Critical Points and the First Derivative Test

You’ll be able to

Locate critical points, including where f′ is undefined rather than zero
Use a sign chart for f′ to classify each critical point as a relative maximum, minimum, or neither
Write a justification that cites the sign change of f′ rather than the value of f

Critical points: two ways, not one

x = c is a critical point of f when c is in the domain of f and either f′(c) = 0 or f′(c) does not exist. The second case is the one that gets dropped. f(x) = x^(2/3) has a minimum at x = 0 that no amount of solving f′(x) = 0 will find, because f′ has a cusp there. When hunting critical points, solve the numerator of f′ for zeros and check the denominator for points where f′ blows up — both lists matter.

The First Derivative Test is about a sign change

At a critical point c, if f′ changes from positive to negative, f has a relative maximum at c; from negative to positive, a relative minimum; if f′ does not change sign, c is neither. Note what the test does not use: the value f(c), the magnitude of f′, or the second derivative. A sign chart for f′ — number line, critical points marked, sign of f′ tested in each interval — is the whole method.

Relative versus absolute

The First Derivative Test finds relative extrema, which are local claims. An absolute extremum on a closed interval is found by the Candidates Test: evaluate f at every critical point and at both endpoints, then compare the values. An endpoint can be the absolute maximum without being a critical point at all, which is why endpoints must appear on the candidate list even though the First Derivative Test never mentions them.

First Derivative Test at a critical point c

f′: + → − ⇒ relative max · f′: − → + ⇒ relative min · no sign change ⇒ neither

The conclusion comes from the sign change of f′ across c, not from the value of f at c.

Worked example

Find and classify the critical points of f(x) = x³ − 6x² + 9x + 1.

1.f′(x) = 3x² − 12x + 9 = 3(x² − 4x + 3) = 3(x − 1)(x − 3). It is defined everywhere, so the only critical points come from f′ = 0.
2.f′(x) = 0 at x = 1 and x = 3.
3.Test the sign of f′ on each interval. At x = 0: 3(−1)(−3) = 9 > 0. At x = 2: 3(1)(−1) = −3 < 0. At x = 4: 3(3)(1) = 9 > 0.
4.Across x = 1, f′ goes from positive to negative, so f has a relative maximum at x = 1.
5.Across x = 3, f′ goes from negative to positive, so f has a relative minimum at x = 3.

Answer: Relative maximum at x = 1 and relative minimum at x = 3, justified by the sign changes of f′ at each.

Watch out

"f(1) = 5 is bigger than f(3) = 1, so x = 1 is the maximum" is not a First Derivative Test justification and earns no credit. Compare values only in the Candidates Test for absolute extrema on a closed interval.

Checkpoint

f′(x) = (x − 2)²(x + 5). At which critical point does f have a relative extremum?

Checkpoint

Which point could be a critical point of f even though f′ is never zero there?

On the exam

Free-response justifications are graded on the sign change. Write "f′ changes from positive to negative at x = 1, so f has a relative maximum there" — the phrase "changes from … to …" is what the rubric looks for.

Answer the 2 checkpoints as you read.