# Function f(x)=(x^3)(e^x) is strictly increasing on the interval of?

Jun 24, 2018

Find the first derivative:

f’(x) = 3x^2e^x +x^3e^x

Now determine the values of $x$ for which f’(x) = 0.

$3 {x}^{2} {e}^{x} + {x}^{3} {e}^{x} = 0$

${x}^{2} {e}^{x} \left(3 + x\right) = 0$

It follows that $x = 0$ and x=-3#

We see that the derivative is positive whenever $x > - 3$, thus $f \left(x\right)$ is strictly increasing on $\left(- 3 , \infty\right)$.

Hopefully this helps!