What are the points of inflection, if any, of #f(x)=x^(1/3)e^(3x) #?

1 Answer
Nov 2, 2016

there are two of them and exactly they are #x=\frac{-1-sqrt{3}}{9}# and #x=\frac{-1+sqrt{3}}{9}#

Explanation:

It is matter of deriving twice #f(x)#
#f'(x)=e^{3x}(\frac{1}{3root3{x^2}}+3root3{x})#
that after a simple algebraic manipulation yelds
#f'(x)=e^{3x}(\frac{1+9x}{3root3{x^2}})#
so that the point of minimum has got abscissa #x=-\frac{1}{9}#
By deriving #f'(x)# we obtain
#f''(x)=\frac{-2}{9root3{x^5}}+\frac{2}{root3{x^2}}+9root3{x}#
that after a simple manipulation
turns to be
#f''(x)=\frac{-2+18x+81x^2}{9root3{x^5}}#
whose roots are just the abscissas of the inflection points