Chapter 14.3, Problem 76AYU

To determine

To find: The continuity of $f$ at $c=-4$ and $c=4$ and Graph $R$.

Answer to Problem 76AYU

Not continuous at $c=4$ but continuous at $c=-4$.

Explanation of Solution

Given:

$R\left(x\right)=\frac{x^2+4x}{x^2-16}$

Calculation:

$R\left(x\right)=\frac{x^2+4x}{x^2-16}$

$R\left(x\right)=\frac{x\left(x+4\right)}{\left(x+4\right)\left(x-4\right)}$

$R\left(x\right)=\frac{x}{x-4}$

At $c=4$; we see that,

${\lim }_{x\to c}R\left(x\right)={\text{ lim}}_{x\to 4}\frac{x}{\left(x-4\right)}=\frac{4}{4-4}=\frac{4}{0}$; does not exist.

Therefore, $R$ is not continuous at $c=4$.

At $c=-4$; we see that,

${\lim }_{x\to c}R\left(x\right)={\text{ lim}}_{x\to -4}\frac{x}{\left(x-4\right)}=\frac{-4}{\left(-4-4\right)}=\frac{-4}{-8}=\frac{1}{2}$

Also, $R\left(-4\right)=\frac{-4}{-4-4}=\frac{-4}{-8}=\frac{1}{2}$

So ${\text{lim}}_{x\to -4}R\left(x\right)=R\left(-4\right)$

Therefore, $R$ is continuous at $c=-4$.

Graph of $R$ is as follows: