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Is it always true that when a function changes concavity from concave-up to concave-down, its derivative at the point of inflection is undefined? And in the reverse order the derivative is Zero?

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    $\begingroup$ Not true. Take $f(x)=x^3$ and look at the inflection point at $x=0$ and then look at $-f(x)$ $\endgroup$ – marwalix Jun 9 '15 at 18:05
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Not at all. The derivative is well defined.

At the point of inflection derivative value or slope is a maximum or minimum. The derivative is constant. At this point the line is curving neither to left nor right but is headed straight without any turning.

For curve $ y = \tan x $ or $ y= \sin x $ the slope or derivative is $1 $ at the origin, $y^{''} (x) $ vanishes. But please look at $ y= x^3 $ only after sometime to avoid the present confusion.

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