# An inflection point where the second derivative doesn't exist?

A point $x=c$ is an inflection point if the function is continuous at that point and the concavity of the graph changes at that point. And a list of possible inflection points will be those points where the second derivative is zero or doesn't exist. But if continuity is required in order for a point to be an inflection point, how can we consider points where the second derivative doesn't exist as inflection points?

Also, an inflection point is like a critical point except it isn't an extremum, correct? So why do we consider points where the second derivative doesn't exist as inflection points?

thanks.

• "But if continuity is required..." The continuity required is the continuity of $f$. – Martín-Blas Pérez Pinilla Jan 20 '14 at 7:49

Take for example $$f(t) = \begin{cases} -x^2 &\text{if x < 0} \\ x^2 &\text{if x \geq 0.} \end{cases}$$

For $x<0$ you have $f''(x) = -2$ while for $x > 0$ you have $f''(x) = 2$. $f$ is continuous as $0$, since $\lim_{t\to0^-} f(t) = \lim_{t\to0^+} f(t) = 0$, but since the second-order left-derivative $-2$ is different from the second-order right-derivative $2$ at zero, the second-order derivative doesn't exist there.

For your second question, maybe things are clearer if stated like this

If the second derivative is greater than zero or less than zero at some point $x$, that point cannot be an inflection point

This is quite reasonable - if the second derivative exists and is positive (negative) at some $x$, than the first derivative is continuous at $x$ and strictly increasing (decreasing) around $x$. In both cases, $x$ cannot be an inflection point, since at such a point the first derivative needs to have a local maximum or minimum.

But if the second derivative doesn't exist, then no such reasoning is possible, i.e. for such points you don't know anything about the possible behaviour of the first derivative.

• This is correct; I'll add that this is also true for the function x^(1/3) (i.e. cube root function). (Lest anyone think this can only happen with "funny functions" that involve different cases.) x^(1/3) has a point of inflection at zero even though not only does the second derivative not exist, the first derivative does not exist either. – Bennett May 2 '19 at 21:15

A function can be continuous but fail to have a second derivative. For example, consider $$f(x)=\cases{ -x^2 & x\le 0 \\ x^2 & x>0 }$$ with second derivative $$f''(x)=\cases{ -2 & x< 0 \\ \text{undefined} & x=0 \\ 2 & x>0 }$$

The statement you give says only that you need to check points without a second derivative or where it's zero. There are examples where

1. the second derivative doesn't exist like $$f(x)=\cases{ x^2 & x\le 0 \\ 2x^2 & x>0 }$$
2. the second derivative does exist and is zero like $f(x)=x^4$

but the function does not have an inflection point.

The function $y=x^{{1/3} }$ has as its second derivative $y''= -\frac{2}{9}\,{x}^{-5/3}$, which is undefined at $x = 0$. The slopes of the tangent lines to the original curve $y$ tend to $\pm \infty$ as $x$ approaches $0$. Despite the second derivative being undefined at the point $x = 0$, it is a true inflection point of $y$ .

A point of inflection exists where the concavity changes. Where the derivative is increasing the graph is concave up; where the derivative is decreasing the graph is concave down. Concavity may change where the second derivative is 0 or undefined. You said that the graph must be continuous. I'm not sure that's true, but if it is then this still works. The graph can be continuous even if the second derivative isn't. In other words if the second derivative is undefined at x=a the undifferentiated f(x) can still exist at x=a. Only the graph must be continuous. The second derivative does not have to be. I'm not sure if I answered all your questions, but I hope I helped.

Take the function $$f(x)=x^{1/3}$$ which has $$0$$ as the point of inflection but derivatives do not exist at that point.In particular double derivative also does not exist.

Not sure if this is exactly what you're looking for, but: the function $$f(x) = x^4$$ has both $$f'(0)=0$$ and $$f''(0)=0$$, and it has a local min at $$0$$.