# How do you find the domain and range of f(x) = sqrt(4 - x²)?

##### 1 Answer
Oct 20, 2015

Domain: $\left[- 2 , 2\right]$.
Range: $\left[0 , 2\right]$.

#### Explanation:

Domain: the root is well defined only if its argument is non-negative.

So, we must solve $4 - {x}^{2} \setminus \ge 0$, which leads to ${x}^{2} \setminus \le 4$. This inequality holds for $x \setminus \in \left[- 2 , 2\right]$.

As for the range, we observe that a root is always positive, so since $f \left(- 2\right) = f \left(2\right) = \sqrt{4 - 4} = 0$, we have that $0$ is the lowest possible value.

Also, we observe that $f \left(0\right) = \sqrt{4} = 2$, and for every other $x$ we will have $f \left(x\right) = \sqrt{4 - {x}^{2}}$, which means the square root of something less than $4$, as thus something less than $2$. So, $2$ is the maximum of the function.