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Question

Find the intervals in which the function $$f$$ given by
$$f\left( x \right) =\sin { x } +\cos { x } ,0\le x\le 2\pi $$ is strictly increasing or strictly decreasing.


Solution

$$f(x)=\sin x+\cos x$$
$$f'(x)=\cos x-\sin x$$
$$f'(x)=0$$
$$\cos x-\sin x=0$$
$$\tan x=1$$
$$x=\dfrac{\pi}{4}, \dfrac{5\pi}{4}, a$$ & $$0\leq x\leq 2\pi$$
The point $$x=\dfrac{\pi}{4}$$ and $$\dfrac{5\pi}{4}$$ divides, the interval $$[0, 2\pi]$$ into $$3$$ disjoint intervals.
i.e., $$\left[\left(0, \dfrac{\pi}{4}\right), \left(\dfrac{\pi}{4}, \dfrac{5\pi}{4}\right),  \left(\dfrac{5\pi}{4}, 2\pi\right)\right]$$
$$f'(x) > 0$$ if $$x\in\left[\left(0, \dfrac{\pi}{4}\right)\cup \left(\dfrac{5\pi}{4}, 2\pi\right)\right]$$
or f is in the intervals.
$$\left[\left(0, \dfrac{\pi}{4}\right), \left(\dfrac{5\pi}{4}, 2\pi\right)\right]$$
Also, $$f'(x) < 0$$, if $$x\in\left(\dfrac{\pi}{4}, \dfrac{5\pi}{4}\right)$$
So fxn is strictly decreasing in $$\left(\dfrac{\pi}{4}, \dfrac{5\pi}{4}\right)$$.

1108059_1197404_ans_3a0a7a633498482e972f5962d526db3d.jpg

Mathematics

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