If m and M are the absolute minimum and absol | vedclass

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(C) Given $f(x) = 2\sqrt{2} \sin x - 	an x$ on $[0, \pi/3]$.First,find the derivative $f'(x) = 2\sqrt{2} \cos x - \sec^2 x$.Set $f'(x) = 0$ to find c

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$1$

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(C) Given $f(x) = 2\sqrt{2} \sin x - 	an x$ on $[0, \pi/3]$.First,find the derivative $f'(x) = 2\sqrt{2} \cos x - \sec^2 x$.Set $f'(x) = 0$ to find critical points:$2\sqrt{2} \cos x = \frac{1}{\cos^2 x} \implies \cos^3 x = \frac{1}{2\sqrt{2}} = \left(\frac{1}{\sqrt{2}}ight)^3$.Thus,$\cos x = \frac{1}{\sqrt{2}}$,which gives $x = \pi/4$.Now,evaluate $f(x)$ at critical points and endpoints:$f(0) = 2\sqrt{2}(0) - 0 = 0$.$f(\pi/4) = 2\sqrt{2}(\frac{1}{\sqrt{2}}) - 1 = 2 - 1 = 1$.$f(\pi/3) = 2\sqrt{2}(\frac{\sqrt{3}}{2}) - \sqrt{3} = \sqrt{6} - \sqrt{3} \approx 2.45 - 1.73 = 0.72$.Comparing values: $m = 0$ and $M = 1$.Therefore,$m + M = 0 + 1 = 1$.

If $m$ and $M$ are the absolute minimum and absolute maximum values of the function $f(x) = 2\sqrt{2} \sin x - \tan x$ in the interval $[0, \pi/3]$,then $m + M =$

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