Chapter 12, Problem 37CLT

To determine

To find the slope of the graph of the given function at the given point.

Answer to Problem 37CLT

  $m=-\frac{1}{16}$

Explanation of Solution

Given:

  $f\left(x\right)=\frac{1}{x+3},\left(1,\frac{1}{4}\right)$

Concept used:

The formula for the slope at any point on the graph of a function $f$ is given by,

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

Calculation:

Given,

  $f\left(x\right)=\frac{1}{x+3},\left(1,\frac{1}{4}\right)$

Now the formula for the slope of the function $f$ is,

  $\begin{array}{l}m=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}\\ \Rightarrow m=\underset{h\to 0}{\lim }\frac{\frac{1}{\left(x+h\right)+3}-\frac{1}{x+3}}{h}\\ \Rightarrow m=\underset{h\to 0}{\lim }\frac{\frac{x+3-\left(x+h+3\right)}{\left(x+h+3\right)\left(x+3\right)}}{h}\\ \Rightarrow m=\underset{h\to 0}{\lim }\frac{-h}{h\left(x+h+3\right)\left(x+3\right)}\\ \Rightarrow m=\underset{h\to 0}{\lim }\frac{-1}{\left(x+h+3\right)\left(x+3\right)}\\ \Rightarrow m=\frac{-1}{\left(x+0+3\right)\left(x+3\right)}\\ \Rightarrow m=\frac{-1}{\left(x+3\right)\left(x+3\right)}\\ \Rightarrow m=\frac{-1}{{\left(x+3\right)}^2}\end{array}$

Now substituting the point $\left(1,\frac{1}{4}\right)$ in the formula for slope, we get

  $\begin{array}{l}m=\frac{-1}{{\left(1+3\right)}^2}\\ \Rightarrow m=-\frac{1}{16}\end{array}$