For the function, use the second derivative test (if possible) to determine if each critical point is a minimum, maximum, or neither. If the second derivative test can't be used, say so.
We use the quotient rule to find the derivative:
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 use the second derivative test, we need to find the second derivative.
Now we evaluate the second derivative at each critical point. At x = -5 we find
which means f is concave down and so has a maximum at x = -5.
At x = -1 we find
which means f is concave up and so has a minimum at x = -1.
Thankfully, these are the same answers we found using the first derivative test.