Find the critical numbers of the function. f(x) = x3 + 3x2 − 9x

$f(x) = x^3 + 3x^2 − 9x$

The derivative of the function is:

$f'(x) = 3x^2 + 6x - 9$

To find the critical numbers, we need to solve the equation $f'(x) = 0$. We can do this by factoring the polynomial on the right-hand side:

$3x^2 + 6x - 9 = 0$

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

$x^2 + (3 - 1)x - 3 = 0$

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

(x + 3)(x - 1) = 0

This equation has two solutions: $x = -3$ and $x = 1$. These are the critical numbers of the function.