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

The inflection is the zeroes of the second derivative.

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

#### Explanation:

Hence we have first derivative

$f ' \left(x\right) = 2 x \ln x + {x}^{2} \cdot \frac{1}{x} = 2 x \ln x + x$

and second derivative

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

Hence the zeroes of f''(x) are

$f ' ' \left(x\right) = 0 \implies 2 \ln x + 3 = 0 \implies \ln {x}^{2} = - 3 \implies x = {e}^{- \frac{3}{2}}$