Maximum of $3x^2e^{-x^3}$

Question

I have a PDF which looks like:

$f(x) = 3x^2e^{-x^3}, \quad x \geq 0 $

I need to find it's maximum (to sample from it using the rejection method), so I differentiate and set the result to $0$:

$(3x^2e^{-x^3})(-3x^2) + 6xe^{-x^3} = 0$ $\qquad$ (chain rule)

$6xe^{-x^3} - 9x^4e^{-x^3} = 0$

Diving by $3xe^{-x3}$ to get:

$2 - 3x^3 = 0$

Now I'm stuck and I'm not sure that I'm even doing the right thing. Plotting a graph of $f(x)$ looks like this. So the value $\approx 0.9$, but I don't know how to get there. Thanks for your help.

Answer 1

The derivative is $$ f'(x)=6xe^{-x^3}+3x^2e^{-x^3}(-3x^2)=3e^{-x^3}x(2-3x^3) $$ which is $0$ if and only if one of terms is $0$. Now, $e^{-x^3}>0$ for all $x$ and hence $f'(x)=0$ if and only if $x=0$ or $2-3x^3=0$. That is if and only if $x=0$ or $\left(\frac23\right)^{1/3}$.

Answer 2

When you took out the factor $3xe^{-x^3}$ the $9$ should have become a $3$. Then you have $x=(\frac 23)^{(1/3)}\approx 0.874$