Question

Show that the function given by f(x)=\(\frac{logx}{x}\) has maximum at x=e

Answer 1

The given function is  f(x)= \(\frac{log x}{x}\)
\(f'(x)=  \frac{x(\frac{1}{x})  -   log x}{x^2}  =\frac{1- logx}{x^2}\)
Now,f'(x)=0
⇒1−logx=0
⇒logx=1
⇒logx=loge
⇒x=e

Now \(f"(x)=\frac{x^2  (\frac{-1}{x})-(1-log x)(2x)}{x^4}   \)
=-\(\frac{-x-2x(1-logx)}{x^4}\)
=\(\frac{-3+2log x}{x^3}\)
Now,\(f"(e)=\frac{-3+2   log e}{e^3}  =   (\frac{-3+2}{e^3})  =  (\frac{1}{e^3}<0)\)
Therefore,by second derivative test,f is maximum at x=e.