# How do you find the domain of f(x) = sqr(25 - x^2)?

The given equation is $f \left(x\right) = \sqrt{25 - {x}^{2}}$ and any value of x>5 will make the radicand negative. Also any value of x<-5 will make the radicand negative, therefore the domain is $- 5 \le x \le 5$.
The range is found by checking the value of f(x) at $x = \pm 5$ and $x = 0$. When $x = \pm 5$ the value of $f \left(x\right) = 0$ and when $x = 0$, the value of $f \left(x\right) = 5$. We can conclude that Range is $0 \le f \left(x\right) \le 5$