- #1
thereddevils
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Homework Statement
y3=6xy-x3-1 , find dy/dx . Prove that the maximm value of y occurs when
x3=8+2sqrt(114) and the minimum value when x3=8-2sqrt(114)
Homework Equations
The Attempt at a Solution
I found dy/dx to be (2y-x2)/(y2-2x)
Then from here , dy/dx=0 for turning points and it happens when y=1/2 x2
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 ..