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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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Can you see the graph is "sad" for negative $x$ and "smiles" for positive $x$? – Henry Mar 16 '12 at 12:34
Yes, but the interval. Is it sad for every negative X? And happy for every positive x? Does this change when the graph crosses the x-axis. How can I know it is not smiling again when negative x becomes big enough. – Algific Mar 16 '12 at 12:37
In this case, it looks like it, but I don't think you can be completely sure without knowing the function's formula. – Javier Mar 16 '12 at 12:38
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. – Johannes Kloos Mar 16 '12 at 12:38
up vote 1 down vote accepted

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 $$f(x)=x^3-3x$$ $$f'(x)=3x^2-3$$ $$f''(x)=6x.$$

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