Lesson 3: Extreme Values and Critical Points
Estimated time: 35-45 minutes
Learning Objectives
By the end of this lesson, you will be able to:
- Define absolute (global) and local (relative) extrema
- State and apply the Extreme Value Theorem
- Find critical points of a function
- Use the Closed Interval Method to find absolute extrema
Absolute and Local Extrema
Absolute Maximum/Minimum
f has an absolute maximum at c if f(c) ≥ f(x) for all x in the domain. Similarly for absolute minimum.
Local Maximum/Minimum
f has a local maximum at c if f(c) ≥ f(x) for all x near c (in some open interval containing c). Similarly for local minimum.
Extreme Value Theorem
If f is continuous on a closed interval [a, b], then f attains both an absolute maximum and an absolute minimum on [a, b].
Critical Points
Critical Point
A number c in the domain of f is a critical point if f'(c) = 0 or f'(c) does not exist.
Fermat's Theorem
If f has a local extremum at c and f'(c) exists, then f'(c) = 0. In other words, local extrema can only occur at critical points.
Example 1: Finding Critical Points
Find the critical points of f(x) = x³ − 3x + 1.
Step 1: f'(x) = 3x² − 3 = 3(x² − 1) = 3(x − 1)(x + 1).
Step 2: Set f'(x) = 0: x = 1 or x = −1.
Critical points: x = −1 and x = 1.
f(−1) = 3 (local max), f(1) = −1 (local min).
Example 2: Critical Point Where f' DNE
Find the critical points of f(x) = |x|.
Step 1: f'(x) = 1 for x > 0, f'(x) = −1 for x < 0, and f'(0) does not exist.
Step 2: f' is never zero, but f'(0) DNE. So x = 0 is a critical point.
f(0) = 0 is an absolute minimum.
Closed Interval Method
Closed Interval Method
To find the absolute max and min of a continuous function f on [a, b]:
- Find all critical points of f in (a, b).
- Evaluate f at each critical point and at the endpoints a and b.
- The largest value is the absolute max; the smallest is the absolute min.
Example 3: Closed Interval Method
Find the absolute extrema of f(x) = x³ − 3x + 1 on [−2, 2].
Step 1: Critical points: x = −1, x = 1 (both in (−2, 2)).
Step 2: Evaluate:
- f(−2) = −8 + 6 + 1 = −1
- f(−1) = −1 + 3 + 1 = 3
- f(1) = 1 − 3 + 1 = −1
- f(2) = 8 − 6 + 1 = 3
Absolute max = 3 at x = −1 and x = 2. Absolute min = −1 at x = −2 and x = 1.
Key Takeaways
- Extreme Value Theorem: continuous on [a, b] guarantees absolute extrema exist.
- Critical points: where f' = 0 or f' DNE. All local extrema occur at critical points.
- Closed Interval Method: compare f-values at critical points and endpoints.
- Not every critical point is an extremum. Further tests (first/second derivative) are needed to classify.
Check Your Understanding
1. Find the critical points of f(x) = 2x³ − 9x² + 12x.
2. Find the absolute max and min of f(x) = x4 − 2x² on [−2, 2].
3. True or false: If f'(c) = 0, then f has a local extremum at c.
4. Find the critical points of g(x) = x2/3.