Find the range of the function given by $f(x)=\sqrt {16-x^2}$ Find the range of the function given by $f(x)=\sqrt {16-x^2}$.
My Attempt:
$$f(x)=\sqrt {16-x^2}$$
$$y=\sqrt {16-x^2}$$
Squaring both sides, 
$$y^2=16-x^2$$
How do I proceed further?
 A: Assuming that $f(x)\in \Bbb R$, we must have:
$$16-x^2\ge 0 \to (x-4)(x+4)\le0\to-4\le x \le 4$$
We also have that $16-x^2$ is continuous on that interval and the maximum is $16$ (when $x=0$) then $0\le 16-x^2\le 16$, so
$$0\le f(x)\le 4 $$
A: You have to resort to the definition. The definition of the range of a function $g: X \to Y$ is the set $\{ y \in Y \mid y = f(x)\ \text{for some}\ x \in X \}$. Clearly the range of $g$ is a subset of $Y$, the codomain of $g$. 
The function $f$ apparently has $[0, +\infty[$ as its codomain. Note that $x < -4$ or $x > 4$ implies that $f(x)$ is not meaningful; so the domain of $f$ can be at most $[-4, 4]$. Let $y \geq 0$. Note that $y = f(x)$ for some $-4 \leq x \leq 4$ if and only if $y^{2} = 16 - x^{2}$, if and only if $x^{2} = 16 - y^{2}$, and if and only if $x = \pm \sqrt{16 - y^{2}}$. Note that $y > 4$ implies that $\sqrt{16-y^{2}}$ is not meaningful. So $y = f(x)$ for some $-4 \leq x \leq 4$ if and only if $0 \leq y \leq 4$; so the range of $f$ is $[0,4]$.
