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?
$\displaystyle y=x^3-2lnx^3$
$\displaystyle y=x^3-6lnx$
$\displaystyle \frac{dy}{dx}=3x^2-\frac{6}{x}$
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 $\displaystyle (1.26, 0.61)$.