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My note says: " If f ''(a)=0 and f ''(x) changes sign at x=a, there is a point of inflection at (a, f(a))."

I was wondering how the original graph would appear if f ''(a)=0 and f ''(x) does not change sign at x=a, would there still be a point of inflection?

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    $\begingroup$ If $f''$ does not change sign, no inflection point occurs. Just like how, if $f'(a)=0$ but $f'$ doesn't change sign at $x=a$, $f$ does not have a relative extremum at $x=a$. $\endgroup$ – Franklin Pezzuti Dyer Mar 19 '18 at 23:26
  • $\begingroup$ My note also says that a point of inflection occurs where f''(x) is undefined but how is this possible? $\endgroup$ – Timmy Mar 19 '18 at 23:44
  • $\begingroup$ Here's how to ask a good question. In particular, note that we use MathJax here. $\endgroup$ – Shaun Mar 20 '18 at 0:28
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Let consider for example $f(x)=x^4$ at the origin

enter image description here

and compare with $f(x)=x^3$ at the origin.

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  • $\begingroup$ I don't think a quartic function can have an inflection point because it never changes concavity $\endgroup$ – Timmy Mar 19 '18 at 23:39
  • $\begingroup$ @Timmy Yes indeed it is a simple example of what occurs if $f''(0)=0$ but $f''$ doesn't change sign at $x=0$. Maybe you are looking for something else? $\endgroup$ – user Mar 19 '18 at 23:43
  • $\begingroup$ I think I understand now, thanks. I was also wondering how a point of inflection can occur where f''(x) is undefined. $\endgroup$ – Timmy Mar 19 '18 at 23:47
  • $\begingroup$ @Timmy Yes think to $x^\frac13$ i.stack.imgur.com/Ljk9m.png the slope at 0 is vertical and it is an inflection point thus f'(x) and f''(x) are undefined. $\endgroup$ – user Mar 19 '18 at 23:49
  • $\begingroup$ That makes sense. Also, I know that f '(x) is undefined at cusps, vertical tangents and at discontinuities, so f''(x) will always be undefined when f'(x) is undefined. Is there a scenario where f(x) has an inflection point when f '(x) is defined and f''(x) is undefined? $\endgroup$ – Timmy Mar 20 '18 at 0:12

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