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Let $$F(x)=-\int_{0}^{x^2}\frac{2}{3+e^t}dt$$ Find all critical points of $F(x)$ and determine whether they are minima, maxima or points of inflection. Prove that $F(300)>F(310)$.

First I differentiated the function and equated to $0$: $$F'(x) = \frac{-4x}{3+e^{x^2}}=0$$ $\implies x=0$ is a critical point. Then differentiating twice: $$F''(x)=\frac{-4(3+e^{x^2})+4x^2(2xe^{x^2})}{(3+e^{x^2})^2}=0$$ But I can't solve this equation (using it for $x=0$ gives that it is a maximum). Is there another way to go? I'm guessing the last part has to do with how the function is decreasing due to some critical points.

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Note that $F(x)$ is an even function and the integrand is positive over all the real line.

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  • $\begingroup$ Does this help for the first or the second part? $\endgroup$ – George Nov 28 '14 at 2:50
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    $\begingroup$ For the whole problem. You've already found the critical point. With this observation you know $F(x)$ negative everywhere except for $0$. $\endgroup$ – hjhjhj57 Nov 28 '14 at 4:08
  • $\begingroup$ But I haven't found out the points of inflexion? $\endgroup$ – George Nov 28 '14 at 12:43

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