Is $f (x) = x cos x$ $f(x)=xcosx$ concave or convex at $x = \frac{π}{2}$ $x=pi/2$ ?

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Answer 1 · 2018-07-27T21:22:26

function is concave at $x = \frac{π}{2}$ $x=\pi/2$

Given function:

$f (x) = x cos x$ $f(x)=x\cos x$

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

$=-x\sin x+\cos x$

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

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

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

$=0-2(1)$

$=-2$

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

Is f(x)=xcosxf(x)=xcosxf(x)=xcosx concave or convex at x=pi/2x=π2x=pi/2?