Chapter 11.3, Problem 8QR

Solution

(a)

To find: The points of continuity for the function $g\left(x\right)=\frac{2x+1}{x^2-4}$ .

The points of continuity are $\left(-\infty ,-2\right)\cup \left(-2,2\right)\cup \left(2,\infty \right)$ .

Given information:

The given function is $g\left(x\right)=\frac{2x+1}{x^2-4}$ .

Formula used:

Continuity of function at a point: The function $f\left(x\right)$ is said to be continuous at point $x=c$ if $\underset{x\to c}{\lim }f\left(x\right)=f\left(c\right)$ .

Calculation:

Domain of $g\left(x\right)$ is $\left(-\infty ,-2\right)\cup \left(-2,2\right)\cup \left(2,\infty \right)$ .

Substitute $c$ for $x$ in $g\left(x\right)$

  $g\left(c\right)=\frac{2c+1}{c^2-4}$

  $\begin{array}{c}\underset{x\to c}{\lim }g\left(x\right)=\underset{x\to c}{\lim }\frac{2x+1}{x^2-4}\\ \underset{x\to c}{\lim }g\left(x\right)=\frac{2c+1}{c^2-4},c\ne \left\{-2,2\right\}\end{array}$

Thus, the points of continuity of $g\left(x\right)=\frac{2x+1}{x^2-4}$ are $\left(-\infty ,-2\right)\cup \left(-2,2\right)\cup \left(2,\infty \right)$ .

(b)

To find: The points of discontinuity for the function $g\left(x\right)=\frac{2x+1}{x^2-4}$ .

The points of discontinuity are $x=-2$ and $x=2$ .

Given information:

The given function is $g\left(x\right)=\frac{2x+1}{x^2-4}$ .

Formula used:

Points of discontinuity: The points of domain where the function is not continuous are called points of discontinuity of that function.

Calculation:

Since $g\left(x\right)=\frac{2x+1}{x^2-4}$ is not defined at $x=-2$ and $x=2$ .

Thus the points of discontinuity of $g\left(x\right)=\frac{2x+1}{x^2-4}$ are $x=-2$ and $x=2$ .