Do inflection points of $f(x)$ give $f'(x)=0$?

Question

I understand that the values of $x$ that allow $f'(x)=0$ are stationary points and therefore potential local maximums and minimums of $f(x)$. When would a stationary point NOT be a local maximum or minimum? Do inflection points also yield $f'(x)=0$?

Answer 1

If x is an inflection point for f then the second derivative, $f″(x)$ is equal to zero if it exists.

Thus, no, inflection points do not give $f'(x)=0$ necessarily, but $f''(x)=0$.

Regarding stationary points and $f'(x)=0$ and stationary points, consider the function :

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

Then, it is :

$$f'(x)=3x^2$$

Which is equal to zero at $x=0$, but this is not a stationary point of the function.

Answer 2

not always, for example $f(x)=x^3+x$ has an inflection point in x=0 but $f’(0)=1$

to be more precise, when derivatives exist:

if $f’(x_0)=0$ and $\exists k \geq 2$ s.t. $f^k(x_0) \neq 0$ then

• when k is even you have a max/min in $x_0$ (depending on the sign)

• when k is odd you have an inflection point in $x_0$

[EDIT] I've added up my 2 answers and corrected a typo

Answer 3

Yes, $f'(x)=0$ implies max, min or inflection point. It is inflection at $x_0$ if: $$f'(x_0-\epsilon)<0 \ \ \text{and} \ \ f'(x_0+\epsilon)<0 \ \ or \ \ f'(x_0-\epsilon)>0 \ \ \text{and} \ \ f'(x_0+\epsilon)>0.$$