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Question

Mathematics Question on Applications of Derivatives

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

Answer

The given function is f(x)= logxx\frac{log x}{x}
f′(x)=x(1x)−logxx2=1−logxx2f'(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)=x2(−1x)−(1−logx)(2x)x4f"(x)=\frac{x^2 (\frac{-1}{x})-(1-log x)(2x)}{x^4}
=-−x−2x(1−logx)x4\frac{-x-2x(1-logx)}{x^4}
=−3+2logxx3\frac{-3+2log x}{x^3}
Now,f"(e)=−3+2logee3=(−3+2e3)=(1e3<0)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.