Chapter 13.5, Problem 54E

To determine

To calculate: The points of discontinuity of the function $f\left(x\right)=\frac{5}{x^2-4}$ .

Answer to Problem 54E

The points of discontinuity of the function $f\left(x\right)=\frac{5}{x^2-4}$ are $x=\pm 2$ .

Explanation of Solution

Given information:

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

Calculation:

Consider the function $f\left(x\right)=\frac{5}{x^2-4}$ .

The function is said to be continuous when it is a smooth curve with no breaks in between.

For the function to be defined the denominator should not be equal to 0.

  $\begin{array}{c}{x}^2-4\ne 0\\ x^2\ne 4\\ x\ne \pm 2\end{array}$

When $x=\pm 2$ the function is not continuous.

The points $x=2\text{ and }x=-2$ are known as the point discontinuities of the function $f\left(x\right)=\frac{5}{x^2-4}$ as the function is defined on all real numbers except for $x=2\text{ and }x=-2$ but the pieces of the graph of the function $f\left(x\right)=\frac{5}{x^2-4}$ match up to sketch the function.

  

Thus, the points of discontinuity of the function $f\left(x\right)=\frac{5}{x^2-4}$ are $x=\pm 2$ .