$f(x)=xe^x$ is such that $f'(x)=e^x+xe^x$ and so $f''(x)=(2+x)e^x$. I know that $f''(x)>0\iff x>-2$ and $f''(x)<0\iff x<-2$, while $x=-2$ is an inflection point. Can I say that in $(-\infty,2)$ the function is STRICTLY concave and in $(2,\infty)$ is STRICTLY convex? Can I include also $x=2$ in both of these intervals?
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2 in the intervals is wrong, should be -2. – user376343 Jan 11 '22 at 20:05
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Assume $f'' \ge 0$, and let $A$ be the zero set of $f''$, then $f$ is strictly convex if $A$ has empty interior. This follows directly from the answer to Monotone functions and non-vanishing of derivative. – dxiv Jan 11 '22 at 20:14
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@dxiv so if I prove that in a right neighbourhood of $x=-2$ I have $f'(x)\neq 0$ then can I conclude that $f$ is strict convex? – Nik Jan 11 '22 at 20:32
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@Nik This has nothing to do with the zeros of $f'$. But you know that the zero set of $f''$ only contains one isolated point $A = {-2}$. That means $A$ has empty interior, which in turn implies that $f'$ is strictly increasing, so $f$ is strictly convex. – dxiv Jan 11 '22 at 20:35
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Sorry there is a typo in my previous comment! I would like to say $f''(x)\neq 0$ in a neighbourhood of $x=-2$...now my idea works? @dxiv – Nik Jan 11 '22 at 20:38
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@Nik Yes, as long as you know (or prove) the result from the linked post. – dxiv Jan 11 '22 at 20:39
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By wikipedia
A differentiable function f is (strictly) concave on an interval if and only if its derivative function f ′ is (strictly) monotonically decreasing on that interval, that is, a concave function has a non-increasing (decreasing) slope.
And for your example $f'$ is strictly increasing (resp. decreasing) on $[-2, \infty)$ and $(-\infty, -2]$ (despite having $f'' = 0$ at $x = -2$) so we can include $x = -2$
To elaborate, this comes about as for a general real valued differentiable function we have: $f' > 0 \implies f$ strictly increasing, but the converse is not always true
For example if $f(x) = x^3$ then $f$ is strictly increasing but $f'(0) = 0$
Red5551
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As it’s currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. – Community Jan 11 '22 at 19:41
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1I don't know what the bot is talking about. The answer is perfectly clear as written. Welcome to the community. – CyclotomicField Jan 11 '22 at 19:49
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So in $[-2,\infty)$ is strict concave and in $(-\infty,2]$ is strict convex? – Nik Jan 11 '22 at 21:11
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Other way round, $f$ is concave on $(-\infty, -2]$ and convex on $[-2, \infty)$ – Red5551 Jan 11 '22 at 21:33