Chapter 12.3, Problem 25E

a.

To determine

ToCalculate:The formula for the slope of the graph of $f\left(x\right)=\frac{1}{x+4}$ at the point $\left(x,f\left(x\right)\right)$ . Then use it to find the slope at the point $\left(0,\frac{1}{4}\right)$

Answer to Problem 25E

The formula for the slope of the graph of $f\left(x\right)=x^3$ at the point $\left(x,f\left(x\right)\right)$ is $m=\frac{-1}{{\left(x+4\right)}^2}$

At $\left(0,\frac{1}{4}\right)$ , the slope is $m=\frac{-1}{16}$

Explanation of Solution

Given:

The function $f\left(x\right)=\frac{1}{x+4}$ and the point $\left(0,\frac{1}{4}\right)$

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 $f\left(x\right)=\frac{1}{x+4}$

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

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

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

  $=\frac{-1}{{\left(x+4\right)}^2}$

Therefore, the formula for the slope of the graph of $f\left(x\right)=x^3$ at the point $\left(x,f\left(x\right)\right)$ is $m=\frac{-1}{{\left(x+4\right)}^2}$

At $\left(0,\frac{1}{4}\right)$ , the slope is

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

Conclusion:

The formula for the slope of the graph of $f\left(x\right)=x^3$ at the point $\left(x,f\left(x\right)\right)$ is $m=\frac{-1}{{\left(x+4\right)}^2}$

At $\left(0,\frac{1}{4}\right)$ , the slope is $m=\frac{-1}{16}$

b.

To determine

ToCalculate: The formula for the slope of the graph of $f\left(x\right)=\frac{1}{x+4}$ at the point $\left(x,f\left(x\right)\right)$ . Then use it to find the slope at the point $\left(-2,\frac{1}{2}\right)$

Answer to Problem 25E

The formula for the slope of the graph of $f\left(x\right)=x^3$ at the point $\left(x,f\left(x\right)\right)$ is $m=\frac{-1}{{\left(x+4\right)}^2}$

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

Explanation of Solution

Given:

The function $f\left(x\right)=\frac{1}{x+4}$ and the point $\left(-2,\frac{1}{2}\right)$

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 $f\left(x\right)=\frac{1}{x+4}$

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

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

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

  $=\frac{-1}{{\left(x+4\right)}^2}$

Therefore, the formula for the slope of the graph of $f\left(x\right)=x^3$ at the point $\left(x,f\left(x\right)\right)$ is $m=\frac{-1}{{\left(x+4\right)}^2}$

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

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

Conclusion:

The formula for the slope of the graph of $f\left(x\right)=x^3$ at the point $\left(x,f\left(x\right)\right)$ is $m=\frac{-1}{{\left(x+4\right)}^2}$

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