# Question #b37dd

Feb 6, 2018

The function is continuous on its entire domain.

#### Explanation:

The domain of $f \left(x\right) = \frac{1}{\sqrt{x}}$
is the open interval $\left(0 , \infty\right)$.
For each point, a, in that interval, f is the quotient of two continuous functions -- with a nonzero denominator -- and is therefore continuous.

Feb 6, 2018

Find the "breaks" in the domain

#### Explanation:

Functions will often have inputs that, for lack of a better word, "break" the function. For functions of the form $\frac{1}{x}$, the denominator cannot equal zero. For functions of the form $\sqrt{x}$, the number under the radical must be greater than or equal to zero.

For your function, $f \left(x\right) = \frac{1}{\sqrt{x}}$, your domain is restricted by both the denominator and the square root.

Since the variable is in the denominator, we can set the denominator equal to zero and find that restriction, in this case $x \ne 0$

But, since the variable is also under a square root, $x$ must be greater than zero as well.

When you look at the domain for your function, $\left(0 , \text{infinity}\right)$, you notice that there are no gaps. Therefore, in its domain, the function $f \left(x\right) = \frac{1}{\sqrt{x}}$ is continuous.