# How do you find the domain and range of  f(a)=3sqrt(4a)?

##### 1 Answer
Jun 30, 2016

Assuming we are restricted to Real numbers:
${\text{Domain}}_{f} = \left[0 , + \infty\right)$
${\text{Range}}_{f} = \left[0 , + \infty\right)$

#### Explanation:

$3 \sqrt{4 a}$ is defined as a Real number for all values of $\left(4 a\right) \ge 0$
which implies for all values of $a \ge 0$
Thus the domain is $\left[0 , + \infty\right)$

The minimum value of $3 \sqrt{4 a}$ is $0$ which occurs when $a = 0$ and increases without bound as $a \rightarrow + \infty$
Thus the range of $f \left(a\right) = 3 \sqrt{4 a}$ is $\left[0 , + \infty\right)$