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I want to check where the function is convex and where concave. For this I need to calculate the second derivative test:

I got $f''(x) = 2*\cos(x^2) -4*x^2(\sin(x^2))$ and this derivative should be equal to 0 in order to reach my goal.

so: $f''(x) = \cos(x^2) - 2*x^2(\sin(x^2)) = 0$

I have simplified this to $\sin^2(x)*(2x^2 - 1) = 1$

But what should I do next? Because all my previous examples i could simplify to something like: $x*(x+1) = 0$

Hetebrij
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user349485
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  • $\cos(x^2)\neq \cos^2(x)$, so your reformulation is incorrect. That said, I am not sure how to handle this yet – b00n heT Jun 22 '16 at 11:25
  • This problem does not have a nice solution: convex for |x|<0.8083, concave for 0.8083<|x|<1.8145, convex for 1.8145<|x|<2.5222, ... If the previous ones were easy, are you sure you have stated it correctly? – almagest Jun 22 '16 at 12:01

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The sine function is convex where negative and concave where positive, since it is a solution of the differential equation $f''=-f$. Almost the same holds for $\sin(x^2)$. We have: $$ \frac{d^2}{dx^2}\,\sin(x)^2 = 2\cos(x^2)-4x^2\sin(x^2) $$ hence the changes of concavity happen at the solutions of: $$ \tan(x^2) = \frac{1}{2x^2} $$ that cannot be written in a explicit way. However, the solutions of $\tan(t)=\frac{1}{2t}$ get closer and closer to $\pi\mathbb{Z}$ as $|t|\to +\infty$. So we may say that $\sin(x^2)$ is convex where negative and concave where positive, provided that $x$ is sufficiently far from the origin and $\sin(x^2)$ is sufficiently positive/negative.

Jack D'Aurizio
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