The absolute maximum value of f(x) = sin x + | vedclass

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(D) To find the absolute maximum value of $f(x) = \sin x + \cos x$ on the interval $[0, \pi]$,we first find the critical points by setting the derivat

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$\sqrt{2}$

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(D) To find the absolute maximum value of $f(x) = \sin x + \cos x$ on the interval $[0, \pi]$,we first find the critical points by setting the derivative to zero.$f'(x) = \cos x - \sin x$.Setting $f'(x) = 0$,we get $\cos x = \sin x$,which implies $	an x = 1$.For $x \in [0, \pi]$,the solution is $x = \pi/4$.Now,we evaluate $f(x)$ at the critical point and the endpoints of the interval:$1$. At $x = 0$: $f(0) = \sin(0) + \cos(0) = 0 + 1 = 1$.$2$. At $x = \pi/4$: $f(\pi/4) = \sin(\pi/4) + \cos(\pi/4) = 1/\sqrt{2} + 1/\sqrt{2} = 2/\sqrt{2} = \sqrt{2}$.$3$. At $x = \pi$: $f(\pi) = \sin(\pi) + \cos(\pi) = 0 - 1 = -1$.Comparing these values,the absolute maximum value is $\sqrt{2}$.

The absolute maximum value of $f(x) = \sin x + \cos x$ for $x \in [0, \pi]$ is . . . . . . .

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