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

Answer 1

Luke Alvarez

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#

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Answer 2

Emily Ammerman

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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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.