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

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 $f'(x)=0$?

• Consider the graph $f(x) = x^3$. Inflection points give $f''(x)=0$ (the second derivative) Nov 26 '17 at 7:24
• Remember: whenever the function is differentiable, an inflection point is when $\;f'(x)\;$ has an isolated extremum, and if $\;f\;$ is differentiable twice this means that if $\;x_0\;$ is an inflection point then $\;f''(x_0)=0\;$ (but not the other way around). Nov 26 '17 at 7:56
• I think this is a normal misconception because I was taught about inflection points in the context of extrema as exceptions to the fact that there is an extreme point when the derivative vanishes. One way of looking at an inflection point is that it is a point where the curvature changes sign. Nov 26 '17 at 9:13
• Please, if you are ok, you can accept the answer and set it as solved. Thanks!
– user
Jan 10 '18 at 9:52

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 $x_0$

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.

• Again: $\;x=0\;$ is a stationary or critical point point of $\;f(x)=x^3\;$ ...! It is just a matter of reading the very definition of "stationary or critical" point... Nov 26 '17 at 8:07

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