What is an example of a function that is differentiable, concave up everywhere, and $f''(0)$ does not exist?

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    $\begingroup$ $f(x)=x^2$ for $x<0$ and $f(x)=x^3$ else. $\endgroup$ – Michael Hoppe Dec 6 '18 at 16:57
  • $\begingroup$ @Eric Maybe only integral of absolute value? $\endgroup$ – Юрій Ярош Dec 6 '18 at 17:01
  • $\begingroup$ @ЮрійЯрош No, I just messed up. The first integral isn't concave up. You can take the first integral of something like$ g(x)=0$ for $x<0$ and $g(x)=x$ for $x \geq 0$ $\endgroup$ – Eric Dec 6 '18 at 17:04

So you need a continuous function, and the first derivative at $0$ must be the same for both $x>0$ and $x<0$, but the second derivative must be different. Start with $f(x)=x^2$ for $x>0$. You have $f(0+)=0$, $f'(0+)=0$, and $f''(0+)=2$. You now need $f(0-)=0$, $f'(0-)=0$, and you can choose $f''(0-)=0$. For example $f(x)=x^4$ for $x<0$

  • $\begingroup$ Not sure why you mention continuity. The desired function should be differentiable, hence it will be continuous. You wrote "the first derivative at $0$ must be the same for both $x>0$ and $x<0$" What does that mean? $\endgroup$ – zhw. Dec 6 '18 at 17:32
  • $\begingroup$ I've started from scratch. So in order to find a differentiable function, the first requirement is that is continuous. In order to be differentiable, the derivative must be continuous, including at $0$. Since I choose to define my function differently for $x>0$ and for $x<0$, I just need $$\lim_{x\to 0}f(x<0)=\lim_{x\to 0}f(x>0)$$ $\endgroup$ – Andrei Dec 6 '18 at 19:22

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