Extreme Value Theorem,
Extrema & Critical Points
Master the Extreme Value Theorem, distinguish global vs. local extrema, and learn how critical points unlock the behavior of any continuous function.
AP Calculus AB
What Are Extrema?
In calculus, extrema are the largest and smallest values a function attains. They come in two flavors: maximum values (peaks) and minimum values (valleys). Extrema tell us about the most extreme behavior of a function on a given domain.
Global vs. Local Extrema
Global (Absolute)
The single largest or smallest value of \( f(x) \) on the entire domain. There can be at most one global max value and one global min value (though they may occur at multiple points).
Local (Relative)
A value \( f(c) \) that is larger or smaller than all nearby values. A function can have many local maxima and minima. Every global extremum is also a local extremum.
Visualizing Extrema on a Graph
The Extreme Value Theorem
This theorem guarantees that extrema exist. It does not tell us where they are or how to find them: that is the job of critical points and endpoint evaluation.
Continuous
No jumps, holes, or asymptotes on \([a,b]\).
Closed Interval
Both endpoints \(a\) and \(b\) are included: \([a,b]\), not \((a,b)\).
Why Both Conditions Matter
Not Continuous?
Consider a function with a jump discontinuity on \([0,2]\). The function may miss achieving its "would-be" maximum.
Not Closed?
Consider \( f(x) = x \) on \((0,1)\). The function approaches 1 but never reaches it: no absolute maximum exists.
What Is a Critical Point?
\( f'(c) = 0 \)
The tangent line is horizontal. The function has a flat spot: a potential peak or valley.
\( f'(c) \) DNE
A cusp, corner, or vertical tangent. The slope is undefined, but extrema can still occur here.
Critical Points on a Graph
Critical Point vs. Extremum
| Critical Point | Extremum | |
|---|---|---|
| Definition | \( f'(c) = 0 \) or \( f'(c) \) DNE | Largest/smallest value in a region |
| Always an extremum? | No: e.g., \( f(x)=x^3 \) at \( x=0 \) | Yes (by definition) |
| Can occur at endpoints? | Typically interior points only | Yes: endpoints can be global extrema |
| Role in the Closed Interval Method | Candidate locations to check | The final answer after comparison |
Fermat's Theorem
Fermat's Theorem tells us where to look. If a local max or min happens inside the interval, the derivative must be zero there (or must not exist). This is why we hunt for critical points: they are the only places interior extrema can hide.
The Closed Interval Method
1
Verify continuity on \([a,b]\). The EVT must apply.
2
Find all critical points in \((a,b)\): solve \( f'(x) = 0 \) and identify where \( f'(x) \) does not exist.
3
Evaluate \( f \) at each critical point and at both endpoints \( a \) and \( b \).
4
Compare 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
Continuous? Yes: polynomial, continuous everywhere.
2
Critical points: \( f'(x) = 3x^2 - 3 = 3(x-1)(x+1) = 0 \). So \( x = -1 \) and \( x = 1 \).
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
Continuous? Yes: \( x^{2/3} \) is continuous on \([-1, 8]\).
2
Derivative: \( g'(x) = \frac{2}{3}x^{-1/3} = \frac{2}{3\sqrt[3]{x}} \). This is never zero, but \( g'(0) \) does not exist. So \( x = 0 \) is a critical point.
3
Evaluate: \( g(-1) = 1 \), \( g(0) = 0 \), \( g(8) = 4 \).
4
Compare: Absolute max = 4 at \( x=8 \). Absolute min = 0 at \( x=0 \) (the cusp).
Example 3: \( h(x) = \sin(x) + \cos(x) \) on \([0, 2\pi]\)
1
Continuous? Yes: trigonometric functions are continuous everywhere.
2
Derivative: \( h'(x) = \cos(x) - \sin(x) = 0 \) when \( \sin(x) = \cos(x) \), i.e., \( x = \frac{\pi}{4} \) and \( x = \frac{5\pi}{4} \).
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: Abs. max = \( \sqrt{2} \) at \( x = \pi/4 \). Abs. min = \( -\sqrt{2} \) at \( x = 5\pi/4 \).
Common Mistakes
✕ Forgetting Endpoints
Absolute extrema can occur at endpoints, not just at critical points. Always evaluate \( f(a) \) and \( f(b) \).
✕ Only Checking \( f'(x)=0 \)
Critical points also include where \( f'(x) \) does not exist (cusps, corners, vertical tangents). Don't skip these.
✕ Assuming All Critical Points Are Extrema
A critical point can be an inflection point (e.g., \( f(x) = x^3 \) at \( x=0 \)). Always compare values to confirm.
Summary Table
| Concept | Key Idea | Remember |
|---|---|---|
| EVT | Continuous + closed interval → extrema guaranteed | Both conditions required |
| Global Extrema | Largest/smallest on entire domain | Check endpoints + critical pts |
| Local Extrema | Largest/smallest in a neighborhood | Occur at critical points (interior) |
| Critical Points | \( f'(c)=0 \) or \( f'(c) \) DNE | Candidates, not guarantees |
| Fermat's Theorem | Interior extremum → \( f'(c)=0 \) | Converse is false |
Question 1 of 5
The Extreme Value Theorem guarantees that a function has both an absolute maximum and an absolute minimum on an interval provided that:
The EVT requires two conditions: the function must be continuous and the interval must be closed \([a,b]\). Differentiability is not required, and an open interval does not guarantee extrema exist.
Question 2 of 5
Find all critical points of \( f(x) = 2x^3 - 9x^2 + 12x \).
\( f'(x) = 6x^2 - 18x + 12 = 6(x^2 - 3x + 2) = 6(x-1)(x-2) \). Setting \( f'(x) = 0 \) gives \( x = 1 \) and \( x = 2 \).
Question 3 of 5
What is the absolute maximum value of \( f(x) = x^3 - 3x \) on \([-2, 3]\)?
\( f'(x) = 3x^2 - 3 = 0 \Rightarrow x = \pm 1 \). Evaluate: \( f(-2) = -2 \), \( f(-1) = 2 \), \( f(1) = -2 \), \( f(3) = 18 \). The largest value is 18 at the endpoint \( x = 3 \).
Question 4 of 5
If \( f(x) = |x - 2| \) on \([0, 5]\), which statement about critical points is true?
The absolute value function \( |x-2| \) has a corner at \( x=2 \) where the derivative does not exist. Since \( x = 2 \) is in the domain and \( f'(2) \) DNE, it is a critical point. Endpoints are not typically classified as critical points.
Question 5 of 5
The function \( f(x) = x^3 \) has \( f'(0) = 0 \). What type of point is \( x = 0 \)?
Since \( f'(0) = 0 \), the point \( x = 0 \) is indeed a critical point. However, \( f(x) = x^3 \) is increasing on both sides of 0 (it changes concavity, not direction). So \( x = 0 \) is an inflection point, not an extremum.