Question

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

Answer 1

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