Browse Questions

# Show that the function given by $f(x) =\large \frac{\log \: x}{x}$ has maximum at $x = e.$

Toolbox:
• $\big(\large\frac{u}{v}\big)'=\frac{u'v-uv'}{v^2}$
• Maxima & Minima $f'(x)=0$
Step 1:
$f(x)=\large\frac{\log x}{x}$
Differentiating with respect to x we get
$f'(x)=\large\frac{\Large\frac{1}{x}.x-\log x.1}{x^2}$
$\qquad=\large\frac{1-\log x}{x^2}$
For maxima and minima $f'(x)=0$