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Let $f: \mathbb{R} \to (0, \infty)$ be two times differentiable function. Does there always exist $x_0 \in \mathbb{R}$ such that $f(x_0) + f'(x_0)(x - x_0) + \frac{f''(x_0)}{2}(x - x_0)^2 \ge 0$ for all $x \in \mathbb{R}$?

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closed as off-topic by Saad, Eevee Trainer, Parcly Taxel, Theo Bendit, mrtaurho Mar 12 at 6:16

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  • $\begingroup$ The condition is roughly asking whether a function is always locally-convex at some point. Have you tried $\ln(x)$? $\endgroup$ – Robert Wolfe Mar 11 at 23:34
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    $\begingroup$ @RobertWolfe The codomain of $f$ is $(0,\infty)$ though. $\endgroup$ – BigbearZzz Mar 11 at 23:35
  • $\begingroup$ Ah, I missed that. Changes it a bit. $\endgroup$ – Robert Wolfe Mar 11 at 23:37
  • $\begingroup$ $q_x(t) = f(x)+t f'(x)+t^2 f''(x)/2$ assuming $f''(x) > 0$ the minimum is attained at $t = -f'(x)/f''(x)$ and $q_x(-f'(x)/f''(x)) = f(x)-f'(x)^2/f''(x)+f'(x)/(2f''(x))$ whose sign is that of $g(x) = f(x)f''(x)+f'(x)(1/2-f'(x))$ you want to say $g$ isn't stricly negative. $\endgroup$ – reuns Mar 11 at 23:52
  • $\begingroup$ This is equivalent to saying that for some $x_0$, $f'(x_0)^2 - 2f(x_0)f''(x_0) \leq 0$ (follows from the quadratic formula). If moreover $f''(x_0)$ is always positive, then there is some function $f$ satisfying these conditions, because take any $f$ with positive second derivative, add an appropriately big constant to make the expression above negative. $\endgroup$ – enedil Mar 12 at 0:43