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Is it possible to draw a function such that $f$ satifies:

  1. $f(0)=1$
  2. $f'(x)<0$ for all real $x \neq0$
  3. $f''(0)=0$
  4. $f'(0)=$ undefined
  5. $f''(x)>0$ for all real $x > 0$
  6. $f''(x)<0$ for all real $x < 0$
vitamin d
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Eric
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  • I was thinking of using a jump or point discontinuity at f(0), but I think that makes the second derivative at 0 also non-differentiable. – Eric Feb 25 '21 at 02:25
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    I don't think this is possible. If $f\left(x_0\right)$ is undefined, certainly $f'\left(x_0\right)$ is undefined, just based on the definition of a derivative. This generalises to higher order derivatives as well. – doobdood Feb 25 '21 at 02:44

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