I found the derivative of a function to be f'(x)=8/x^3 and thus its second derivative as f''(x)=0/3x^2. After setting the second derivative to zero and doing the substitution into the parent function, it seems that there's a point of inflection at (0,0).

The only issue is that the graph has a vertical asymptote at x=0. Does this mean I messed up somewhere?

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    $\begingroup$ If the first derivative is as given, the second is not. $\endgroup$ – André Nicolas Jul 26 '14 at 5:12
  • $\begingroup$ The derivative of a quotient is not the quotient of the derivatives. $\endgroup$ – André Nicolas Jul 26 '14 at 5:20
  • $\begingroup$ Are you saying I divided the two? I thought I'd just use the power rule on both the numerator and denominator because it was evident that it would equal zero. Is that incorrect? $\endgroup$ – Louzzy56 Jul 26 '14 at 5:27
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    $\begingroup$ Yes, very incorrect. The derivative of $\frac{8}{x^3}$ is $-\frac{24}{x^4}$. $\endgroup$ – André Nicolas Jul 26 '14 at 5:29
  • $\begingroup$ Thanks for letting me know. I'm going to try using the quotient rule and see if my new answer for the second derivative matches yours. Either way, the point of inflection would still equal to (0,0) and my dilemma would still exist. $\endgroup$ – Louzzy56 Jul 26 '14 at 5:32

Others have pointed out your derivation mistake. To answer the title question: yes. Consider $f(x)=\sqrt[3]{x}$. This is continuous at $x=0$, has a vertical asymptote there, and also changes inflection there. It is nice when the derivatives exist at a would-be inflection point, as we have nice tests then, but technically all we need is for the function to be continuous there and the concavity to change there.

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    $\begingroup$ I would not call that a "vertical asymptote". I would call it a "vertical tangent". $\endgroup$ – GEdgar Jan 9 '19 at 14:57

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