$\displaystyle f(x)={ln(2x)}/{x} ;[1,e]$

Find the critical points

I used the quotient rule and got

$\displaystyle f'(x)= {-ln(2x)}/{x^{2}}$

and now i equal the equation to zero and them im stuck cant solve it

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- Feb 23rd 2010, 12:11 PMvinson24[SOLVED] Find Critical Points
$\displaystyle f(x)={ln(2x)}/{x} ;[1,e]$

Find the critical points

I used the quotient rule and got

$\displaystyle f'(x)= {-ln(2x)}/{x^{2}}$

and now i equal the equation to zero and them im stuck cant solve it - Feb 23rd 2010, 12:15 PMicemanfan
This is because

$\displaystyle f'(x) = \frac{1 - \ln (2x)}{x^2}$.

Try solving it now. - Feb 23rd 2010, 12:26 PMvinson24
Still not sure. Not sure how what to do with the ln(2x) and the $\displaystyle x^{2}$ on the bottom

- Feb 23rd 2010, 12:30 PMicemanfan
The only way to get $\displaystyle \frac{1 - \ln (2x)}{x^2} = 0$ is to have $\displaystyle 1 - \ln (2x) = 0$ or $\displaystyle \ln (2x) = 1$.

- Feb 23rd 2010, 12:38 PMvinson24
So then I would only have on critical point on that interval?