prove that $[0;36]$ is the range of $f(x) = 3 \sqrt{144 - x^2}$ I need to prove that $[0;46]$ is the range of $f(x) = 3 \sqrt{144 - x^2}$.
At first I found that the domain of this function is $x\in [-12;12]$. So to prove that the set $[0;46]$ = $f([-12;12])$ (the image of x) I proceed to the following.
Prove that $f([-12;12]) \subset [0;46]$ => For every $x \in D = [-12;12]$, $f(x) = [0;46]$
So I wrote $-12 \leqslant x \leqslant 12$ and continued modifing $x$ in order to get the form $3 \sqrt{144 - x^2}$
Problem Now i'm unsure what to write next, $-144 \leqslant x^2 \leqslant 144$ or $0 \leqslant x^2 \leqslant 144$? I found the second one in my notes.
My aim is  to continue step by step writing the same for  $x^2, 144 - x^2, \sqrt{144 - x^2}$ and so on so I can finally get the form $3 \sqrt{144 - x^2}$. I'm particularly interested in proving by this approach.
 A: Its rather very easy to sketch the graph using derivatives, thereby finding the extrema:


Following OP's approach:
$$-12 \leqslant x \leqslant 12$$
$$\implies 0 \leqslant x^2 \leqslant 144 $$ Note that $x^2$ is always non-negative
Multiply by $-1$, thereby reversing the inequality
$$\implies 0 \geqslant -x^2 \geqslant -144$$
Add $144$
$$\implies 144 \geqslant 144-x^2 \geqslant 0$$
Take square root
$$\implies 12 \geqslant \sqrt{144-x^2} \geqslant 0$$
Multiply by $3$
$$\implies 36 \geqslant 3\sqrt{144-x^2} \geqslant 0$$
A: Domain of $f(x)=3\sqrt{144-x^2}$ is $x \in [-12,12]$ in this domain $min(f(x))=f(\pm 12)=0$
and $max(f(x))=f(0)=36$. Hence, the range of $f(x)$ is $[0,36].$
A: If we consider the graph of y=f(x), we can rearrange to get the ellipse $$\frac{x^2}{12^2}+\frac{y^2}{36^2}=1.$$ The function itself is only the top half since we only have the positive square root. In this form, the ellipse has center $(0,0)$, minor axis of 12 in the x-direction, and major axis of 36 in the y-direction. The domain is thus $[-12,12]$ and the range $[0,36]$.
