Maths question ! (1 Viewer)

hollaholla · Jun 10, 2009

Umm can anyone do this one??

Find the stationary point on the curve y= Ln x and determine its nature.
x

kurt.physics · Jun 10, 2009

hollaholla said:
Umm can anyone do this one??

Find the stationary point on the curve y= Ln x and determine its nature.
x

$\frac{d}{dx}\left(log_{e} x \right) = \frac{1}{x}$

For stationary points,

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

ie:

$\frac{1}{x} = 0$

multiplying by $x$ gives 1 = 0, which is impossible,

$\therefore$ There are no stationary points.

hollaholla · Jun 10, 2009

kurt.physics said:
$\frac{d}{dx}\left(log_{e} x \right) = \frac{1}{x}$

For stationary points,

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

ie:

$\frac{1}{x} = 0$

multiplying by $x$ gives 1 = 0, which is impossible,

$\therefore$ There are no stationary points.

lol ohkay my bad

the X below the ln went onto the next line...so i cant blame you for missing it lol!

so its ln X by X .

kurt.physics · Jun 10, 2009

$y = \frac{\log_{e} x}{x}$

Letting $u = \log_{e} x$ and $v = x$ :

$u' = \frac{1}{x}$

$v' = 1$

$\frac{dy}{dx} = \frac{vu' - uv'}{v^2}$

$= \frac{x \cdot \frac{1}{x} - \log_{e} x \cdot 1}{x^2}$

$= \frac{1 - log_{e} x}{x^2}$

$= \frac{1 - ln x}{x^2}$

kurt.physics · Jun 10, 2009

whoops, i forgot to finnish the question!

For stationary points, dy/dx is zero

$\frac{1 - ln x}{x^2} = 0$

$1 - ln x = 0$

$ln x = 1$

$log_{e} x = 1$

$e^1 = x$

$\therefore x = e$

Testing both sides of the derivative with e

On the left side of e:

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

On the right side of e:

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

$\therefore$ x = e is a maximum

<e[ tex],="" $="" \frac{dy}{dx}=""></e[>$

hollaholla · Jun 11, 2009

kurt.physics said:
whoops, i forgot to finnish the question!

For stationary points, dy/dx is zero

$\frac{1 - ln x}{x^2} = 0$

$1 - ln x = 0$

$ln x = 1$

$log_{e} x = 1$

$e^1 = x$

$\therefore x = e$

Testing both sides of the derivative with e

On the left side of e:

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

On the right side of e:

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

$\therefore$ x = e is a maximum

<e[ tex],="" $="" \frac{dy}{dx}=""></e[>[/quote] lol thanks plenty. (still not doNE XD)$

kurt.physics · Jun 11, 2009

when $x = e, y = \frac{1}{e}$

$\therefore \left( e, \frac{1}{e} \right)$

hollaholla · Jun 11, 2009

kurt.physics said:
when $x = e, y = \frac{1}{e}$

$\therefore \left( e, \frac{1}{e} \right)$

There we are!

swiftshift · Aug 29, 2009

kurt.physics said:
when $x = e, y = \frac{1}{e}$

$\therefore \left( e, \frac{1}{e} \right)$

hey, can u gimme your email.. i got a load of questions to be answered..

Thanks

Maths question ! (1 Viewer)

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