Chapter 11.3, Problem 7QR

Solution

(a)

To find: The points of continuity for the function $f\left(x\right)=\sqrt{x+2}$ .

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

Given information:

The given function is $f\left(x\right)=\sqrt{x+2}$ .

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 the function is $\left[-2,\infty \right)$ .

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

  $f\left(c\right)=\sqrt{c+2}$

  $\begin{array}{c}\underset{x\to c}{\lim }f\left(x\right)=\underset{x\to c}{\lim }\sqrt{x+2}\\ \underset{x\to c}{\lim }f\left(x\right)=\sqrt{c+2},\forall x\in \left[-2,\infty \right)\end{array}$

Thus, the points of continuity of $f\left(x\right)=\sqrt{x+2}$ are $\left[-2,\infty \right)$ .

(b)

To find: The points of discontinuity for the function $f\left(x\right)=\sqrt{x+2}$ .

There are no points of discontinuity.

Given information:

The given function is $f\left(x\right)=\sqrt{x+2}$ .

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 $f\left(x\right)=\sqrt{x+2}$ is continuous at every point of its domain. So, there is no point of discontinuity of $f\left(x\right)=\sqrt{x+2}$ .