Chapter 14.3, Problem 74AYU

To determine

To find: The continuity of $f$ at $c=-2$ and $c=2$ and Graph $R$.

Answer to Problem 74AYU

Not continuous at $c=2$ but continuous at $c=-2$.

Explanation of Solution

Given:

$R\left(x\right)=\frac{3x+6}{x^2-4}$

Calculation:

$R\left(x\right)=\frac{3x+6}{x^2-4}$

$R\left(x\right)=\frac{3\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}$

$R\left(x\right)=\frac{3}{x-2}$

At $c=2$; we see that,

${\lim }_{x\to c}R\left(x\right)={\text{ lim}}_{x\to 2}\frac{3}{\left(x-2\right)}=\frac{3}{2-2}=\frac{3}{0}$; does not exist.

Therefore, $R$ is not continuous at $c=2$.

At $c=-2$; we see that,

${\lim }_{x\to c}R\left(x\right)={\text{ lim}}_{x\to -2}\frac{3}{\left(x-2\right)}=\frac{3}{\left(-2-2\right)}=-\frac{3}{4}$

Also, $R\left(-2\right)=\frac{3}{-2-2}=-\frac{3}{4}$

So ${\lim }_{x\to -2}R\left(x\right)=R\left(-2\right)$

Therefore, $R$ is continuous at $c=-2$.

Graph of $R$ is as follows: