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I'm trying to find a value of x that will make the second derivative of $f(x)$ , $f''(x) = \frac{6(x^3-6x^2+3x-2)}{(x^2-1)^3}$ , equal to $0$.

so on my own, I have come this far:

$f"(x)=0$

$\rightarrow \frac{6(x^3-6x^2+3x-2)}{(x^2-1)^3} = 0$

$\rightarrow 6(x^3-6x^2+3x-2) = 0$

$\rightarrow x^3-6x^2+3x-2 = 0$

and this is where I have gotten stuck, I used a graphing tool to look at it's graph, and it tells me it's $0$ at $x \approx 5.522$

But I cant figure out how it got to that solution at all. I tried the grouping method which didn't exactly go well either.

I tried to look into it a bit and found something called the "cubic formula", which makes it just more confusing.

for context's sake, I'm a high-school senior and my knowledge of math goes that far, so any explanation on terms I might not understand will be appreciated.

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  • $\begingroup$ what is $f(x)$? $\endgroup$ – Dr. Sonnhard Graubner Jul 28 '17 at 14:25
  • $\begingroup$ If you just want to approximate, try Newton's Method: en.wikipedia.org/wiki/Newton's_method $\endgroup$ – Franklin Pezzuti Dyer Jul 28 '17 at 14:25
  • $\begingroup$ @Dr.SonnhardGraubner it's $f(x)=\frac{3(x-2)}{x^2-1}$ , I'm pretty sure I haven't made mistakes in taking the derivative , but I might have not noticed something $\endgroup$ – Dahen Jul 28 '17 at 14:30
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You can use the explicit formule to solve the cubic equation, also, you can find an aproximated solution use this theorem: IVT. For example, $f(6) > 0$ and $f(5) < 0$, then there is a solution $ a \in (5, 6)$ you can refinate the interval.

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  • $\begingroup$ thanks, I'll try both methods and the newton's method that @Nilknarf suggested above as well $\endgroup$ – Dahen Jul 28 '17 at 14:36
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HINT: $$x=2 + 3^{1/3} + 3^{2/3}$$ is one solution of the equation $$f''(x)=0$$

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  • $\begingroup$ how did you get there? $\endgroup$ – Dahen Jul 28 '17 at 14:36
  • $\begingroup$ i'm solved the cubic equation after the Cartano formulas $\endgroup$ – Dr. Sonnhard Graubner Jul 28 '17 at 14:40
  • $\begingroup$ I looked it up, and sadly that's way beyond my scope ( not to mention that we skipped complex numbers like $i$ ), but anyways, thanks and i'll make due with the other mentioned methods and hope for the best $\endgroup$ – Dahen Jul 28 '17 at 14:51

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