How do you find the inflection point of the function #f(x) = x^2ln(x)#?

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Answer 1 · 2015-09-13T22:40:13

The inflection is the zeroes of the second derivative.

The inflection point is #x=e^(-3/2)#

Hence we have first derivative

#f'(x)=2xlnx+x^2*1/x=2xlnx+x#

and second derivative

#f''(x)=2lnx+2x*1/x+1=> f''(x)=2lnx+3#

Hence the zeroes of f''(x) are

#f''(x)=0=>2lnx+3=0=>lnx^2=-3=>x=e^(-3/2)#