Relative minimum question calculus

Can someone please help with steps, thanks in advance!
If g is a differentiable function such that g(x) < 0 for all real numbers x and f ' (x) = (x^2 - 4) g (x), what is true ?
The answer was f has a relative minimum at x = -2 and relative maximum at x = 2 but I do not get why.

at a minimum, f ' (x) = 0 and f '' (x) > 0
at a maximum, f ' (x) = 0 and f '' (x) < 0

g(x) is not 0 for all x ==> f ' (x) = 0 if and only if x^2 - 4 = 0 if and only if x = 2 or -2

f '' (x) = 2x g(x) + (x^2 - 4) [g ' (x)] by the product rule

so
f '' (2) = 2(2) g(2) + 0 < 0 so x = 2 is local max
f '' (-2) = 2(-2) g(-2) + 0 > 0 so x = -2 is local min

when you differentiate f ' (x) with respect to x using the product rule, you get
f '' (x) = [d/dx(x^2-4)] g(x) + (x^2-4) [d/dx(g(x))]
= 2x g(x) + (x^2-4) g'(x)

at x = 2,
f''(x) = 2(2) g(2) + (2^2-4) g'(2)
= 4 g(2)
we only need that g(2)<0
this tells us f '' (2)<0

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