# Number of inflection points of $(x-2)^6(x-3)^9$

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

Attempt:

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.

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

It should be

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

and

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

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

$$x_2=3,\,x_{3,4}=\frac{12}5\pm\frac{\sqrt{3/7}}{5}$$

• From where do you know that $x=3$ is an inflection point? – Dr. Sonnhard Graubner Aug 2 '18 at 20:11
• 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 $x=3$!!! – Dr. Sonnhard Graubner Aug 2 '18 at 20:12
• I think it must go on..let me see! – Dr. Sonnhard Graubner Aug 2 '18 at 20:27
• It is only one inflection point,$$x=3,f(3)$$ no others., it is horizental turning Point. – Dr. Sonnhard Graubner Aug 2 '18 at 20:48
• What is this?,From where does it come? – 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).

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 $x = 3$.

• But he was searching for horizontal turning points, and there is only one! – Dr. Sonnhard Graubner Aug 2 '18 at 21:29
• @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"? – Brian Tung Aug 2 '18 at 21:39
• Thanks a lot for the graph! Very helpful. – Abcd Aug 2 '18 at 21:47
• However, the curve is so smooth that it is really hard to clearly spot the inflection points even in this graph. – Abcd Aug 2 '18 at 21:49
• @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. – 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.