Chapter 1.3, Problem 37E

To determine

To Verify: The given function is continuous and state the domain. State the theorem

used and the function that assumed to be continuous.

Answer to Problem 37E

The given function is continuous and it is verified by using the sum theorem,

composite theorem and quotient theorem.

The domain is $\left(-2,\infty \right)$

Explanation of Solution

Given:

  $y=\frac{1}{\sqrt{x+2}}$

Calculation:

Given the function $y=\frac{1}{\sqrt{x+2}}$ :

Here $f\left(x\right)=x$ , $g\left(x\right)=2$ and $h\left(x\right)=\sqrt{x}$ are continuous by using various theorem:

By using the sum theorem: $k\left(x\right)=f\left(x\right)+g\left(x\right)$

Substitute $f\left(x\right)=x$ and $g\left(x\right)=2$

  $\begin{array}{c}k\left(x\right)=f\left(x\right)+g\left(x\right)\\ =x+2\end{array}$ which is continuous

Thus, the function $y=\frac{1}{\sqrt{x+2}}$ is continuous.

By using the composite theorem, $\left(h\circ k\right)\left(x\right)=h\left(k\left(x\right)\right)$

Substitute $k\left(x\right)=x+2$ :

  $\begin{array}{c}\left(h\circ k\right)\left(x\right)=h\left(k\left(x\right)\right)\\ =\sqrt{x+2}\end{array}$ which is continuous

Thus, the function $y=\frac{1}{\sqrt{x+2}}$ is continuous.

And by using the quotient theorem: $\frac{1}{\left(h\circ k\right)\left(x\right)}$

Substitute $\left(h\circ k\right)\left(x\right)=\sqrt{x+2}$

  $\frac{1}{\left(h\circ k\right)\left(x\right)}=\frac{1}{\sqrt{x+2}}$ which is continuous

Thus, the function $y=\frac{1}{\sqrt{x+2}}$ is continuous.

Find Domain:

To find the domain, set $x+2>0$ and solve for $x$ :

  $\begin{array}{c}x+2>0\\ x>-2\end{array}$

Thus, the domain of the function is $\left(-2,\infty \right)$ .