# What is the domain and range if the function f(x)= sqrt(4-x^2)?

Jun 7, 2015

Your domain is all the legal (or possible) values of $x$, while the range is all the legal (or possible) values of $y$.

Domain
The domain of a function includes every possible value of $x$ that will not involve division by zero or make a complex number. You can only get complex numbers if you can turn the stuff inside of the square root negative. Because there is no denominator, you will never divide by zero. What about complex numbers? You have to set the inside of the square root to less than zero and solve:

$4 - {x}^{2} < 0$
$\left(2 + x\right) \left(2 - x\right) < 0$ or when
$2 + x < 0$ and $2 - x < 0$. That is, when
$x < - 2$ and $x > 2$

So your domain is $\left[- 2 , 2\right]$. Both the $2$ and $- 2$ are included, because the stuff inside the square root is allowed to be zero.

Range
Your range is partly determined by your legal values of $x$. It is best to look at the graph to see the smallest and the greatest value of $y$ that falls within the domain.

graph{sqrt(4-x^2) [-2.1,2.1,-1,2.5]}

This is the top half a circle and the range is $\left[0 , 2\right]$.

Jun 7, 2015

{x$\in$R : $- 2 \le x \le 2$} and
{y$\in$R: $0 \le y \le 2$}

Because of the radical sign, for f(x) to be a real function, $4 \ge {x}^{2}$, that implies $2 \ge \pm x$. Stated more simply, it is $- 2 \le x \le 2$. The domain is therefore, [-2,2] and within this domain the Range would be [0,2]. In set builder notation {x$\in$R : $- 2 \le x \le 2$} and
{y$\in$R: $0 \le y \le 2$}