stationary point

**furor celtica** · August 2nd 2009, 09:10 PM

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?

**alexmahone** · August 2nd 2009, 09:24 PM

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 $(1.26, 0.61)$ .

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