Find the intervals in which f(x)= - , where 0<x<2 is increasing o… - Vidyadip

Question

Find the intervals in which $\text{f}(\text{x})=\sin\text{x}-\cos\text{x},$ where $0<\text{x}<2\pi$ is increasing or decreasing.

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Answer

$\text{f}(\text{x})=\sin\text{x}-\cos\text{x},\text{x}\in(0,2\pi)$ $\text{f}'(\text{x})=\cos\text{x}+\sin\text{x}$ For f(x) to be increasing, we must have $\text{f}'(\text{x})>0$ $\Rightarrow\cos\text{x}+\sin\text{x}>0$ $\Rightarrow\sin\text{x}>-\cos\text{x}$ $\Rightarrow\tan\text{x}>-1$ $\Rightarrow\text{x}\in\Big(0,\frac{3\pi}{4}\Big)\cup\Big(\frac{7\pi}{4},2\pi\Big)$So, f(x) is increasing on $\Big(0,\frac{3\pi}{4}\Big)\cup\Big(\frac{7\pi}{4},2\pi\Big).$
For f(x) to be decreasing, we must have $\text{f}'(\text{x})<0$ $\Rightarrow\cos\text{x}+\sin\text{x}<0$ $\Rightarrow\sin\text{x}<-\cos\text{x}$ $\Rightarrow\tan\text{x}<-1$ $\Rightarrow\text{x}\in\Big(\frac{3\pi}{4},\frac{7\pi}{4}\Big)$ So, f(x) is decreasing on $\Big(\frac{3\pi}{4},\frac{7\pi}{4}\Big).$

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