An inflection point is a point on a curve at which the sign of the curvature (i.e. the concavity) changes.

According to Wikipedia,

"If x is an inflection point for f then the second derivative, f″(x), is equal to zero if it exists, but this condition does not provide a sufficient definition of a point of inflection. One also needs the lowest-order (above the second) non-zero derivative to be of odd order (third, fifth, etc.). If the lowest-order non-zero derivative is of even order, the point is not a point of inflection, but an undulation point. . . . An example of such an undulation point is y = x^4 for x = 0."

"It follows from the definition that the sign of f′(x) on either side of the point (x,y) must be the same. If this is positive, the point is a rising point of inflection; if it is negative, the point is a falling point of inflection."

Q. 1) How does the second part 'follow from the definition' in the first part?

Q. 2) According to my textbook,

"A curve y = f(x) has one of its points x = c as an inflection point if f"(c) = 0 or is not defined; and f"(x) changes sign as x increases through c. The latter condition may be replaced by f'''(c) ≠ 0 when f'''(c) exists."

I didn't understand how the 'latter condition may be replaced'. Has it got anything to do with Wikipedia's 'sufficient definition' part? What if the 'lowest-order (above the second) non-zero derivative' is not the third but, say, the fifth derivative? In that case, is my textbook wrong? (Please give an example of such a function!)


Your textbook needed an editor who could understand that the punctuation of the definition was rotten. It's be better to say

The point $x = c$ of a curve $y = f(x)$ defined by a function $f$ that's twice differentiable almost everywhere is an inflection point if either

(a) $f''(c) = 0$ and $f'$ changes sign as $x$ increases through $c$, or

(b) $f''(c)$ is undefined and $f'$ changes sign as $x$ increases through $c$.

The final clause ("The latter condition") is just plain wrong, as the example $y = x^4$ shows, because although $f''(0) = 0$, the concavity of the function is "up" on both sides of $x = 0$.

It also doesn't handle cases like $y = x^4 \sin (1/x)$ for $x \ne 0$ and $ y = 0$ for $x = 0$, which have second derivative zero, but for which the curvature changes sign infinitely often in any neighborhood of the origin -- one might call that an inflection point, or might not (I'd say "not", given the choice), but the authors' "as $x$ increases through $c$" suggests that they expect $f''$ to have one sign to the left of $c$ and the opposite sign to the right of $c$, at least locally, and for this function, that's just not true.

  • $\begingroup$ I'm not sure what you meant by "The final clause ("The latter condition") is just plain wrong, as the example $y=x^4$ shows". The quote from the textbook says: "The latter condition may be replaced by f'''(c) ≠ 0 when f'''(c) exists." $y=x^4$ is not a counterexample to this, because f'''(0) = 0, so it does not meet the textbook's condition implying that it's an inflection point. $\endgroup$
    – Bennett
    May 2 '19 at 22:32

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