(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

y^{3}=6xy-x3-1 , find dy/dx . Prove that the maximm value of y occurs when

x^{3}=8+2sqrt(114) and the minimum value when x^{3}=8-2sqrt(114)

2. Relevant equations

3. The attempt at a solution

I found dy/dx to be (2y-x^{2})/(y^{2}-2x)

Then from here , dy/dx=0 for turning points and it happens when y=1/2 x^{2}

Substitute this back to the original equation and find x from there using the quadratic formula

[tex]x^3=\frac{16\pm\sqrt{16^2-4(1)(8)}}{2}[/tex] which simplifies to

[tex]x^3=8\pm \sqrt{56}[/tex]

First off , the x coordinate of my turning points is wrong but where is that mistake ? Also , How do i test whether this is a max or min ?

Second order differentiation doesn't help ..

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# Homework Help: Maximums and minimums

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