f'(x) is undefined only at x = -3, in which case f is also undefined so this is not a critical point. f'(x) is zero when
x2 + 6x + 5 = 0.
Happily, this quadratic factors as
x2 + 6x + 5 = (x + 5)(x + 1),
so f'(x) is zero at x = -5 and x = -1. Here's the numberline so far:
To figure out whether f has a maximum or a minimum at each of these critical points, we need to find the sign of the first derivative so we can fill in the numberline.
f'(x) = (x + 5)(x + 1)
is negative when exactly one of the quantities (x + 5) or (x + 1) is negative. This is only possible when
-5 < x < -1,in which case (x + 1) is negative but (x + 5) is positive. So the numberline looks like this:
Since f' is positive to the left and negative to the right of x = -5, the function f has a maximum at x = -5. Since f' is negative to the left and positive to the right of x = -1, the function f has a minimum at x = -1.