• Determine whether the Extreme Value Theorem applies to a given function

The Extreme Value Theorem

When analyzing functions, we often need to find the largest and smallest values. In multivariable calculus, this becomes crucial for optimization problems involving surfaces and regions in space.

absolute extrema definitions

Let [latex]f[/latex] be a function defined over an interval [latex]I[/latex] and let [latex]c \in I[/latex].

• [latex]f[/latex] has an absolute maximum on [latex]I[/latex] at [latex]c[/latex] if [latex]f(c) \geq f(x)[/latex] for all [latex]x \in I[/latex]
• [latex]f[/latex] has an absolute minimum on [latex]I[/latex] at [latex]c[/latex] if [latex]f(c) \leq f(x)[/latex] for all [latex]x \in I[/latex]

Important: The extremum is the [latex]y[/latex]-value [latex]f(c)[/latex], while the location is the [latex]x[/latex]-value [latex]c[/latex].

Not all functions have absolute extrema. Consider [latex]f(x) = x^3[/latex] over [latex](-\infty, \infty)[/latex]—as [latex]x[/latex] increases or decreases without bound, so does [latex]f(x)[/latex]. This function has neither an absolute maximum nor an absolute minimum.

However, when certain conditions are met, we can guarantee that both an absolute maximum and absolute minimum exist. The extreme value theorem guarantees that a continuous function [latex]f[/latex] over a closed, bounded interval [latex][a,b][/latex] has both an absolute maximum and an absolute minimum.

extreme value theorem

If [latex]f[/latex] is continuous over a closed, bounded interval [latex][a,b][/latex], then [latex]f[/latex] has both:

• An absolute maximum at some point in [latex][a,b][/latex]
• An absolute minimum at some point in [latex][a,b][/latex]

This theorem provides a powerful guarantee, but only when both conditions are satisfied:

1. Continuity: The function [latex]f[/latex] must be continuous over the entire interval [latex][a,b][/latex]
2. Closed, bounded interval: The interval must include both endpoints and have finite length

If either condition fails, the theorem does not apply, and the function may or may not have absolute extrema.

Why Both Conditions Matter

• Continuity is essential. A single discontinuity can prevent absolute extrema from existing. Consider a piecewise function on [latex][0,4][/latex] that jumps at [latex]x = 2[/latex]. Even though the interval is closed and bounded, the discontinuity means the Extreme Value Theorem doesn’t apply.
• The interval must be closed and bounded. Open intervals like [latex](a,b)[/latex] or half-open intervals like [latex][a,b)[/latex] don’t guarantee extrema exist. Similarly, unbounded intervals like [latex][0,\infty)[/latex] fail this requirement.

Quick Check for the Extreme Value Theorem:

Ask yourself two questions:

1. Is [latex]f[/latex] continuous on the entire interval [latex][a,b][/latex]?
2. Does the interval include both endpoints (closed) and have finite length (bounded)?

If yes to both → EVT applies → absolute max and min are guaranteed to exist

If no to either → EVT does not apply → extrema may or may not exist

Determine whether the Extreme Value Theorem applies to each function. Explain your reasoning.

1. [latex]f(x) = (x-2)^2[/latex] on [latex][1, 4][/latex]

2. [latex]f(x) = \begin{cases} 2-x^2 & \text{if } 0 \leq x < 2 \ x-3 & \text{if } 2 \leq x \leq 4 \end{cases}[/latex] on [latex][0, 4][/latex]

3. [latex]f(x) = \frac{x}{2-x}[/latex] on [latex][0, 2)[/latex]

Show Answer

1. The Extreme Value Theorem applies.
   
   • [latex]f(x) = (x-2)^2[/latex] is a polynomial, so it’s continuous everywhere, including on [latex][1,4][/latex]
   • The interval [latex][1,4][/latex] is closed (includes both endpoints) and bounded (finite length)
   • Both conditions are satisfied, so [latex]f[/latex] is guaranteed to have an absolute maximum and absolute minimum on [latex][1,4][/latex]
2. The Extreme Value Theorem does not apply.
   
   • The interval [latex][0,4][/latex] is closed and bounded
   • However, at [latex]x = 2[/latex]: [latex]\lim_{x \to 2^-} f(x) = 2-2^2 = -2[/latex] but [latex]f(2) = 2-3 = -1[/latex]
   • The function is discontinuous at [latex]x = 2[/latex], so the continuity condition fails
   • Without EVT, we cannot guarantee both extrema exist (in this case, the function has an absolute maximum but no absolute minimum)
3. The Extreme Value Theorem does not apply.
   
   • [latex]f(x) = \frac{x}{2-x}[/latex] is continuous on [latex][0,2)[/latex] (the denominator is never zero on this interval)
   • However, [latex][0,2)[/latex] is not a closed interval—it doesn’t include the right endpoint [latex]x = 2[/latex]
   • The interval condition fails, so EVT doesn’t apply
   • Without EVT, we cannot guarantee both extrema exist (in this case, the function has an absolute minimum but no absolute maximum)

Common Reasons EVT Fails:

• Discontinuity: Jumps, vertical asymptotes, removable discontinuities, or piecewise functions with mismatched values at transition points
• Open interval: Missing one or both endpoints, like [latex](a,b)[/latex], [latex](a,b][/latex], or [latex][a,b)[/latex]
• Unbounded interval: Intervals extending to infinity, like [latex][a, \infty)[/latex] or [latex](-\infty, b][/latex]

Always verify both continuity and the interval type before concluding EVT applies.