The endpoints are candidates too AB & BC

Let $f$ be continuous on the closed interval $[a, b]$. Then:

1. Find every critical point of $f$ in the open interval $(a, b)$, from both clauses of the definition.
2. Add $a$ and $b$ to the list.
3. Evaluate $f$ at every entry on the list.
4. The largest of those values is the absolute maximum, and the smallest is the absolute minimum.

No sign chart, no First Derivative Test, no classification of anything. The reason this works is Topic 5.2's one-way street: an absolute extremum on $[a, b]$ either sits at an interior point, where it must also be a relative extremum and therefore a critical point, or it sits at an endpoint. There is nowhere else for it to be, so a list containing all of both is complete.

Work $f(x) = x^{3} - 3x^{2} - 9x + 5$ on $[-2, 6]$. The derivative is $3(x + 1)(x - 3)$, so the interior critical points are $-1$ and $3$, and the full list is $-2$, $-1$, $3$, $6$. Evaluating:

$$f(-2) = 3, \qquad f(-1) = 10, \qquad f(3) = -22, \qquad f(6) = 59.$$

The absolute maximum is $59$, at $x = 6$, and the absolute minimum is $-22$, at $x = 3$. Notice what happened to $x = -1$: it is a genuine relative maximum, the only one this function has, and it lost by a wide margin to an endpoint that has no local claim to anything.

Step 3 says evaluate $f$. The derivative did its work in step 1 and is finished.

Comparing values of $f'$ instead is a surprisingly durable error, probably because $f'$ is the expression most recently written down. It answers a different question: $f'$ is smallest where the function is falling fastest, which has nothing to do with where the function is lowest. In the worked example, $f'(6) = 63$ is the largest derivative on the list and $f(6) = 59$ is also the largest value, which is a coincidence of this problem, and $f'(-2) = 15$ exceeds $f'(-1) = 0$ while $f(-2) = 3$ is well below $f(-1) = 10$.

One consequence worth stating: at every interior critical point where $f'$ exists, $f'$ is zero, so comparing derivative values there compares a list of zeros. Whatever that procedure produces, it is not an answer.

A related habit is comparing the sizes of values rather than the values themselves. The absolute minimum of a function that runs from $-13$ to $12$ is $-13$, not the value closest to zero. Minimum means smallest, and smallest on the number line means furthest left.

A relative extremum is a fact about the function near a point. An absolute extremum is a fact about the function on an interval, and changing the interval changes the answer even though the function has not moved.

The same cubic on $[-2, 2]$: the interior critical point is $-1$ only, since $3$ is outside, so the list is $-2$, $-1$, $2$, giving $3$, $10$, $-17$. The absolute maximum is $10$ at $x = -1$ and the absolute minimum is $-17$ at $x = 2$.

On $[-2, 6]$ that same $x = -1$ is still exactly as much of a relative maximum as it was, and it is no longer the absolute maximum. Nothing about the point changed. The competition changed.

Two working rules follow.

1. Discard critical points outside the interval. They are not candidates, and including them can manufacture an extreme value the function never reaches on the interval you were asked about.
2. Never carry an answer across intervals. Redo the list.

When the interval is not closed, the test does not apply as stated, and the honest answer is often that no absolute extremum exists. On $(0, 3)$, $f(x) = x^{2}$ has no absolute maximum and no absolute minimum: the values it would need are at endpoints the interval does not contain.

Every absolute extremum problem has two numbers attached to it, and they are almost never the same number.

1. The value: $59$. This is what "the absolute maximum" means on its own, and it is an output of $f$.
2. The location: $x = 6$. This is what "where the absolute maximum occurs" means, and it is an input.

"Find the absolute maximum of $f$ on $[-2, 6]$" wants $59$. "Find the value of $x$ at which $f$ attains its absolute maximum" wants $6$. Answering the first with $6$ is wrong for the same reason that answering "how tall is the mountain" with "in Nepal" is wrong.

Two more distinctions in the same family, both worth saying out loud before writing anything down:

1. Relative or absolute. A question asking for relative extrema wants a classification of critical points. A question asking for absolute extrema wants a comparison of values.
2. Maximum or minimum. On a list where every candidate value is negative, the absolute maximum is still the largest of them, which will also be negative.

A good habit for the free-response section: write the candidate list as an explicit table of inputs and outputs, then circle the row you were asked about. The table also constitutes the justification, since the comparison is the argument.