Critical Points and Extrema

Definitions

Critical Point

A point $c$ in the domain of $f$ where:

• $f'(c) = 0$, or
• $f'(c)$ does not exist

Local (Relative) Extrema

• Local maximum: $f(c) \geq f(x)$ for all $x$ near $c$
• Local minimum: $f(c) \leq f(x)$ for all $x$ near $c$

Absolute (Global) Extrema

• Absolute maximum: $f(c) \geq f(x)$ for all $x$ in the domain
• Absolute minimum: $f(c) \leq f(x)$ for all $x$ in the domain

Finding Critical Points

1. Find $f'(x)$
2. Solve $f'(x) = 0$
3. Find where $f'(x)$ is undefined (but $f(x)$ is defined)

Tests for Local Extrema

First Derivative Test

At a critical point $c$:

• If $f'$ changes from $+$ to $-$ at $c$: local maximum
• If $f'$ changes from $-$ to $+$ at $c$: local minimum
• If $f'$ doesn't change sign: neither (inflection point)

Second Derivative Test

At a critical point $c$ where $f'(c) = 0$:

• If $f''(c) > 0$: local minimum (concave up)
• If $f''(c) < 0$: local maximum (concave down)
• If $f''(c) = 0$: inconclusive (use first derivative test)

Finding Absolute Extrema on [a, b]

Closed Interval Method

1. Find all critical points in $(a, b)$
2. Evaluate $f$ at critical points and endpoints
3. Compare values:
   • Largest = absolute maximum
   • Smallest = absolute minimum

Examples

Example 1: Finding critical points

$f(x) = x^3 - 3x^2 - 9x + 5$

$f'(x) = 3x^2 - 6x - 9 = 3(x^2 - 2x - 3) = 3(x - 3)(x + 1)$

$f'(x) = 0 \text{ when } x = 3 \text{ or } x = -1$

Critical points: $x = -1$, $x = 3$

Example 2: First derivative test

For $f(x) = x^3 - 3x^2 - 9x + 5$:

Interval	Test value	$f'(x)$	Conclusion
$x < -1$	$x = -2$	$3(-5)(-1) = 15 > 0$	increasing
$-1 < x < 3$	$x = 0$	$3(-3)(1) = -9 < 0$	decreasing
$x > 3$	$x = 4$	$3(1)(5) = 15 > 0$	increasing

• At $x = -1$: changes $+$ to $-$ → local max
• At $x = 3$: changes $-$ to $+$ → local min

Example 3: Second derivative test

$f(x) = x^3 - 3x^2 - 9x + 5$, $f''(x) = 6x - 6$

$f''(-1) = -12 < 0 \rightarrow \text{local maximum at } x = -1$

$f''(3) = 12 > 0 \rightarrow \text{local minimum at } x = 3$

Example 4: Absolute extrema

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

$f'(x) = 3x^2 - 3 = 3(x^2 - 1) = 0$

$x = \pm 1$

Evaluate at critical points and endpoints:

$x$	$f(x)$
$-2$	$(-8) + 6 = -2$
$-1$	$-1 + 3 = 2$
$1$	$1 - 3 = -2$
$3$	$27 - 9 = 18$

• Absolute maximum: 18 at $x = 3$
• Absolute minimum: $-2$ at $x = -2$ and $x = 1$

Example 5: Critical point where derivative undefined

$f(x) = x^{2/3}$

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

$f'(0)$ is undefined, but $f(0) = 0$ is defined.

Critical point at $x = 0$ (cusp).

Since $f'(x) < 0$ for $x < 0$ and $f'(x) > 0$ for $x > 0$: $x = 0$ is a local minimum.