0

$$ F(x) = \begin{cases} x^2 & x \le 0 \\ 0 & 0 \le x \le 3 \\ -(x-3)^2 & x>3 \end{cases} $$

My question is does this function have any points of inflection? Double Derivative at $x=0,3$.Thus the necessary condition is satisfied for all points such that $0 \le x \le 3$.

But, i have also read that the third derivative at x is not equal to 0 is also regarded as a sufficient condition, which is fulfilled at $x=0,3$. Thus, which sufficient condition should hold and which points should be regarded as inflection points?

caverac
  • 19,345
ghghghghghg
  • 23
  • f"(x)=0 is necessary but not sufficient condition for an inflection point. See https://en.m.wikipedia.org/wiki/Inflection_point – user600016 Nov 08 '19 at 14:59

1 Answers1

1

An inflection point occurs at points of the domain at which the function changes concavity. The second derivative at an inflection point may be zero but it also may be undefined. If we look at the second derivative for your function

$$ F''(x) = \begin{cases} 2 & x < 0 \\ 0 & 0 \le x \le 3 \\ -2 & x>3 \end{cases} $$

then $F''(x) > 0$ when $x < 0$ and $F''(x)<0$ when $x>3$. We know that the function $F(x)$ is concave up when $F''(x) > 0$ and concave down when $F''(x) < 0$. Therefore, it is clear that $F(x)$ is concave up on $(-\infty,0)$ and concave down on $(3,\infty)$.

But, what about the interval $[0,3]$? Since $F(x)\equiv 0$ when $0\le x \le 3$, the second derivative is also zero. As $F(x)$ changes concavity at the end points we know that there must be at least one inflection point. Therefore, we must decide whether there is one inflection point or two inflection points.

Suppose that we assume that there are two inflection points. Then, we would need to show that the function changes concavity twice - there would be two intervals where it was concave up and one interval where it is concave down. Alternatively, the function could be concave down on two intervals and then concave up on one interval.

Your function is concave up on one interval and then concave down on another interval. In the interval between these two intervals, the function is neither concave up nor concave down. Hence, your function only changes concavity once.

So, $F(x)$ has one inflection point. If we plot the function on Desmos, then we see that this point of inflection occurs in the middle of the interval $[0,3]$.

enter image description here

Interestingly, the function $f(x)=x^3$ exhibits similar behavior. This function is concave up on $(0,\infty)$ and concave down on $(-\infty,0)$. It has one inflection point at $(0,0)$.

enter image description here

Axion004
  • 10,056
  • Yes, also that all straight lines a considered both concave and convex. So if ur talking about the change in concavity, then both 0,3 are inflection points. – ghghghghghg Nov 08 '19 at 22:55
  • So, what ur saying, does that imply that any function that is concave in some interval and convex in some interval will surely have atleast 1 inflection point? – ghghghghghg Nov 09 '19 at 13:38
  • Intuitively, yes. The inflection point must occur when the function changes concavity. When this happens, the function would go from a parabola that opens upward to a parabola that opens downward (or vice versa). I have never seen a proof of such a result. Although, I cannot think of a counterexample and therefore believe that this is a theorem. – Axion004 Nov 09 '19 at 14:43