Problem: Determine maximal intervals of monotonicity and local extrema for

Solution: As usual, we start with the domain. There is just one condition, the denominator cannot be zero, so

D( f ) = (−∞,1) ∪ (1,∞).

We have two intervals to start with, now we will find out whether they need to be split in order to get intervals of monotonicity. We find potential splitting points by finding critical points. First we find the derivative using the usual rules.

There are no points of the domain where the derivative would not exist, therefore the only critical points are the zero points of this derivative, x = −1 and x = 3. They split the starting two intervals into four intervals of strict monotonicity. We now use the table method to determine this monotonicity. Note that we will use closed endpoints when the function is continuous there.

We have two adjacent intervals of the same monotonicity. Can they be connected? Actually, we do not have to worry here, since they are separated by a hole in the domain and thus they can not form an interval together. The conclusion is that the given function is increasing on (−∞,−1] and on [3,∞) and it is decreasing on [−1,1) and on (1,3]. It remains to determine local extrema.

The function f has a local maximum f (−1) = −2 and a local minimum f (3) = 6.

We will sketch the data that we just found. It might help if we determine limits at endpoints of intervals from the domain.

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