Extreme Value Theorem
Extrema & Critical Points
Master the tools for finding maximum and minimum values of functions: the Extreme Value Theorem, the distinction between global and local extrema, and how critical points guide the search.
AP Calculus BC
Why Do Extrema Matter?
Optimization is everywhere: engineers minimize cost, physicists maximize efficiency, economists find optimal production levels. All these problems reduce to one question: where does a function reach its highest or lowest value?
What Are Extrema?
Global (Absolute) Maximum
\( f(c) \geq f(x) \) for every \( x \) in the domain. The single highest point overall.
Global (Absolute) Minimum
\( f(c) \leq f(x) \) for every \( x \) in the domain. The single lowest point overall.
Local (Relative) Maximum
\( f(c) \geq f(x) \) for all \( x \) in some open interval around \( c \). A hilltop in the neighborhood.
Local (Relative) Minimum
\( f(c) \leq f(x) \) for all \( x \) in some open interval around \( c \). A valley in the neighborhood.
Visualizing Global vs Local Extrema
The Extreme Value Theorem
Continuous
No holes, jumps, or vertical asymptotes on the interval. The graph is an unbroken curve.
Closed Interval
Both endpoints \( a \) and \( b \) are included. Written \( [a, b] \), not \( (a, b) \).
When the EVT Does Not Apply
What Is a Critical Point?
Fermat's Theorem: The Bridge
Contrapositive: If \( f'(c) \neq 0 \), then \( f \) does not have a local extremum at \( c \). This is why we focus on critical points: they are the only candidates for interior extrema.
Global vs Local: Key Distinction
| Global (Absolute) | Local (Relative) | |
|---|---|---|
| Scope | Entire domain or interval | Some open neighborhood |
| How many? | At most one max, one min value | Can have many |
| Where found? | Critical points or endpoints | Only at critical points (interior) |
| EVT guarantees? | Yes (on \( [a, b] \), continuous) | No |
| Relationship | Every global extremum is also a local extremum (if interior) | |
The Closed Interval Method
To find the absolute maximum and minimum of a continuous function \( f \) on \( [a, b] \):
1
Find all critical points of \( f \) in \( (a, b) \): solve \( f'(x) = 0 \) and identify where \( f'(x) \) does not exist.
2
Evaluate \( f \) at each critical point and at both endpoints \( a \) and \( b \).
3
Compare all values. The largest is the absolute maximum; the smallest is the absolute minimum.
Example 1: \( f(x) = x^3 - 3x + 1 \) on \( [-2, 2] \)
1
Differentiate: \( f'(x) = 3x^2 - 3 = 3(x^2 - 1) = 3(x-1)(x+1) \).
2
Critical points: \( f'(x) = 0 \) when \( x = -1 \) and \( x = 1 \). Both lie in \( (-2, 2) \).
3
Evaluate: \( f(-2) = -1 \), \( f(-1) = 3 \), \( f(1) = -1 \), \( f(2) = 3 \).
4
Compare: Absolute max = 3 at \( x = -1 \) and \( x = 2 \). Absolute min = \(-1\) at \( x = -2 \) and \( x = 1 \).
Example 1: Graph
Example 2: \( g(x) = x^{2/3} \) on \( [-1, 8] \)
1
Differentiate: \( g'(x) = \frac{2}{3}x^{-1/3} = \frac{2}{3\sqrt[3]{x}} \).
2
Critical points: \( g'(x) \) is never zero, but \( g'(0) \) does not exist (division by zero). So \( x = 0 \) is the only critical point.
3
Evaluate: \( g(-1) = 1 \), \( g(0) = 0 \), \( g(8) = 4 \).
4
Compare: Absolute max = 4 at \( x = 8 \) (endpoint). Absolute min = 0 at \( x = 0 \) (critical point where \( f' \) DNE).
Example 3: \( h(x) = \sin(x) + \cos(x) \) on \( [0, 2\pi] \)
1
Differentiate: \( h'(x) = \cos(x) - \sin(x) \).
2
Critical points: \( \cos(x) - \sin(x) = 0 \Rightarrow \tan(x) = 1 \Rightarrow x = \frac{\pi}{4}, \frac{5\pi}{4} \).
