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

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Answer 1 · 2018-06-24T09:31:07

Find the first derivative:

$f’(x) = 3x^2e^x +x^3e^x$

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

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

$x^2e^x(3+ x) = 0$

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

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

Hopefully this helps!