Is #f(x)=xe^x-x^2# concave or convex at #x=0#?

Answer 1
You need to compute the second derivative in the point #x=0#, and check its sign. If it's positive, the function is convex, otherwise it's concave. So, we have
#f(x) = xe^x-x^2# #f'(x) = e^x + xe^x-2x# #f''(x) = e^x + e^x + xe^x-2 = e^x(2+x)-2#
So, #f''(0)=e^0(2+0)-2 = 2-2=0#.
Since the second derivative is negative immediatly before #0#, and positive after, #x=0# is an inflection point, in which the function passes from being concave to be convex, as you can see in this graph, in which the scales of #x# and #y# axes are changed in order to emphatize the behaviour of the curve.
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Answer 2

To determine whether ( f(x) = xe^x - x^2 ) is concave or convex at ( x = 0 ), we need to examine the second derivative of ( f(x) ) at ( x = 0 ).

( f''(x) = e^x + xe^x - 2 )

Substituting ( x = 0 ) into the second derivative:

( f''(0) = e^0 + 0 \cdot e^0 - 2 = 1 - 2 = -1 )

Since the second derivative is negative at ( x = 0 ), the function is concave at that point.

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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.

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.

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.

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.

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