Problem: Determine maximal intervals of monotonicity and local extrema for

Solution: As usual, we start with the domain. There is no problem in this expression, so the domain is the whole real line. We have one interval to start with, now we will find out whether it needs to be split in order to get intervals of monotonicity. We find potential splitting points by finding critical points. First we find the derivative; however, because of the absolute value, we cannot use rules. The standard trick is to get rid of this absolute value by splitting the function.

Now we can differentiate each expression using the usual rules, but we have to remember that such derivatives only work on open sets.

We do not know whether we have a derivative at 3. We could find out using one-sided derivatives, but it is not really necessary, since we are now looking for critical points in order to find intervals of monotonicity. At x = 3 the derivative might not exist, so we include it in our list of splitting points. Other critical points can be found by asking where is the derivative equal to zero.

We disregarded x = 3/2; while it is a solution to the equation e^x(2x − 3) = 0, this expression has nothing to do with f ′ for x < 3.

We have two splitting points, which makes three intervals and we check on monotonicity on them using a table. We include the endpoints when the function is continuous there. As usual with functions defined by cases, we have to be careful to use each formula only within regions where it is valid.

We see that monotonicity really changes at 3, so we were right including it in our list. The conclusion is that the given function is increasing on (−∞,5/2] and on [3,∞) and it is decreasing on [5/2,3]. It remains to determine local extrema.

The function f has a local maximum f (5/2) = 2e^5/2 and a local minimum f (3) = e³.

Next problem
Back to Solved Problems - Graphing