3
Evaluate: \( h(0) = 1 \), \( h(\pi/4) = \sqrt{2} \approx 1.414 \), \( h(5\pi/4) = -\sqrt{2} \approx -1.414 \), \( h(2\pi) = 1 \).
4
Compare: Absolute max = \( \sqrt{2} \) at \( x = \frac{\pi}{4} \). Absolute min = \( -\sqrt{2} \) at \( x = \frac{5\pi}{4} \).
Common Mistakes to Avoid
✕ Forgetting Endpoints
The absolute extremum can occur at an endpoint, not just at critical points. Always evaluate \( f(a) \) and \( f(b) \).
✕ Assuming \( f'(c)=0 \) Means Extremum
A critical point is only a candidate. For example, \( f(x) = x^3 \) has \( f'(0) = 0 \) but no extremum at \( x = 0 \).
✕ Missing DNE Critical Points
Points where \( f' \) does not exist (cusps, corners) are critical points too. Don't only solve \( f'(x) = 0 \).
✕ Applying EVT on Open Intervals
The EVT requires a closed interval \( [a, b] \). On \( (a, b) \) or \( (-\infty, \infty) \), absolute extrema may not exist.
✕ Ignoring Domain Restrictions
Critical points outside the given interval are irrelevant. Only consider \( c \in (a, b) \).
Summary of Key Concepts
| Concept | Definition | Key Fact |
|---|---|---|
| EVT | Continuous on \( [a,b] \) \(\Rightarrow\) extrema exist | Guarantees existence, not location |
| Critical Point | \( f'(c) = 0 \) or \( f'(c) \) DNE | Only candidates for interior extrema |
| Global Extrema | Largest/smallest on entire domain | Found via Closed Interval Method |
| Local Extrema | Largest/smallest in a neighborhood | Must occur at critical points |
| Fermat's Thm | Local ext + \( f' \) exists \(\Rightarrow\) \( f'(c) = 0 \) | Converse is false |
Question 1 of 5
The Extreme Value Theorem guarantees that a function \( f \) has both an absolute maximum and an absolute minimum on \( [a, b] \) provided that:
The EVT requires continuity on a closed interval \( [a, b] \). Differentiability is a stronger condition that is not necessary, and an open interval does not satisfy the theorem's hypothesis.
Question 2 of 5
Find all critical points of \( f(x) = 3x^4 - 4x^3 \).
We have \( f'(x) = 12x^3 - 12x^2 = 12x^2(x - 1) \). Setting \( f'(x) = 0 \): \( 12x^2(x-1) = 0 \), so \( x = 0 \) or \( x = 1 \). There is no solution at \( x = -1 \).
Question 3 of 5
Find the value of the absolute maximum of \( f(x) = x^3 - 6x^2 + 9x + 2 \) on the interval \( [0, 4] \).
\( f'(x) = 3x^2 - 12x + 9 = 3(x-1)(x-3) \), so critical points are \( x = 1 \) and \( x = 3 \). Evaluating: \( f(0) = 2 \), \( f(1) = 6 \), \( f(3) = 2 \), \( f(4) = 6 \). The absolute maximum value is 6, which occurs at both \( x=1 \) (a critical point) and \( x=4 \) (an endpoint).
Question 4 of 5
If \( f \) is continuous on \( [1, 5] \) and \( f'(3) = 0 \), which statement must be true?
By definition, \( f'(3) = 0 \) makes \( x = 3 \) a critical point. However, a critical point is not necessarily an extremum (local or absolute). Consider \( f(x) = (x-3)^3 \): \( f'(3) = 0 \) but \( x = 3 \) is an inflection point, not an extremum.
Question 5 of 5
Let \( g(x) = |x - 2| \) on \( [0, 5] \). What is the absolute minimum and where does it occur?
\( g(x) = |x - 2| \) has a cusp at \( x = 2 \) where \( g'(2) \) does not exist, making it a critical point. Evaluating: \( g(0) = 2 \), \( g(2) = 0 \), \( g(5) = 3 \). The absolute minimum is 0 at \( x = 2 \): a critical point where the derivative does not exist.