`f(x) = x^3 - 3x^2` Find the critical numbers of the function.
You need to evaluate the critical numbers of the function and for this reason, you must differentiate the function with respect to x, such that:
`f'(x) = (x^3 - 3x^2)'`
`f'(x) = 3x^2 - 6x`
You need to solve for x the equation f'(x) = 0:
`3x^2 - 6x = 0`
Factoring out 3x yields:
`3x(x - 2) = 0=> 3x = 0` or `x - 2 = 0`
`3x = 0=> x = 0`
`x - 2 = 0 => x = 2`
Hence, evaluating the critical values of the given function, yields x = 0 and x = 2.