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Local extrema can be very useful as we’ve discussed, but often you want the best of the best - the absolute extrema. Fortunately these are found in almost the exact same way as the local extrema... almost.

For most applications, absolute extrema are simply the local extrema that is the largest (or smallest, depending on what you are looking for) and so finding absolute extrema is simply the process of finding all the local extrema and determining their values, then determining which one is the largest maximum, or lowest minimum. There are two exceptions to this however.

First, we need to determine if there actually are absolute extrema. This is the easiest part to overlook or forget. Indeed a common error is to get so focused on finding the critical points and local extrema values, that one forgets to check if it is even necessary.

To determine if there are absolute extrema, we need to determine the range of the function (or more accurately, whether there is an upper or lower bound). This is accomplished by looking for vertical asymptotes, and horizontal asymptotes of the function (if applicable). Note that horizontal asymptotes can only occur in a direction where the domain goes infinitely far.

If, on the other hand, we have a bounded domain like $[a,b]$ for some real numbers $a$ and $b$, then we don’t have to worry about horizontal asymptotes, i.e. end term behavior. Indeed, if the interval is closed and the function is continuous, it turns out we must have an absolute extrema, which is the extreme value theorem - discussed in another segment.

If we don’t have any vertical asymptotes, and if we can determine that the horizontal asymptotes are finite (or the domain is bounded) then we can go ahead and look for the local extrema and determine their $y$-values. This is where the second exception comes into play - whether or not the function is defined on a closed interval.

If the function’s domain is indeed a closed interval, then we need to test the endpoints of the domain. Let’s see an example: consider the function $f(x)=x^2-4$, but suppose we restrict the domain to the interval $[-3, 1]$. Then the graph would look like this:

If we take the derivative we can see that we get $f'(x) = 2x$ which has a single critical point at $x=0$. Testing this one point gives us a local minimum at $(0,-4)$ which is also the absolute minimum. But... our graph shows that we should have found two potential maximums, but they didn’t show up? Why not? The answer lies in seeing a little more of the graph.

Here we’ve added a little more of the function in red to show why the derivative didn’t detect the endpoints of the original graph as critical points. Indeed, mathematically there is no reason to think there should be maximums at those values because the function would normally continue on right through them; it’s the restricted domain that causes us to have potential extrema that we need to check. For this reason, whenever you have a closed endpoint, you should check that value as well against your list of local extrema to see if it is potentially the absolute extrema you want. Notice that an open endpoint need not be checked because an open endpoint is, by definition, not included in the original graph... so it’s not a valid value attained by the function on it’s defined domain and thus can’t be an extrema.

So, as we’ve seen, the process of finding absolute extrema is essentially the same as finding local extrema, with the exception that you need to check two things first. The first is whether or not the function has absolute extrema, which you check by looking for vertical asymptotes and horizontal asymptotes (to determine if the range of the function is infinite in one or both directions). The second thing to remember is that, if the function has a closed endpoint, you need to check that endpoint’s value as if it were an extrema because it almost certainly won’t register as a critical point.