$f(c)$ 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.

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)

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.

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

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 ✓

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 $(-2, 2)$ ✓

Step 3: Evaluate at critical points and endpoints

Step 4: Identify extrema

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

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

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!

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 ✓

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 $0 \in (-1, 8)$) ✓

Step 3: Evaluate function

Step 4: Compare

Largest: $f(8) = 4$

Smallest: $f(0) = 0$

Answer:

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

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 ✓

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 $0 < x < \frac{\pi}{2}$)

Step 3: Evaluate at endpoints only

Step 4: Compare

Largest: $f(0) = 1$

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

Answer:

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

Both extrema occur at endpoints!

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

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 $\infty$)

⚠️ 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!

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

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

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

📝 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?

Step 2: Find critical points

Use product rule: $g'(x) = (1)\sqrt{4-x^2} + x \cdot \frac{-2x}{2\sqrt{4-x^2}}$

$= \sqrt{4-x^2} - \frac{x^2}{\sqrt{4-x^2}}$

$= \frac{4-x^2 - x^2}{\sqrt{4-x^2}} = \frac{4-2x^2}{\sqrt{4-x^2}}$

$g'(x) = 0$ when numerator = 0: $4-2x^2 = 0 \implies x^2 = 2 \implies x = \pm\sqrt{2}$

$g'(x)$ undefined when denominator = 0: $4-x^2 = 0 \implies x = \pm 2$

But $x = \pm 2$ are endpoints, not in $(- 2, 2)$

Critical points in $(-2, 2)$: $x = \sqrt{2}$ and $x = -\sqrt{2}$

Step 3: Evaluate function

$g(-2) = (-2)\sqrt{4-4} = (-2)(0) = 0$

$g(-\sqrt{2}) = (-\sqrt{2})\sqrt{4-2} = (-\sqrt{2})(\sqrt{2}) = -2$

$g(\sqrt{2}) = (\sqrt{2})\sqrt{4-2} = (\sqrt{2})(\sqrt{2}) = 2$

$g(2) = (2)\sqrt{4-4} = (2)(0) = 0$

Step 4: Compare values

Answer:

• Absolute maximum: $g(\sqrt{2}) = 2$ at $x = \sqrt{2}$
• Absolute minimum: $g(-\sqrt{2}) = -2$ at $x = -\sqrt{2}$

Step 2: Interpret $h'(x) < 0$

Since $h'(x) < 0$ everywhere on $(0, 4)$:

• The function is strictly decreasing throughout the interval
• There are no critical points in $(0, 4)$ (because $f'(x) \neq 0$ anywhere)

Step 3: Implications for extrema

For a strictly decreasing function:

• The highest point must be at the left endpoint
• The lowest point must be at the right endpoint

Step 4: Identify extrema

Since $h$ is decreasing from left to right:

At $x = 0$ (left endpoint): $h(0) = 3$ is the highest value

At $x = 4$ (right endpoint): $h(4) = 1$ is the lowest value

Answer:

• Absolute maximum: $h(0) = 3$ at $x = 0$
• Absolute minimum: $h(4) = 1$ at $x = 4$

Key insight: When a function is strictly monotonic (always increasing or always decreasing) with no critical points, the absolute extrema MUST occur at the endpoints!