# How do you find the domain of f(x)=sqrt(4-x^2)?

Aug 31, 2017

Assuming we are restricted to Real values,
Domain of $f \left(x\right)$ is $x \in \left[- 2 , + 2\right]$

#### Explanation:

For $\sqrt{4 - {x}^{2}}$ to be a valid Real number

$4 - {x}^{2} \ge 0$ (the square root is not defined in Real numbers for negative arguments)

$\rightarrow 4 \ge {x}^{2}$

$\rightarrow - 2 \le x \le + 2$