Find the local maximum and local minimum valu | vedclass

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(A) Given function: $h(x) = \sin x + \cos x$ for $x \in (0, \frac{\pi}{2})$.First,find the derivative: $h'(x) = \cos x - \sin x$.Set $h'(x) = 0$ to fi

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Local maximum value is $\sqrt{2}$ and no local minimum value exists.

Vedclass · Answer

(A) Given function: $h(x) = \sin x + \cos x$ for $x \in (0, \frac{\pi}{2})$.First,find the derivative: $h'(x) = \cos x - \sin x$.Set $h'(x) = 0$ to find critical points:$\cos x - \sin x = 0 \Rightarrow \sin x = \cos x \Rightarrow 	an x = 1 \Rightarrow x = \frac{\pi}{4}$.Find the second derivative: $h''(x) = -\sin x - \cos x$.Evaluate $h''(x)$ at $x = \frac{\pi}{4}$:$h''(\frac{\pi}{4}) = -\sin(\frac{\pi}{4}) - \cos(\frac{\pi}{4}) = -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = -\frac{2}{\sqrt{2}} = -\sqrt{2}$.Since $h''(\frac{\pi}{4}) The local maximum value is $h(\frac{\pi}{4}) = \sin(\frac{\pi}{4}) + \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2}$.Since the function is strictly concave down on the interval $(0, \frac{\pi}{2})$,there is no local minimum value in the given open interval.

Find the local maximum and local minimum values for the function $h(x) = \sin x + \cos x$ in the interval $0 < x < \frac{\pi}{2}$.

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