\begin{array}{c}m=\frac{-1}{{\left(x+4\right)}^2}\\ =\frac{-1}{{\left(-2+4\right)}^2}\\ =\frac{-1}{4}\end{array}$

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