AP® Calculus AB review sheet from Aim for Five (aimforfive.com/calc-ab/units/5/5-2)
Unit 5 · Topic 5.2
5.2 Extreme Value Theorem, Global Versus Local Extrema, and Critical Points
The Extreme Value Theorem guarantees that a continuous function on a closed interval has a highest and a lowest value. Critical points, where f′ is 0 or doesn't exist, are the only interior places those extremes (or any local max or min) can occur.
Key terms
- Extreme Value Theorem
- critical point
- absolute (global) extremum
- relative (local) extremum
- closed interval
Absolute vs. relative extrema
Two kinds of highest and lowest points:
- Absolute (global) maximum: the largest value f takes on its whole domain or interval. Absolute minimum: the smallest.
- Relative (local) maximum: a value at least as large as every nearby value. Relative minimum: at least as small as every nearby value.
- The y-value is the max or min value; the x-value is where it occurs. Questions may ask for either, so read carefully.
The Extreme Value Theorem
If f is continuous on the closed interval [a, b], then f has at least one absolute maximum value and at least one absolute minimum value on [a, b].
Both conditions matter. On an interval that isn't closed, the extremes can be missing: f(x) = 1/x on (0, 1] has no maximum, because it shoots up near 0. A function with a jump or an asymptote inside the interval can also miss its maximum or minimum. Like IVT and MVT, EVT tells you the extremes exist but not where they are.
Critical points
A critical point of f is an x-value in the domain of f where f′(x) = 0 or f′(x) does not exist. On a graph, these are horizontal tangents, corners, cusps and vertical tangents.
If f has a relative max or min at an interior point c, then c must be a critical point. That's because at a smooth peak or valley the tangent is horizontal, and the only other possibility is a point where the derivative fails to exist.
Endpoints of a closed interval are not called critical points, but they still matter: an absolute max or min can happen at an endpoint, which is why the Candidates Test (5.5) checks them.
Not every critical point is an extremum
The rule only works one way. f(x) = x³ has f′(0) = 0, but the graph keeps rising through x = 0. It's a critical point but neither a max nor a min. To decide what happens at a critical point, you need a test: the First Derivative Test (5.4) or the Second Derivative Test (5.7).
Finding critical points
Find f′(x), then solve f′(x) = 0 and find where f′(x) is undefined. Keep only x-values in the domain of f. Factoring f′ completely helps; write negative exponents as fractions so you can see where the denominator is 0.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Critical points including an undefined derivative
Find all critical points of f(x) = x^(2/3)(x − 5).
Show the solutionHide the solution
- Step 1: Rewrite: f(x) = x^(5/3) − 5x^(2/3).
- Step 2: f′(x) = (5/3)x^(2/3) − (10/3)x^(−1/3).
- Step 3: Factor out (5/3)x^(−1/3): f′(x) = (5/3)x^(−1/3)(x − 2) = 5(x − 2)/(3∛x).
- Step 4: f′(x) = 0 when x = 2. f′(x) is undefined when x = 0.
- Step 5: Both x = 0 and x = 2 are in the domain of f (all real numbers).
Answer: Critical points at x = 0 (f′ undefined) and x = 2 (f′ = 0).
- Example 2
Does EVT apply?
For each, does the Extreme Value Theorem guarantee an absolute max and min? (a) f(x) = 1/x on [1, 3]; (b) f(x) = 1/x on (0, 1]; (c) f(x) = 1/x on [−1, 1].
Show the solutionHide the solution
- Step 1: (a) 1/x is continuous on [1, 3], a closed interval. EVT applies.
- Step 2: (b) The interval isn't closed. Also, near 0 the values grow without bound, so there's no max. EVT doesn't apply.
- Step 3: (c) The interval is closed, but 1/x isn't continuous at 0, which is inside it. EVT doesn't apply.
Answer: (a) Yes. (b) No: the interval isn't closed. (c) No: f isn't continuous on [−1, 1].
- Example 3
Trap: a critical point that isn't an extremum
A student says f(x) = x³ + 2 has a relative minimum at x = 0 because f′(0) = 0. Is that right?
Show the solutionHide the solution
- Step 1: f′(x) = 3x², so f′(0) = 0. x = 0 is a critical point.
- Step 2: But f′(x) = 3x² > 0 on both sides of 0, so f is increasing on both sides.
- Step 3: A function that increases through a point has no peak or valley there.
Answer: No. x = 0 is a critical point, but f has neither a relative max nor a relative min there.
Common mistakes
- Forgetting critical points where f′ is undefined, like x = 0 for x^(2/3).
- Calling every critical point a max or min. You need a test to decide.
- Including x-values where f itself is undefined. Critical points must be in the domain of f.
On the exam
- Free-response questions often ask you to find critical points from a formula for f′, or from a graph of f′. Look for zeros and points where f′ doesn't exist.
- When asked to justify that an absolute max exists, cite EVT with continuity on a closed interval.
Connected topics
Videos
Check yourself
4 questions on 5.2 Extreme Value Theorem, Global Versus Local Extrema, and Critical Points. Pick an answer to see if you got it, and why.
What are all the critical points of f(x) = x^(2/3)(x − 5)?
The function f is continuous on the closed interval [1, 5], with f(1) = 3 and f(5) = −2. Which of the following must be true? I. f attains an absolute maximum value on [1, 5]. II. f(c) = 0 for some c in (1, 5). III. f′(c) = −5/4 for some c in (1, 5).
How many critical points does f(x) = x cos x have on the open interval 0 < x < 7?
What are all the critical points of f(x) = x − 3x^(1/3) ?
0 of 4 answered