AP Calculus AB and BC

Critical Points That Are Not Extrema

The Interior Extremum Theorem says a local extremum at an interior point where the derivative exists forces that derivative to be zero. The converse is false: a zero derivative does not produce an extremum, as the cube function at the origin shows.

Interior Extremum Theorem

f has a local extremum at c∈(a,b) and f′(c) exists  ⟹  f′(c)=0f \text{ has a local extremum at } c \in (a,b) \text{ and } f'(c) \text{ exists} \implies f'(c) = 0

If f has a local maximum or minimum at an interior point and the derivative exists there, then the derivative is zero at that point.

The hypotheses, and what each one buys

A theorem is only as strong as its conditions. Below, each hypothesis is dropped on its own while every other one is kept, so you can see precisely what it was holding up.

  1. 1

    There really is a local extremum at c

    f(c)≥f(x) near c, or f(c)≤f(x) near cf(c) \ge f(x) \text{ near } c, \text{ or } f(c) \le f(x) \text{ near } c

    This is the hypothesis, and the useful failure is running the theorem backwards. A vanishing derivative is a necessary condition for an interior extremum, never a sufficient one, and the whole business of first and second derivative tests exists because of that gap.

    Drop it and the theorem fails

    f(x)=x3,f′(0)=0f(x) = x^{3}, \qquad f'(0) = 0

    f(x) = x cubed at the origin

    The derivative 3x23x^2 is zero at the origin, so this is a critical point in the fullest sense. But the check confirms the function rises above f(0)=0f(0) = 0 immediately to the right and falls below it immediately to the left, at every scale examined. A point that is beaten from both sides in every neighbourhood is neither a maximum nor a minimum.

  2. 2

    c is an interior point of the interval

    c∈(a,b)c \in (a, b)

    At an endpoint the function only has one side to be compared against, so an extremum there says nothing about the derivative. f(x)=xf(x) = x on [0,1][0,1] has its maximum at x=1x = 1 with f′(1)=1f'(1) = 1, which is not zero and does not need to be. That is exactly why the candidates test checks endpoints separately rather than trusting the critical points alone.

    No counterexample built from an elementary formula breaks this one on its own, so none is claimed here.

  3. 3

    The derivative exists at c

    f′(c) existsf'(c) \text{ exists}

    If the derivative does not exist, the theorem's conclusion is not false, it is unstated. f(x)=∣x∣f(x) = |x| has a genuine minimum at 0 and no derivative there. This is why the definition of a critical point includes points where the derivative fails to exist, not just points where it is zero.

    No counterexample built from an elementary formula breaks this one on its own, so none is claimed here.

The counterexample above is checked numerically on every build: each function is evaluated and the conclusion is confirmed to fail. A counterexample that stopped working would fail the build rather than sit here misleading you.

Why it is true

  1. Suppose ff has a local maximum at the interior point cc and f′(c)f'(c) exists.
  2. Approaching from the right, the difference quotient f(c+h)−f(c)h\frac{f(c+h)-f(c)}{h} has a non-positive numerator and a positive denominator, so its limit is at most 0.
  3. Approaching from the left, the same numerator is still non-positive but the denominator is now negative, so that limit is at least 0.
  4. The derivative exists, so both one-sided limits equal f′(c)f'(c). A number that is at most 0 and at least 0 is 0. The local minimum case is identical with the inequalities reversed.

What it does not say

Critical point means extremum.

It does not. The cube function above is the standard counterexample. A critical point is a candidate, and the first or second derivative test is what decides.

Every extremum is a critical point.

Every INTERIOR extremum is, which is the theorem. Endpoint extrema are not critical points and still have to be checked, which is the entire reason the candidates test exists.

This is the same as the first derivative test.

It is the theorem the first derivative test is built on. This one says where extrema can be; the first derivative test says which candidates actually are extrema, by looking at the sign change.

Frequently asked questions

Is this the same as Fermat's theorem?

Yes, it is often called Fermat's theorem on stationary points, distinct from Fermat's Last Theorem. AP materials usually call it the interior extremum theorem or fold it into the definition of a critical point.

What is a saddle point in single variable calculus?

The usual name for a critical point that is not an extremum, like the origin for x3x^3. The function flattens momentarily and then carries on in the same direction, so the derivative touches zero without changing sign.