$\displaystyle f(x)={ln(2x)}/{x} ;[1,e]$ 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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This is because $\displaystyle f'(x) = \frac{1 - \ln (2x)}{x^2}$. Try solving it now.
Still not sure. Not sure how what to do with the ln(2x) and the $\displaystyle x^{2}$ on the bottom
The only way to get $\displaystyle \frac{1 - \ln (2x)}{x^2} = 0$ is to have $\displaystyle 1 - \ln (2x) = 0$ or $\displaystyle \ln (2x) = 1$.
So then I would only have on critical point on that interval?
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