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

function is concave at

Given function:

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To determine the concavity or convexity of ( f(x) = x \cos(x) ) at ( x = \frac{\pi}{2} ), we need to examine the second derivative of the function.

The first derivative of ( f(x) ) is ( f'(x) = \cos(x) - x \sin(x) ).

The second derivative of ( f(x) ) is ( f''(x) = -2 \sin(x) - x \cos(x) ).

Plugging ( x = \frac{\pi}{2} ) into the second derivative yields ( f''\left(\frac{\pi}{2}\right) = -2 \sin\left(\frac{\pi}{2}\right) - \frac{\pi}{2} \cos\left(\frac{\pi}{2}\right) = -2(1) - 0 = -2 ).

Since the second derivative is negative at ( x = \frac{\pi}{2} ), the function ( f(x) = x \cos(x) ) is concave at ( x = \frac{\pi}{2} ).

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