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f(t) = t√(16 − t^2) on interval [−1, 4]. I was able to find the critical points which are 2√2 and -2√2. All was left for me to do was to evaluate both my critical and end points into the given function. I ended up with the points: (-1,-3.87) (4,0) (2√2,33.94) and (-2√2,-8). I thought the absolute minimum value had to be (-2√2,-8) and the absolute maximum value was (2√2,33.94) but apparently this is incorrect. Where did I go wrong?

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The first issue is that $-2 \sqrt{2}$ is not in the interval [-1,4]. However, $2\sqrt{2}$ is indeed a critical point, as $$\frac{\mathrm{d}f}{\mathrm{d}t} = \sqrt{16-t^2} -\frac{t^2}{\sqrt{16-t^2}} = \frac{16-2t^2}{\sqrt{16-t^2}}$$ which is zero at $t=2\sqrt{2}$. Now our possible candidates for the absolute minima and maxima are $t = -1, 2\sqrt{2}, 4$. Evaulating, we obtain $f(-1) = -\sqrt{16-1} = -\sqrt{15}$, $f(2\sqrt{2}) = 2\sqrt{2}(\sqrt{16-8}) = 8$ and $f(4) = 4\sqrt{16-16} = 0$. Thus, our absolute minimum value is $-\sqrt{15}$ at $t=-1$ and our absolute maximum value is $8$ at $t=2\sqrt{2}$.

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