Find the number of inflection points of $(x-2)^6(x-3)^9$


If $f(x)$ has $n$ critical points then even $f(x+a)$ will also have $n$ critical points.

So we can simplify it to finding the number of inflection points of $(x)^6(x-1)^9$.

I have evaluated the double derivative to be:

$$6(x-1)^7x^4(35x^2-28x +5)$$

Clearly, the double derivative is $0$ at $4$ points, $0$, $1$ and the roots of the quadratic. But curvature is not changing around $x=0$ so it's not an inflection point.

Thus, there should be $3$ inflection points.

But answer given is $1$ inflection point only.

Please let me know what concept am I missing on.

I tried my level best to zoom into the graph and catch three critical points but there appears to be only one.

  • $\begingroup$ Try to plug in your possible inflection point values into the third derivative. $\endgroup$ – mrtaurho Aug 2 '18 at 20:05

It should be

$$f'(x)=3(x-3)^8(x-2)^5(5x-12)=0 \implies x=3,2,\frac{12}5$$


$$f''(x)=6(x-3)^7(x-2)^4(35x^2-168x+201)=0 \\\implies x_1=2,\,x_2=3,\,x_{3,4}=\frac{12}5\pm\frac{\sqrt{3/7}}{5}$$

and since $p(x)=35x^2-168x+201\implies p(12/5)\neq 0$ the unique stationary inflection point is at $x=3$ otherwise, if we look at general inflection points, by the sign of $f''(x)$ the inflection points are three at


  • 1
    $\begingroup$ From where do you know that $x=3$ is an inflection point? $\endgroup$ – Dr. Sonnhard Graubner Aug 2 '18 at 20:11
  • 1
    $\begingroup$ also the third derivative $$f'''(x)=6\, \left( x-2 \right) ^{3} \left( x-3 \right) ^{6} \left( 455\,{x}^{3 }-3276\,{x}^{2}+7839\,x-6234 \right) $$ is Zero for $x=3$!!! $\endgroup$ – Dr. Sonnhard Graubner Aug 2 '18 at 20:12
  • 2
    $\begingroup$ I think it must go on..let me see! $\endgroup$ – Dr. Sonnhard Graubner Aug 2 '18 at 20:27
  • 1
    $\begingroup$ It is only one inflection point,$$x=3,f(3)$$ no others., it is horizental turning Point. $\endgroup$ – Dr. Sonnhard Graubner Aug 2 '18 at 20:48
  • 1
    $\begingroup$ What is this?,From where does it come? $\endgroup$ – Dr. Sonnhard Graubner Aug 2 '18 at 20:51

It may help to see a very vertically magnified plot of the second derivative (of the original untranslated function).

enter image description here

Given that the inflection points can only be at $x = 2$, $x = 3$, or the roots of the quadratic term $35x^2-168x+201$ (i.e., $r_{1, 2} = \frac{12}{5} \pm \frac{\sqrt{21}}{35}$), the plot above shows that there must be three inflection points: at $x = r_1 \approx 2.269$, $x = r_2 \approx 2.531$, and $x = 3$.

  • $\begingroup$ But he was searching for horizontal turning points, and there is only one! $\endgroup$ – Dr. Sonnhard Graubner Aug 2 '18 at 21:29
  • $\begingroup$ @Dr.SonnhardGraubner: I looked in the OP statement, and I don't see that. The quoted problem says "inflection points." Where was the clarification to "horizontal turning point"? $\endgroup$ – Brian Tung Aug 2 '18 at 21:39
  • $\begingroup$ Thanks a lot for the graph! Very helpful. $\endgroup$ – Abcd Aug 2 '18 at 21:47
  • $\begingroup$ However, the curve is so smooth that it is really hard to clearly spot the inflection points even in this graph. $\endgroup$ – Abcd Aug 2 '18 at 21:49
  • $\begingroup$ @Abcd: Keep in mind that the inflection points are those at which the above graph (the second derivative) changes sign. So the inflection points at $x = r_1$ and $x = r_2$ should be pretty easy to see. The one at $x = 3$ is tough, admittedly. $\endgroup$ – Brian Tung Aug 3 '18 at 0:01

It should be obvious from the degree of the factors:

$f(2) = 0, f(3) = 0\\ f'(2) = 0, f'(3)= 0\\ f''(2) = 0, f''(3)= 0$

Since $f(x) \le 0$ when $x< 3$ and $f(x) > 0$ when $x>3,$ $f(2)$ is a maximum (and not an inflection point, and $f(3)$ is an inflection point and not a min or a max.

There must be a point $c\in (2,3)$ where $f'(c) = 0$ and it must be a local minimum.

There is a point of inflection in $(2,c)$ and one in $(c,3)$

That is 3 points of inflection.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.