Finding Points of Inflection on a Curve

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Hi guys... I have this problem that i need to verify. I did it but i just want one of you guys to maybe check to see if it is correct.

problem: find the points of inflection on the curve

y = -3x^4 + 4x^3 +36x^2

so first i found y'' = -36x^2 + 24x + 72 = 0
since b^2 - 4ac is not a perfect square i had to use the formula
x = [ -b + sqrt(b^2 - 4ac) ]/2a
and
x = [ -b - sqrt(b^2 - 4ac) ]/2a

by using these i get the values of x as
x = -1.1196
and x = 1.7863

so the points of inflection are (-1.1196, 34.79) and (1.7863, 107.12)

Thanks in advance
 
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Perfect :).
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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