How Do You Determine Turning Points and Concavity for y=(x^3)/(x^2-1)?

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Homework Statement


need to find the turning points an concavity of this equation. y=(x^3)/(x^2-1),


Homework Equations


i know y'=(x^2(x^2-3))/((x^2-1)^2)
and y''=(2x(x^2+3))/((x^2-1)^3)

need to know the turning points and concavity

The Attempt at a Solution

 
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Hi bruno87 ! Welcome to PF! :smile:

(try using the X2 tag just above the Reply box :wink:)
bruno87 said:
i know y'=(x^2(x^2-3))/((x^2-1)^2)
and y''=(2x(x^2+3))/((x^2-1)^3)

need to know the turning points and concavity]

The turning points should be easy …

what's the test for turning points? :smile:​
 
just need to know what the answer is
 
Wrong answer.

Thread locked
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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