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Question

Show that the function given by f(x)=log xx has maximum at x = e.


Solution

Let f(x)=log xx
On differentiating w.r.t.x, we get f(x)=x(1x)(log x).1x2=1log xx2
Again differentiating, we get f"(x)=x2(1x)(1log x)2x(x2)2
=x2x+2x log xx4=x(2 log x3)x4=2 log x3x3
For maximum put f(x)=01log xx2=0log x=1x=e
At x=e,f"(e)2 log e3e3=2.13e3=1e3<0
Therefore, by second derivative test, f is the maximum at x = e.


Mathematics

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