Q2 of 168 Page 242

Show that the function given by has maximum at x = e.

It is given that f(x) =

Then, f’(x) =


Now, f’(x) = 0


⇒ 1 - logx =0


⇒ log x =1


⇒ log x = log e


⇒ x = e


Now, f’’(x) =




Now, f’’(e)=


Therefore, by second derivative test, f is the maximum at x = e.


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