Question

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

Answer 1

Domain: `[-2, 2]`

Explanation:

Start by solving the equation

`4 - x^2 = 0`

Then

`(2 + x)(2 -x) = 0`

`x = +- 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 `[-2, 2]`.

Hopefully this helps!

Answer 2

Range: `0lt=ylt=2`

Explanation:

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

`y=sqrt(4-x^2)=(4-x^2)^(1/2)`

`dy/dx=1/2(4-x^2)^(-1/2)d/dx(4-x^2)=1/2(4-x^2)^(-1/2)(-2x)=(-x)/sqrt(4-x^2)`

`dy/dx=0` when `x=0` and is undefined when `x=pm2`.

`y(-2)=0`, `y(2)=0` and `y(0)=2`.

Thus the range is `0lt=ylt=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 `(0,0)` with radius `2`.

Note that solving for `y` gives `y=pmsqrt(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 `(-2,0)`, rises to `(0,2)`, then descends to `(2,0)`, showing its range of `0lt=ylt=2`.