1
$\begingroup$

Point of inflection literally means that the slope does not change at that point so shouldn't all points of inflection be compulsarily differentiable? I don't quite understand point of inflection. It's a point where the concavity of graph changes, but that seems too vague. The slope is constant at that point, but how can a slope be constant at a 'POINT'?? Don't we need nearby points for the slope? What exactly is so special about inflection point when all the points that are differentiable already have same left hand and right hand derivative? When x is tending to 0, don't all differentiable points have equal slope on left and right side? So why is inflection point any different?

Hope you got my question!

$\endgroup$
4
  • 1
    $\begingroup$ $y=x|x|$ has a point of inflection (a change in concavity) at $x=0$. Is it differentiable there? $y=x^{1/3}$ also has a point of inflection at $x=0$, and is certainly not differentiable there. $\endgroup$ Nov 3, 2020 at 6:01
  • $\begingroup$ Then it is not necessary that at point of inflection the double differentiation of function is zero right? $\endgroup$
    – Qwerty
    Nov 3, 2020 at 6:20
  • $\begingroup$ I would say that if there's a change in concavity then it doesn't matter whether there's a second derivative or not. $\endgroup$ Nov 3, 2020 at 6:22
  • $\begingroup$ Okay got it. Thanks $\endgroup$
    – Qwerty
    Nov 3, 2020 at 7:25

1 Answer 1

1
$\begingroup$

As also illustrated in this answer, it is incorrect to assume that the condition $f''(x) = 0$ is the sole criteria to determine whether a point is a point of infection. I will quote the same example just to make this point.

Consider the function: $$f(x)=\cases{ -x^2 & $x\le 0$ \\ x^2 & $x>0$ }$$

If you sketch the graph of the function, you can note quite clearly that the concavity of the above function changes at $x = 0$. However, the second derivative is not continuous at that point.

Therefore, your argument that a point of inflection implies that the slope "does not change" is also quite incorrect. While it is true that if the second derivative is continuous at a point $x = a$ and the second derivative exists at the point, then $f''(a) = 0$, we cannot simply conclude a point is a point of inflexion by noting the double derivative vanishes. You can also consider the graphs of the functions $y = x^{2n}$ for $2 \le n$ where the double derivative vanishes at $x=0$ but the concavity does not change.

$\endgroup$
3
  • $\begingroup$ Okay so we just have to see the concavity while second derivative may or may not be zero. Thanks a lot $\endgroup$
    – Qwerty
    Nov 3, 2020 at 7:27
  • $\begingroup$ Don't want to confuse you too much, but you may also want to note that it is not necessary for the curvature to change only at a point of inflection! Think of $y = \frac{1}{x}$ at $x=0$ :) $\endgroup$
    – Scilife
    Nov 3, 2020 at 7:30
  • $\begingroup$ Yupp in this case it's changing. Interesting! $\endgroup$
    – Qwerty
    Nov 3, 2020 at 9:50

This site is temporarily in read-only mode and not accepting new answers.

Not the answer you're looking for? Browse other questions tagged .