Find the absolute maximum value and the absol | vedclass

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(A) The given function is $f(x) = \sin x + \cos x$. First,we find the derivative: $f'(x) = \cos x - \sin x$. To find the critical points,set $f'(x) =

Vedclass · Accepted Answer

Absolute maximum: $\sqrt{2}$,Absolute minimum: $-1$

Vedclass · Answer

(A) The given function is $f(x) = \sin x + \cos x$. First,we find the derivative: $f'(x) = \cos x - \sin x$. To find the critical points,set $f'(x) = 0$: $\cos x - \sin x = 0 \Rightarrow \sin x = \cos x \Rightarrow 	an x = 1$. Since $x \in [0, \pi]$,the only critical point is $x = \frac{\pi}{4}$. Now,we evaluate $f(x)$ at the critical point and the endpoints of the interval $[0, \pi]$: $f\left(\frac{\pi}{4}ight) = \sin\left(\frac{\pi}{4}ight) + \cos\left(\frac{\pi}{4}ight) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2}$. $f(0) = \sin(0) + \cos(0) = 0 + 1 = 1$. $f(\pi) = \sin(\pi) + \cos(\pi) = 0 - 1 = -1$. Comparing these values,the absolute maximum value is $\sqrt{2}$ at $x = \frac{\pi}{4}$ and the absolute minimum value is $-1$ at $x = \pi$.

Find the absolute maximum value and the absolute minimum value of the function given by $f(x) = \sin x + \cos x$ for $x \in [0, \pi]$.

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