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

Apr 12, 2015

Answer: $- 2 \le x \le 2$

The domain of any function is the set of values of $x$ that can produce a real output $y$ or $f \left(x\right)$

So you would bear with me that if $4 - {x}^{2}$ is negative then you have you would have the square root of a negative number which is imaginary and not real

Hence the function returns a real value when $4 - {x}^{2}$ is positive or zero

That is $4 - {x}^{2} \ge 0$

Hence we find the range of value that the above inequality represents,

Here we go,

$4 - {x}^{2} \ge 0 \implies \left(2 - x\right) \left(2 + x\right) \ge 0$

$\implies - 1 \cdot \left(x - 2\right) \left(x + 2\right) \ge 0$

$\implies \left(x - 2\right) \left(x + 2\right) \le 0$

$\implies - 2 \le x \le 2$ is the required domain!