locate the stationary point(s) on this curve:

y=x^3 - 2lnx^3

i keep on finding (1,1), which apparently is so wrong. helpful hints?

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- Aug 2nd 2009, 09:10 PMfuror celticastationary point
locate the stationary point(s) on this curve:

y=x^3 - 2lnx^3

i keep on finding (1,1), which apparently is so wrong. helpful hints? - Aug 2nd 2009, 09:24 PMalexmahone
$\displaystyle y=x^3-2lnx^3$

$\displaystyle y=x^3-6lnx$

$\displaystyle \frac{dy}{dx}=3x^2-\frac{6}{x}$

At a stationary point,

$\displaystyle \frac{dy}{dx}=0$

$\displaystyle 3x^2-\frac{6}{x}=0$

$\displaystyle x^2=\frac{2}{x}$

$\displaystyle x^3=2$

$\displaystyle x=2^\frac{1}{3}=1.26$

$\displaystyle y=x^3-2lnx^3$

$\displaystyle y=2-2ln2=0.61$

So the stationary point is $\displaystyle (1.26, 0.61)$.