🏔️ Absolute Extrema on Closed Intervals

What Are Absolute Extrema?

Absolute extrema are the highest and lowest values of a function over an entire interval.

💡 Key Idea: On a closed interval, a continuous function MUST have an absolute maximum and minimum - and they occur either at critical points or endpoints!

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Definitions

Absolute Maximum

$f(c)$ is an absolute maximum on $[a, b]$ if: $f(c) \geq f(x) \text{ for all } x \in [a, b]$

$f(c)$ is the highest value on the entire interval.

Absolute Minimum

$f(c)$ is an absolute minimum on $[a, b]$ if: $f(c) \leq f(x) \text{ for all } x \in [a, b]$

$f(c)$ is the lowest value on the entire interval.

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Extreme Value Theorem

Statement: If $f$ is continuous on a closed interval $[a, b]$, then $f$ has both an absolute maximum and an absolute minimum on $[a, b]$.

Why It Matters

This theorem guarantees that absolute extrema exist!

Key requirements:

1. Function must be continuous
2. Interval must be closed (includes endpoints)

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Where Do Absolute Extrema Occur?

Absolute extrema can ONLY occur at:

1. Critical points in $(a, b)$ (where $f'(x) = 0$ or $f'(x)$ undefined)
2. Endpoints $x = a$ or $x = b$

That's it! These are the only candidates.

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The Closed Interval Method

This is your foolproof strategy for finding absolute extrema:

Step-by-Step Process

Step 1: Verify that $f$ is continuous on $[a, b]$

Step 2: Find all critical points in the open interval $(a, b)$

• Solve $f'(x) = 0$
• Find where $f'(x)$ is undefined

Step 3: Evaluate $f$ at:

• Each critical point
• Both endpoints $a$ and $b$

Step 4: Compare all these values:

• Largest value → absolute maximum
• Smallest value → absolute minimum

Step 5: State your answer with both $x$-value and $y$-value

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Example 1: Basic Application

Find the absolute extrema of $f(x) = x^3 - 3x + 1$ on $[-2, 2]$.

Step 1: Verify continuity

$f(x)$ is a polynomial → continuous everywhere ✓

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Step 2: Find critical points

$f'(x) = 3x^2 - 3 = 3(x^2 - 1) = 3(x-1)(x+1)$

$f'(x) = 0$ when $x = 1$ or $x = -1$

Both are in $(-2, 2)$ ✓

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Step 3: Evaluate at critical points and endpoints

| $x$ | $f(x) = x^3 - 3x + 1$ | Type | |-----|---------------------|------| | $-2$ | $(-2)^3 - 3(-2) + 1 = -8 + 6 + 1 = -1$ | Endpoint | | $-1$ | $(-1)^3 - 3(-1) + 1 = -1 + 3 + 1 = 3$ | Critical pt | | $1$ | $(1)^3 - 3(1) + 1 = 1 - 3 + 1 = -1$ | Critical pt | | $2$ | $(2)^3 - 3(2) + 1 = 8 - 6 + 1 = 3$ | Endpoint |

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Step 4: Identify extrema

Largest value: $f(-1) = 3$ and $f(2) = 3$

Smallest value: $f(-2) = -1$ and $f(1) = -1$

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Answer:

• Absolute maximum: $f(-1) = 3$ and $f(2) = 3$
• Absolute minimum: $f(-2) = -1$ and $f(1) = -1$

Note: It's possible to have absolute extrema at multiple points!

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Example 2: With Undefined Derivative

Find the absolute extrema of $f(x) = x^{2/3}$ on $[-1, 8]$.

Step 1: Continuity

$f(x) = x^{2/3}$ is continuous everywhere ✓

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Step 2: Find critical points

$f'(x) = \frac{2}{3}x^{-1/3} = \frac{2}{3x^{1/3}}$

$f'(x) = 0$: No solution (numerator never zero)

$f'(x)$ undefined: When $x = 0$ (denominator zero)

Critical point: $x = 0$ (and $0 \in (-1, 8)$) ✓

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Step 3: Evaluate function

| $x$ | $f(x) = x^{2/3}$ | |-----|------------------| | $-1$ | $(-1)^{2/3} = 1$ | | $0$ | $(0)^{2/3} = 0$ | | $8$ | $(8)^{2/3} = (\sqrt[3]{8})^2 = 2^2 = 4$ |

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Step 4: Compare

Largest: $f(8) = 4$

Smallest: $f(0) = 0$

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Answer:

• Absolute maximum: $f(8) = 4$
• Absolute minimum: $f(0) = 0$

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Example 3: Extrema at Endpoints

Find the absolute extrema of $f(x) = \cos x$ on $[0, \frac{\pi}{2}]$.

