`y = ln(x)/x` Locate any relative extrema and points of inflection.

Wednesday, October 8, 2014

Locate extrema and inflection points for $\displaystyle{y}=\frac{{{\ln{{\left({x}\right)}}}}}{{x}}$ :

The domain is x>0.

Extrema can only occur at critical points, or points where the first derivative is zero or fails to exist.

$\displaystyle{y}=\frac{{1}}{{x}}\cdot{\ln{{\left({x}\right)}}}$

$\displaystyle{y}'=\frac{{1}}{{x}}\cdot{\left(\frac{{1}}{{x}}\right)}+{\left(-\frac{{1}}{{{x}}^{{2}}}\right)}{\ln{{x}}}$

$\displaystyle{y}'=\frac{{1}}{{{x}}^{{2}}}{\left({1}-{\ln{{x}}}\right)}$

Setting equal to zero we get:

$\displaystyle\frac{{1}}{{{x}}^{{2}}}=\frac{{{\ln{{x}}}}}{{{x}}^{{2}}}=\Rightarrow{\ln{{x}}}={1}=\Rightarrow{x}={e}$

For 0<x<e the first derivative is positive and for x>e it is negative, so there is a maximum at x=e and this is the only extrema.

Inflection points can only occur if the second derivative is zero:

$\displaystyle{y}''=-\frac{{2}}{{{{x}}^{{3}}}}{\left({1}-{\ln{{x}}}\right)}+\frac{{1}}{{{x}}^{{2}}}{\left(-\frac{{1}}{{x}}\right)}$

$\displaystyle{y}''=-\frac{{1}}{{{x}}^{{3}}}{\left({2}-{2}{\ln{{x}}}+{1}\right)}=-\frac{{1}}{{{x}}^{{3}}}{\left({3}-{2}{\ln{{x}}}\right)}$

So $\displaystyle{3}-{2}{\ln{{x}}}={0}=\Rightarrow{\ln{{x}}}=\frac{{3}}{{2}}=\Rightarrow{x}={{e}}^{{\frac{{3}}{{2}}}}\approx{4.482}$

There is an inflection point at $\displaystyle{x}={{e}}^{{{1.5}}}$ where the graph changes from concave down to concave up.

The graph:

Ninenotes