Question

How would I find the critical numbers of
f'(x)=(x-3)^3(x+4)(x^2+1)? The answer is x=-4, x=3

Answer 1

I'm assuming that's
f'(x) = (x-3)^3 * (x + 4) * (x^2 + 1)
instead of
f'(x) = (x-3) ^ 3(x + 4)(x^2 + 1)

Critical numbers occur at f'(x) = 0 or does not exist.
Let f'(x) = 0:
0 = (x-3)^3 * (x + 4) * (x^2 + 1)
Will be true if the terms inside the parenthesis are equal to zero (any among them).
x - 3 = 0
x = 3 (critical number)

x + 4 = 0
x = -4 (critical number)

x^2 + 1 = 0
x^2 = -1
x = ±i (imaginary)

Answer 2

thanks!