[Math] Find the absolute maximum and absolute minimum values of f on the given interval

Question

Find the absolute maximum and absolute minimum values of f on the given interval.
$f(t) = t\sqrt{9 – t^2}$ on the interval $[-1,3]$. So $f'(x)=\frac{t}{2\sqrt{9-t^2}}+t\sqrt{9-t^2}$ and that is as far as I got simplifying. The website I'm doing this on simplifies to $f'(x)=\frac{9-t^2}{\sqrt{9-t^2}}$ And after that it is simple to get the answer, but I am not currently able to simplify enough to get $f'(x)$ to that and it saddens me… So if someone could show me step by step on how to properly answer this question, I would really appreciate it, thanks!

Answer 1

$f'(t) = \sqrt{9-t^2} - \dfrac{t^2}{\sqrt{9-t^2}} = \dfrac{9-2t^2}{\sqrt{9-t^2}} = 0$ if $ 9-2t^2=0$ or $t = \pm\dfrac{3}{\sqrt{2}}$. But only $t = \dfrac{3}{\sqrt{2}}$ can be accepted since $-\dfrac{3}{\sqrt{2}} < -1$. Thus:

$f_{max} = \max\{f(-1), f(\frac{3}{\sqrt{2}}), f(3)\} = \max\{-2\sqrt{2},0, \frac{9}{2}\} = \dfrac{9}{2}$, and this maximum value occurs at $t = \dfrac{3}{\sqrt{2}}$, and $f_{min} = -2\sqrt{2}$ which occurs when $t = -1$.