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 $$\left(\dfrac{\pi}{4}, \dfrac{5\pi}{4}\right)$$.Mathematics

Suggest Corrections

0

Similar questions
View More

People also searched for
View More