Chapter 14.3, Problem 69AYU

$f\left(x\right)=\frac{2x+5}{x^2-4}$

To determine

To find: The numbers at which $f$ is continuous.

Answer to Problem 69AYU

Continuous for all real values of $x$ excluding $x=\pm 2$.

Explanation of Solution

Given:

$f\left(x\right)=\frac{2x+5}{x^2-4}$

Calculation:

$f\left(x\right)=\frac{2x+5}{x^2-4}$ is a rational function.

The denominator of $f\left(x\right)=\frac{2x+5}{x^2-4}$ is zero for $x^2-4=0$ ie for $x=\pm 2$.

Since $f\left(x\right)=\frac{2x+5}{x^2-4}=\frac{2x+5}{\left(x+2\right)\left(x-2\right)}$ is undefined for $x=\pm 2$; therefore $f\left(x\right)=\frac{2x+5}{x^2-4}$ is continuous for all real numbers excluding $x=\pm 2$.