Is #f(x)=xcosx# concave or convex at #x=pi/2#?

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Answer 1 · 2018-07-27T21:22:26

function is concave at #x=\pi/2#

Given function:

#f(x)=x\cos x#

#f'(x)=x(-\sin x)+\cosx#

#=-x\sin x+\cos x#

#f''(x)=-x(\cos x)-\sin x-\sin x#

#f''(x)=-x\cos x-2\sin x#

#f''(pi/2)=-\pi/2\cos (\pi/2)-2\sin (\pi/2)#

#=0-2(1)#

#=-2#

Since #f(\pi/2)<0# hence the given function is concave at #x=\pi/2#