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f(x) it self has roots aprox. -1.75 and 1.75. And 0. It's respectively is neg, pos, neg, pos. That is obvious from the graph.

f´(x) has roots at -1 and 1. And is pos, neg, pos.

How can I view the double derivative f´´(x)? I know if the graph "smiles" f´´(x) is positive and if it is sad f´´(x) < 0.

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    Another way to put this: If you run along the graph, it has positive second derivative if it turns left, and negative if it turns right. In this example, you will find that its second derivative switches signs exactly once.2012-03-16

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You can see the graph looks "sad" for negative $x$ and "smiles" for positive $x$.

I would expect you to find here that f''(x) is negative for negative $x$ and positive for positive $x$ with a root at $x=0$.

In general, if $f(x)$ is a cubic then f''(x) is a linear function of $x$. If f'(x) is a parabola symmetric about the origin then f''(x) is proportional to $x$ and f''(0)=0.

Your curve is very likely to be $f(x)=x^3-3x$ f'(x)=3x^2-3 f''(x)=6x.