Find the intervals in which f(x)= - , where 0<x<2 is increasing or decreasing.

f(x)= - ,x (0,2 ) f'(x)= + For f(x) to be increasing, we must have f'(x)>0 + >0 >- >-1 (0,3 4 ) (7 4 ,2 ) So, f(x) is increasing on (0,3 4 ) (7 4 ,2 ). For f(x) to be decreasing, we must have f'(x)<0 + <0 <- <-1 (3 4 ,7 4 ) So, f(x) is decreasing on (3 4 ,7 4 ).

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

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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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