Chapter 2.3, Problem 37E

To determine

To find: The reason for the continuity of the function $y=\frac{1}{\sqrt{x+2}}$ , check the domain of the function, also mention the theorems used in the continuity of function with the functions assumed as continuous.

Answer to Problem 37E

The domain of the function $y=\frac{1}{\sqrt{x+2}}$ is equal to interval $\left(-2,\infty \right)$ and the quotient, composition theorem is used and the functions $\sqrt{x}$ and $x+2$ are assumed as continuous.

Explanation of Solution

Given information:

The function is $y=\frac{1}{\sqrt{x+2}}$ .

Calculation:

The function $y=\frac{1}{\sqrt{x+2}}$ is the quotient of two functions $f\left(x\right)=1$ and $g\left(x\right)=\sqrt{x+2}$ .

The function $y=\frac{1}{\sqrt{x+2}}$ is equal to the function $\frac{f\left(x\right)}{g\left(x\right)}$ .

The function $f\left(x\right)=1$ is a continuous function.

The function $g\left(x\right)=\sqrt{x+2}$ is the composition of two functions $h\left(x\right)=\sqrt{x}$ and $j\left(x\right)=x+2$ .

Assume that the function $h\left(x\right)=\sqrt{x}$ and $j\left(x\right)=x+2$ are continuous functions.

So, by the composition theorem the function $g\left(x\right)=\sqrt{x+2}$ is continuous function.

So, by the help of quotient theorem the function $y=\frac{1}{\sqrt{x+2}}$ is continuous as the functions $f\left(x\right)$ and $g\left(x\right)$ both are continuous.

The function $y=\frac{1}{\sqrt{x+2}}$ is defined for $x>-2$ . So, the domain of the function $y=\frac{1}{\sqrt{x+2}}$ is equal to interval $\left(-2,\infty \right)$ .

Therefore, the domain of the function $y=\frac{1}{\sqrt{x+2}}$ is equal to interval $\left(-2,\infty \right)$ and the quotient, composition theorem is used and the functions $\sqrt{x}$ and $x+2$ are assumed as continuous.