AP Calculus BC 5.2 Extreme Value Theorem Study Notes - New Syllabus
AP Calculus BC 5.2 Extreme Value Theorem Study Notes- New syllabus
AP Calculus BC 5.2 Extreme Value Theorem Study Notes – AP Calculus BC- per latest AP Calculus BC Syllabus.
LEARNING OBJECTIVE
- A function’s derivative can be used to understand some behaviors of the function.
Key Concepts:
- Extreme Value Theorem, Global Versus Local Extrema, and Critical Points
Extreme Value Theorem, Global Versus Local Extrema, and Critical Points
Extreme Value Theorem (EVT):
If a function \( f \) is continuous on a closed interval \([a, b]\), then \( f \) must attain both a global maximum and a global minimum value on that interval. That is, there exist points \( c \) and \( d \) in \([a, b]\) such that:
\( f(c) \geq f(x) \text{ and } f(d) \leq f(x) \text{ for all } x \in [a, b] \)
Important: The Extreme Value Theorem does not say where the extrema occur only that they must exist if the function is continuous on the closed interval.
Local vs. Global Extrema:
- Local (Relative) Maximum: A point \( x = c \) where \( f(c) \geq f(x) \) for all \( x \) near \( c \).
- Local (Relative) Minimum: A point \( x = c \) where \( f(c) \leq f(x) \) for all \( x \) near \( c \).
- Global (Absolute) Maximum: A point where the function reaches its highest value on the entire domain.
- Global (Absolute) Minimum: A point where the function reaches its lowest value on the entire domain.
Consider the curve on the figure below:
On the closed interval $[a,e]$ we see that the function has a local maximum at $x = b$, a local minimum at $x = c$, a global maximum at $x = d$, and a global minimum at $x = a$.
Consider another curve below:
On the closed interval $[a,b]$ we see that the function has a global maximum at $x = b$, and a global minimum at $x = c$. Even though the function is not continuous at $x = c$, we still have a minimum value at that location.
Critical Points:
A critical point of a function \( f \) is a point \( c \) in the domain of \( f \) such that either:
- \( f'(c) = 0 \) (horizontal tangent)
- or \( f'(c) \) does not exist
Critical points are candidates for local maxima, local minima, or points of inflection. They are not guaranteed to be extrema.
Example:
Consider the function \( f(x) = x^3 – 3x^2 + 1 \) on the interval \([0, 3]\). Find the critical points and determine its type.
▶️Answer/Explanation
Consider the function \( f(x) = x^3 – 3x^2 + 1 \) on the interval \([0, 3]\).
Step 1: Find the derivative:
\( f'(x) = 3x^2 – 6x \)
Step 2: Set derivative equal to 0 to find critical points:
\( 3x^2 – 6x = 0 \Rightarrow x(x – 2) = 0 \Rightarrow x = 0, 2 \)
Step 3: Evaluate \( f \) at endpoints and critical points:
- \( f(0) = 0^3 – 3(0)^2 + 1 = 1 \)
- \( f(2) = 8 – 12 + 1 = -3 \)
- \( f(3) = 27 – 27 + 1 = 1 \)
So the global minimum is \( -3 \) at \( x = 2 \), and the global maximum is \( 1 \), which occurs at both \( x = 0 \) and \( x = 3 \).
This example illustrates the EVT: the function is continuous on \([0, 3]\), and it attains both a max and min value. The critical point at \( x = 2 \) yields a local and global minimum. The endpoints \( x = 0 \) and \( x = 3 \) are where the global maximum occurs. Even though \( x = 0 \) is also a critical point, it doesn’t give a minimum.
Example :
Let \( f(x) = \frac{1}{x} \) on the interval \((0, 2]\). Does the Extreme Value Theorem guarantee a global maximum or minimum on this interval?
▶️Answer/Explanation
No. The Extreme Value Theorem requires that the function be continuous on a closed interval. Here, the interval is not closed on the left (\( x = 0 \) is not included), so EVT does not apply.
Additionally, as \( x \to 0^+ \), \( f(x) \to \infty \), so there is no global maximum. The global minimum occurs at \( x = 2 \):
\( f(2) = \frac{1}{2} \), which is the minimum on this interval.
Example :
Let \( f(x) = x^4 – 4x^2 \). Find the critical points and determine if they are local extrema.
▶️Answer/Explanation
First, take the derivative:
\( f'(x) = 4x^3 – 8x \)
Set derivative to 0:
\( 4x(x^2 – 2) = 0 \Rightarrow x = 0, \pm\sqrt{2} \)
Use the second derivative test:
\( f”(x) = 12x^2 – 8 \)
- At \( x = 0 \): \( f”(0) = -8 \) → local maximum
- At \( x = \pm\sqrt{2} \): \( f”(\pm\sqrt{2}) = 12(2) – 8 = 16 > 0 \) → local minima
So the function has:
- a local max at \( x = 0 \)
- local minima at \( x = \pm\sqrt{2} \)
Example :
Let \( f(x) = \sqrt{x(4 – x)} \) on the interval \([0, 4]\). Find all critical points and determine the global maximum and minimum.
▶️Answer/Explanation
Let’s find the critical points:
Let \( f(x) = \sqrt{x(4 – x)} = \sqrt{4x – x^2} \)
Use the chain rule:
\( f'(x) = \frac{1}{2\sqrt{4x – x^2}} \cdot (4 – 2x) \)
Set \( f'(x) = 0 \):
\( 4 – 2x = 0 \Rightarrow x = 2 \)
Evaluate function at endpoints and critical point:
- \( f(0) = \sqrt{0} = 0 \)
- \( f(4) = \sqrt{0} = 0 \)
- \( f(2) = \sqrt{2(4 – 2)} = \sqrt{4} = 2 \)
So, global maximum is 2 at \( x = 2 \), global minimum is 0 at both endpoints.