question_answer

Find the local maxima and local minima of the function \[f(x)=(\sin x-\cos x),\] where \[0<x<2\pi .\]

Answer:

Given,   \[f(x)=\sin x-\cos x\]             \[\therefore \]      \[f'(x)=\cos x+sinx\]             and       \[f''(x)=-\sin x+\cos x\] Now, for local maxima or local minima, put f?(x) = 0 \[\Rightarrow \] \[\cos x+\sin x=0\] \[\Rightarrow \]   \[\tan x=-\,1\] \[\Rightarrow \]   \[x=\frac{3\pi }{4}\] or \[x=\frac{7\pi }{4};\] \[x\in (0,\,\,2\pi )\] \[\therefore \] The points at which extremum may occur are \[\frac{3\pi }{4}\] and \[\frac{7\pi }{4}.\] At \[x=\frac{3\pi }{4},\] \[f''\left( \frac{3\pi }{4} \right)=\left( -\sin \frac{3\pi }{4}+\cos \frac{3\pi }{4} \right)\]             \[=\left( \frac{-1}{\sqrt{2}}-\frac{1}{\sqrt{2}} \right)=\frac{-\,2}{\sqrt{2}}=-\,\sqrt{2}<0\] So, \[x=\frac{3\pi }{4}\] is a point of local maxima. At \[x=\frac{7\pi }{4},\] \[f''\left( \frac{7\pi }{4} \right)=\left( -\sin \frac{7\pi }{4}+\cos \frac{7\pi }{4} \right)\]             \[=\sin \frac{\pi }{4}+\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}>0\] So, \[x=\frac{7\pi }{4}\] is a point of local minima.