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I'm trying to determine this function concavity: enter image description here

2=0 is a contradiction, so we have no inflection points on this function, so ¿How could I determine the concavity if I have no inflection points?

This function's graph: enter image description here

Should I take the "0" as a refered point, then evaluate the f''(x) (for example) with f''(-1) and f''(1) to determine the concavity?

Because on that case, I can effectively determinate the concavity, but is this legal? ¿why I'd take the "0"? On this case, I've take the "0" just because I've seen the graph previously.

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  • $\begingroup$ The fact that $x=0$ is not in the domain of the given function (and its derivatives), does not mean that there cannot be a sign change in the derivatives... $\endgroup$ – imranfat Dec 4 '16 at 20:42
  • $\begingroup$ How do you define "concave function"? For example, en.wikipedia.org/wiki/Concave_function defines this only for the case of a convex domain and $\Bbb R\setminus\{0\}$ is not convex. $\endgroup$ – Hagen von Eitzen Dec 4 '16 at 20:48
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    $\begingroup$ But in the intervals $(-\infty,0)$ and $(0,\infty)$ a sign-change of $f''(x)$ cannot occur. So, in fact, it is sufficient to calculate $f''(-1)$ and $f''(1)$ $\endgroup$ – Peter Dec 4 '16 at 20:48
  • $\begingroup$ so, a inflection point could be the number where the f'(x) is indeterminated? ("0" on this case) $\endgroup$ – Roberto Sepúlveda Bravo Dec 4 '16 at 20:53
  • $\begingroup$ No, there is no inflection point at $0$, and there isn't any one anywhere else either. The "level of concavity" (for the lack of better words) gets greater as $x$ get closer to zero $\endgroup$ – imranfat Dec 4 '16 at 21:21
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Here $x=0$ is the critical value since $f^{\prime \prime} (0) $ is undefined.

Now use this to divide out your intervals into two intervals.

$(-\infty, 0)$ and $(0, \infty)$.

Pick a test point on each interval and see whether the $f^{\prime \prime}(test value)$ is positive or negative. If it's positive then that mean $f$ is concave up in that interval, and if it's negative then it's concave down.

For example, on the interval, $(-\infty, 0)$ , pick $x=-1$ then $f^{\prime \prime}(-1) = -2$, hence concave down.

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