Rises then falls is a peak. Read it left to right AB & BC

Let $c$ be a critical point of $f$, with $f$ continuous at $c$ and differentiable on both sides of it. Read the sign of $f'$ from left to right across $c$:

1. $f'$ goes from positive to negative: $f$ rises then falls, so $f(c)$ is a relative maximum.
2. $f'$ goes from negative to positive: $f$ falls then rises, so $f(c)$ is a relative minimum.
3. $f'$ keeps the same sign: $f$ passes straight through, so $c$ gives neither.

Nothing here is a rule to be memorized in the abstract. Rising then falling is a peak because that is what a peak is. If the two words ever come apart, draw the two little arrows: up then down.

Worked all the way through, on $f(x) = 3x^{4} - 4x^{3} - 12x^{2} + 5$:

$$f'(x) = 12x^{3} - 12x^{2} - 24x = 12x(x + 1)(x - 2),$$

with critical points $-1$, $0$, and $2$. Testing one value in each of the four intervals gives $f'(-2) = -48$, $f'(-0.5) = 7.5$, $f'(1) = -24$, and $f'(3) = 144$: negative, positive, negative, positive. So the sign changes are $-$ to $+$ at $x = -1$, $+$ to $-$ at $x = 0$, and $-$ to $+$ at $x = 2$. That is a relative minimum, a relative maximum, and a relative minimum, at heights $0$, $5$, and $-27$.

Three critical points, three sign changes, three verdicts. That will not always happen, which is the next section.

The third case is the one students hesitate over, because the derivative did something at $c$ and the instinct is that something must follow. Nothing follows.

$f(x) = (x - 1)^{3} + 2$ has $f'(x) = 3(x - 1)^{2}$, which is zero at $x = 1$ and positive on both sides. The tangent flattens for an instant and the function keeps climbing. There is no relative extremum at $x = 1$, and the correct answer to "classify the critical point" is exactly that.

This is the same fact Topic 5.2 introduced with $f(x) = x^{3}$, and it recurs because the visual is misleading: a flat tangent looks like a landing. A flat tangent means the rate is momentarily zero. Whether the function turns is a separate question, answered only by the signs on either side.

Two consequences worth stating plainly.

1. A function can have many critical points and no extrema. $f(x) = x^{3}$ has one critical point and no extrema at all.
2. The number of extrema is the number of sign changes, which is at most the number of critical points and is often fewer.

So the chart is not a list of answers. It is a list of places to look, and part of looking is being willing to find nothing.

Read the statement again: the sign of $f'$ just before and just after. Neither of those is the value at $c$. So the test applies perfectly well at a corner or a cusp, where $f'(c)$ does not exist at all.

1. $f(x) = |x|$ at $x = 0$. To the left $f' = -1$; to the right $f' = +1$. Negative to positive, so a relative minimum, which is obviously correct and is a conclusion the second derivative test in 5.7 cannot reach.
2. $f(x) = x^{2/3}$ at $x = 0$. To the left $f' < 0$; to the right $f' > 0$. Again a relative minimum, at a cusp.
3. $f(x) = x^{1/3}$ at $x = 0$. Positive on both sides, so no extremum, even though the tangent there is vertical.

What the test does require is that $f$ be continuous at $c$. A sign change across a jump proves nothing, because the two sides are not connected and the value at $c$ can be anything at all.

This is the practical advantage of the First Derivative Test over the one coming in 5.7: it works at every critical point, of both kinds, and it never returns an inconclusive answer.

An answer to a relative extremum question has three parts, and questions ask for different ones.

1. Which kind. Maximum or minimum, from the direction of the sign change.
2. Where. The value of $x$, which is the critical point itself.
3. What. The value of $f$ there, which requires going back to the original function and substituting. $f'$ cannot supply it.

"The relative maximum is at $x = 0$" and "the relative maximum is $5$" are different sentences about the quartic above, and both are true of it. Reading the last clause of the prompt is the whole discipline: find the relative maximum asks for a value, find where the relative maximum occurs asks for a location.

One more piece of vocabulary that will matter in 5.5. A relative extremum needs the function to beat its neighbours on both sides, so it lives at an interior point. If $f' > 0$ across all of $(0, 4)$, then $f(4)$ is the largest value on $[0, 4]$ and there is no relative maximum anywhere, because at $x = 4$ there is no right-hand side to compare with. Largest and relative maximum are not synonyms, and the next topic is built on the gap between them.