# Question #04a42

Nov 25, 2017

$2 \ge y \ge 0$

#### Explanation:

The range of a function is the range of values which the function can take given it's domain.
As the domain is not given, assume that it is all values of x which give a real y (in this case $| \setminus x | \le 2$).

Three important facts to remember in this question are that:
a) you can only take the square root of a positive number
b) $y = \sqrt{f \left(x\right)}$ only plots the positive square root (principal root)
c) any real number squared $\ge 0$

Using a), you know $4 - {x}^{2} \ge 0$. Because of c), the maximum value $4 - {x}^{2}$ can take is 4, when $x = 0$, so the maximum value of y is $\sqrt{4} = 2$.
Using a) the minimum value $4 - {x}^{2}$ can take is 0, when $x = \pm 2$, and $\sqrt{0} = 0$.

Given the minimum and maximum it follows that $0 \le x \le 2$