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

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Answer 1 · 2015-09-13T22:40:13

The inflection is the zeroes of the second derivative.

The inflection point is $x = e^{- \frac{3}{2}}$ $x=e^(-3/2)$

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

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

How do you find the inflection point of the function f(x) = x^2ln(x)f(x)=x2ln(x)f(x) = x^2ln(x)?