Step 1: Continuity

$\cos x$ is continuous everywhere ✓

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Step 2: Critical points

$f'(x) = -\sin x$

$f'(x) = 0$ when $\sin x = 0$

On $(0, \frac{\pi}{2})$: No solutions (since $\sin x > 0$ for $0 < x < \frac{\pi}{2}$)

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Step 3: Evaluate at endpoints only

| $x$ | $f(x) = \cos x$ | |-----|------------------| | $0$ | $\cos 0 = 1$ | | $\frac{\pi}{2}$ | $\cos(\frac{\pi}{2}) = 0$ |

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Step 4: Compare

Largest: $f(0) = 1$

Smallest: $f(\frac{\pi}{2}) = 0$

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Answer:

• Absolute maximum: $f(0) = 1$
• Absolute minimum: $f(\frac{\pi}{2}) = 0$

Both extrema occur at endpoints!

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Local vs. Absolute Extrema

Local (Relative) Extrema

• Local maximum: Highest in some neighborhood
• Can occur at any critical point
• There can be multiple local maxima

Absolute (Global) Extrema

• Absolute maximum: Highest on entire interval
• Only one value (but can occur at multiple points)
• Must be at critical point or endpoint

Relationship

• Every absolute extremum is also a local extremum
• NOT every local extremum is an absolute extremum

Example: $f(x) = x^3 - 3x + 1$ on $[-2, 2]$ has:

• Local max at $x = -1$, which is also absolute max
• Local min at $x = 1$, which is also absolute min

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What If the Interval Isn't Closed?

Open Interval $(a, b)$

• No guarantee of absolute extrema
• May or may not exist

Example: $f(x) = x$ on $(0, 1)$

• No absolute max (approaches 1 but never reaches it)
• No absolute min (approaches 0 but never reaches it)

Infinite Interval $(-\infty, \infty)$

• No guarantee of extrema
• Often no absolute extrema

Example: $f(x) = x^2$ on $(-\infty, \infty)$

• Absolute min at $x = 0$: $f(0) = 0$
• No absolute max (grows to $\infty$)

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⚠️ Common Mistakes

Mistake 1: Forgetting Endpoints

Always check the endpoints! They're often where absolute extrema occur.

Mistake 2: Not Checking if Critical Points Are in the Interval

If you find $f'(x) = 0$ at $x = 5$, but the interval is $[0, 3]$, don't include $x = 5$!

Mistake 3: Only Listing $x$-values

Give BOTH the location ($x$-value) AND the value ($f(x)$)!

WRONG: "Absolute max at $x = 2$"

RIGHT: "Absolute maximum is $f(2) = 7$ at $x = 2$"

Mistake 4: Confusing Local and Absolute

A local maximum might NOT be the absolute maximum!

Mistake 5: Not Verifying Continuity

If $f$ is not continuous on $[a, b]$, the Extreme Value Theorem doesn't apply!

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Special Cases

Case 1: Constant Function

If $f(x) = c$ (constant), then:

• Every point is both an absolute max and min
• Value is $c$ everywhere

Case 2: Linear Function

$f(x) = mx + b$ on $[a, b]$:

• Absolute extrema always at endpoints
• If $m > 0$: min at $a$, max at $b$
• If $m < 0$: max at $a$, min at $b$

Case 3: No Critical Points

If $f'(x) \neq 0$ and $f'$ always defined on $(a, b)$:

• Function is strictly monotonic (always increasing or decreasing)
• Absolute extrema MUST be at endpoints

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Quick Decision Tree

Is $f$ continuous on $[a, b]$?

• NO → Extreme Value Theorem doesn't apply
• YES → Continue

Find critical points in $(a, b)$:

1. Solve $f'(x) = 0$
2. Find where $f'(x)$ undefined

Evaluate $f$ at:

• All critical points
• Both endpoints

Compare values:

• Largest → Absolute max
• Smallest → Absolute min

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Real-World Applications

Optimization Problems

• Maximizing profit over a time period
• Minimizing cost over a production range
• Finding best dimensions within constraints

Physics

• Maximum height of projectile
• Minimum potential energy
• Optimal angle for range

Economics

• Maximum revenue over demand range
• Minimum average cost in production interval

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📝 Practice Strategy

1. Check continuity first - is Extreme Value Theorem applicable?
2. Find all critical points - solve $f'(x) = 0$ and check where $f'$ undefined
3. Make a table with columns: $x$, $f(x)$, Type
4. Evaluate systematically - critical points AND endpoints
5. Identify largest and smallest values
6. Write complete answers - include both $x$ and $f(x)$
7. Double-check - did you evaluate at ALL candidates?