# What is the domain and range of f(x) = 1/(1 + sqrtx)?

Feb 1, 2018

The domain is $x \in \left[0 , + \infty\right)$ and the range is $\left(0 , 1\right]$

#### Explanation:

What's under the square root sign is $\ge 0$

Therefore,

$x \ge 0$

So , the domain is $x \in \left[0 , + \infty\right)$

To calculate the range, proceed as follows :

Let $y = \frac{1}{1 + \sqrt{x}}$

When $x = 0$, $\implies$, $y = 1$

And

${\lim}_{\to + \infty} \frac{1}{1 + \sqrt{x}} = {0}^{+}$

Therefore the range is $\left(0 , 1\right]$

graph{1/(1+sqrtx) [-2.145, 11.9, -3.52, 3.5]}