# Math Help - stationary point

1. ## stationary 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?

2. Originally Posted by furor celtica
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?
$y=x^3-2lnx^3$

$y=x^3-6lnx$

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

At a stationary point,

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

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

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

$x^3=2$

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

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

$y=2-2ln2=0.61$

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