# What is the domain and range of y=sqrt(4-x^2) ?

Jun 9, 2017

Domain: $\left[- 2 , 2\right]$

#### Explanation:

Start by solving the equation

$4 - {x}^{2} = 0$

Then

$\left(2 + x\right) \left(2 - x\right) = 0$

$x = \pm 2$

Now select a test point, let it be $x = 0$. Then $y = \sqrt{4 - {0}^{2}} = 2$, so the function is defined on [-2, 2[.

Thus, the graph of $y = \sqrt{4 - {x}^{2}}$ is a semicircle with radius $2$ and domain $\left[- 2 , 2\right]$.

Hopefully this helps!

Jun 12, 2017

Range: $0 < = y < = 2$

#### Explanation:

The domain has already been determined to be $- 2 < = x < = 2$. To find the range, we should find any absolute extrema of $y$ on this interval.

$y = \sqrt{4 - {x}^{2}} = {\left(4 - {x}^{2}\right)}^{\frac{1}{2}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} {\left(4 - {x}^{2}\right)}^{- \frac{1}{2}} \frac{d}{\mathrm{dx}} \left(4 - {x}^{2}\right) = \frac{1}{2} {\left(4 - {x}^{2}\right)}^{- \frac{1}{2}} \left(- 2 x\right) = \frac{- x}{\sqrt{4 - {x}^{2}}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 0$ when $x = 0$ and is undefined when $x = \pm 2$.

$y \left(- 2\right) = 0$, $y \left(2\right) = 0$ and $y \left(0\right) = 2$.

Thus the range is $0 < = y < = 2$.

We could also arrive at this conclusion by considering the graph of the function:

${y}^{2} = 4 - {x}^{2}$

${x}^{2} + {y}^{2} = 4$

Which is a circle centered at $\left(0 , 0\right)$ with radius $2$.

Note that solving for $y$ gives $y = \pm \sqrt{4 - {x}^{2}}$, which is a set of two functions, since a circle by itself does not pass the vertical line test, so a circle is not a function but can be described by a set of $2$ functions.

Thus $y = \sqrt{4 - {x}^{2}}$ is the top half of the circle, which starts at $\left(- 2 , 0\right)$, rises to $\left(0 , 2\right)$, then descends to $\left(2 , 0\right)$, showing its range of $0 < = y < = 2$.