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