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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!

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    $\begingroup$ A correction: $f'(x)=\sqrt{9-t^2}-\dfrac{t^2}{\sqrt{9-t^2}}=\dfrac{9-2t^2}{\sqrt{9-t^2}}$. $\endgroup$ – DiegoMath Jul 12 '14 at 1:45
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$f'(t) = \sqrt{9-t^2} - \dfrac{t^2}{\sqrt{9-t^2}} = \dfrac{9-2t^2}{\sqrt{9-t^2}} = 0$ 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 $t = -1$.

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  • $\begingroup$ Thanks a lot for the help $\frac{d}{dx}4\pi r^2$! $\endgroup$ – Kenshin Jul 12 '14 at 2:19
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    $\begingroup$ @Kenshin: you are close to the secret.... $\endgroup$ – DeepSea Jul 12 '14 at 2:19
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Given is

$$ f(t) = t \sqrt{9-t^2}. $$

Put the $t$ inside the square root, then we get

$$ f(t) = \frac{|t|}{t} \sqrt{9t^2 - t^4} = \frac{|t|}{t} \sqrt{ \frac{81}{4} - \Big(t^2 - \frac{9}{2}\Big)^2}, $$

so we obtain

$$ \begin{eqnarray} f_{max} &=& \frac{9}{2},\ \textrm{for}\ t = + \frac{\sqrt{3}}{2},\\ f_{min} &=& -\frac{9}{2},\ \textrm{for}\ t = - \frac{\sqrt{3}}{2}. \end{eqnarray} $$

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