Find the absolute min and max value on given interval
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 $$\frac{\mathrm{d}f}{\mathrm{d}t} = \sqrt{16t^2} \frac{t^2}{\sqrt{16t^2}} = \frac{162t^2}{\sqrt{16t^2}}$$ 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 $f(1) = \sqrt{161} = \sqrt{15}$, $f(2\sqrt{2}) = 2\sqrt{2}(\sqrt{168}) = 8$ and $f(4) = 4\sqrt{1616} = 0$. 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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user106039
Updated on January 29, 2020Comments

user106039 almost 4 years
$f(t) = t\sqrt{16 − t^2}$ 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: $(1,3.87) ; (4,0) ; (2\sqrt2,33.94)$ and $(2\sqrt2,8)$
I thought the absolute minimum value had to be $(2\sqrt2,8)$ and the absolute maximum value was $(2\sqrt2,33.94)$ but apparently this is incorrect. Where did I go wrong?