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.

Answer

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

*x*^{2} + 6*x* + 5 = 0.

Happily, this quadratic factors as

*x*^{2} + 6*x* + 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.