What are the local extrema of f(x)=x^2/lnxf(x)=x2lnx?

1 Answer
Nov 3, 2016

f(x)f(x) has local extrema at x=0x=0 and x=sqrt(e)x=e

Explanation:

f(x) = x^2/lnxf(x)=x2lnx

f'(x) = (lnx*2x - x^2*1/x)/(lnx)^2

f(x) will have local extrema where f'(x)=0

(lnx*2x - x^2*1/x)/(lnx)^2 = 0

2xlnx-x=0

x(2lnx-1)=0

x=0 or lnx=1/2 -> x=e^(1/2) = sqrt(e)

Hence f(x) has local extrema at x=0, (0,0) and x=sqrt(e), (sqrt(e), 2e)

This can be seen from the graph of f(x) below:

graph{x^2/lnx [-11.91, 33.68, -8.58, 14.24]}