Where extrema can hide, and where they do AB & BC

If $f$ is continuous on a closed, bounded interval $[a, b]$, then $f$ attains an absolute maximum value and an absolute minimum value somewhere on $[a, b]$.

Every word is load-bearing, and the theorem says nothing about where. It is a pure existence guarantee, the licence that makes the search in 5.5 worth starting.

Watch each condition fail on its own.

1. Not closed. $f(x) = x$ on $(0, 1)$. Continuous, bounded, and it attains neither extreme: values climb toward $1$ and sink toward $0$ without ever arriving.
2. Not bounded. $f(x) = x^{2} - 6x + 5$ on $[1, \infty)$. Continuous and closed, but there is no absolute maximum, because the values grow without limit.
3. Not continuous. $f(x) = x$ for $0 \le x < 2$ with $f(2) = 1$, on $[0, 2]$. Closed and bounded, and there is no absolute maximum: the values climb toward $2$, and the one point that could have caught them was reassigned to $1$.

One direction only. When the conditions hold, the extrema exist. When a condition fails, the theorem is silent, which is not the same as saying nothing exists. The third example above still has an absolute minimum of $0$ at $x = 0$, and the second still has an absolute minimum of $-4$ at $x = 3$. A failed hypothesis withdraws a guarantee; it does not supply the opposite one.

A relative (or local) maximum at $c$ means $f(c) \ge f(x)$ for all $x$ in some open interval around $c$. It is a claim about a neighborhood. A relative minimum is the same with the inequality reversed.

An absolute (or global) maximum at $c$ means $f(c) \ge f(x)$ for every $x$ in the whole domain under discussion. It is a claim about everything.

Three consequences worth holding onto:

1. Because a relative extremum needs room on both sides, it happens at interior points. An endpoint can hold the absolute maximum without being a relative maximum.
2. The absolute maximum need not be any of the relative maxima. On a closed interval it is often at an endpoint, which is exactly the omission 5.5 is built around.
3. The value and the location are different answers. "The maximum is $7$" and "the maximum occurs at $x = 7$" are different sentences, and a question asks for one of them. Read the last clause of the prompt before writing anything down.

A single function can carry several relative maxima, one absolute maximum, and have the absolute one sitting at a place that is not relative at all.

A number $c$ is a critical point of $f$ when $c$ is in the domain of $f$ and either

1. $f'(c) = 0$, or
2. $f'(c)$ does not exist.

The second clause is where the misses happen. Setting the derivative equal to zero and solving is the whole of most students' search, and it silently skips every corner, cusp, and vertical tangent.

$f(x) = x^{2/3}$ has $f'(x) = \frac{2}{3x^{1/3}}$, which is never zero. Solve $f' = 0$ and the answer is that there are no critical points. There is one, at $x = 0$, where the derivative fails to exist and the graph has a cusp, and it is the absolute minimum of the function.

The membership rule cuts the other way. $f(x) = \frac{x}{x - 2}$ has $f'(x) = \frac{-2}{(x - 2)^{2}}$, which is never zero and fails only at $x = 2$. But $x = 2$ is not in the domain of $f$, so it is not a critical point, and this function has none at all. A vertical asymptote is not a candidate for an extremum, because $f$ has no value there to be extreme.

So the search is: differentiate, find where the derivative is zero, find where it is undefined, then throw out anything the original function does not reach. For $f(x) = |x^{2} - 4|$ that yields $x = 0$ from the first list and $x = \pm 2$ from the second, three critical points where solving $f' = 0$ alone would have found one.

The theorem underneath all of this, sometimes called Fermat's Theorem, runs in exactly one direction:

$$\text{if } f \text{ has a relative extremum at an interior } c \text{ and } f'(c) \text{ exists, then } f'(c) = 0.$$

Read forward it is a search strategy: every interior extremum is a critical point, so the critical points are a complete list of candidates and nothing can hide outside it. Read backward it is false, and the counterexamples are the two most familiar functions in the unit.

1. $f(x) = x^{3}$ at $x = 0$. Here $f'(0) = 0$, so it is a critical point. But $f'(x) = 3x^{2}$ is positive on both sides, so $f$ is increasing straight through. The tangent flattens and the function never turns.
2. $f(x) = x^{1/3}$ at $x = 0$. Here $f'$ does not exist, so it is a critical point. The tangent is vertical, and $f$ is increasing on both sides. Again no extremum.

Compare those with $f(x) = 4 - x^{2}$ at $x = 0$, where $f'$ is zero and the sign changes from positive to negative, and with $f(x) = |x|$ at $x = 0$, where $f'$ fails and the sign changes from negative to positive. Both of those are extrema.

All four are critical points. Two are extrema. What separates them is not whether the derivative vanished or failed, but whether $f'$ changed sign. That test is Topic 5.4, and everything between here and there is machinery for applying it.