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Sketch the graph of a function that does not have a point of inflection at $(c,f(c))$ even though $f''(c)=0$. Does anybody have any insight on what function satisfies this condition? My thought was that a possible answer is the $f(x)=x^4$ since the second derivative of the function is $12x^2$, and $f''(0)=0$

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    That's a good example. You still need to note that $f''(x)=12x^2$ does not change sign at $x=0\,$, therefeore $x=0$ is not a point of inflection. And of course you can "shift" it to $f(x)=(x-c)^4$ for an arbitrary $c\,$. – 2017-01-03
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    Yes, that's the easiest that I can think of. – 2017-01-03

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$f(x) = x^4$ works, and this is why.

$f'(x) = 4x^3$

$f''(x) = 12x^2$

$f''(x) = 0$, means that $x = 0$

However, we must consider the concavity to the right and left of $x = 0$, and see if it changes.

Concavity:

$x < 0, f''(x) > 0$

$x > 0, f''(x) > 0$

As you can see the concavity does not change, and thus this is valid because the second derivative does equal $0$, but it is $not$ an inflection point.