Verified answer

Local maxima and minima can be figured out using the second derivative test. We first have to find the first two derivatives:

f(x) = x^4 + 12x^2 + 36

f'(x) = 4x^3 + 24x = 4x(x^2 + 6)

f''(x) = 12x^2 + 24

We need to solve the first derivative for f'(x) = 0 to find candidates for maximum and minimum. We can see that the only real solution would be x = 0. Next we find the sign of f'(0), which is positive, which implies a local minimum at x = 0.

Now to find the complex zeros:

f(x) = x^4 + 12x^2 + 36 = (x^2 + 6)^2

If f(x) = 0, then x^2 + 6 = 0, and x = +/- i * sqrt(6)