1. $f'(c) = 0$ (horizontal tangent), OR
2. $f'(c)$ does not exist (sharp corner, vertical tangent, discontinuity)

⚠️ Important: Not all critical points are extrema! But all local extrema (that occur in the interior of the domain) ARE critical points.

Fermat's Theorem

Fermat's Theorem: If $f$ has a local extremum at $x = c$, and $f'(c)$ exists, then $f'(c) = 0$.

What this means:

• Local maxima and minima can only occur where the derivative is zero or undefined
• To find extrema, we must check critical points!

BUT: Not every critical point is an extremum. We need to test them!

Finding Critical Points

Step-by-Step Process

Step 1: Find $f'(x)$

Step 2: Solve $f'(x) = 0$

Step 3: Find where $f'(x)$ is undefined

Step 4: Check that these $x$-values are in the domain of $f$

All these points are critical points!

Example 1: Basic Polynomial

Find all critical points of $f(x) = x^3 - 3x^2 - 9x + 5$

Step 1: Find the derivative

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

Step 2: Set equal to zero

$3x^2 - 6x - 9 = 0$

$3(x^2 - 2x - 3) = 0$

$3(x - 3)(x + 1) = 0$

So $x = 3$ or $x = -1$

Step 3: Check if undefined

$f'(x)$ is a polynomial, so it's defined everywhere ✓

Answer: Critical points at $x = -1$ and $x = 3$

Example 2: Rational Function

Find critical points of $f(x) = \frac{x^2}{x - 2}$

Step 1: Find derivative (Quotient Rule)

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

$= \frac{2x^2 - 4x - x^2}{(x-2)^2}$

$= \frac{x^2 - 4x}{(x-2)^2}$

$= \frac{x(x-4)}{(x-2)^2}$

Step 2: Set equal to zero

$\frac{x(x-4)}{(x-2)^2} = 0$

Numerator = 0: $x(x-4) = 0$

So $x = 0$ or $x = 4$

Step 3: Check if undefined

$f'(x)$ is undefined when $(x-2)^2 = 0$, i.e., when $x = 2$

But $f(2)$ is also undefined (not in domain), so $x = 2$ is NOT a critical point.

Answer: Critical points at $x = 0$ and $x = 4$

The Extreme Value Theorem

Extreme Value Theorem (EVT): 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 This Matters

On a closed interval, extrema can occur at:

1. Critical points in the interior $(a, b)$
2. Endpoints $x = a$ or $x = b$

So to find absolute extrema on $[a, b]$:

• Find all critical points in $(a, b)$
• Evaluate $f$ at critical points AND endpoints
• The largest value is the absolute max
• The smallest value is the absolute min

Finding Absolute Extrema on a Closed Interval

The Closed Interval Method

Step 1: Find all critical points in $(a, b)$

Step 2: Evaluate $f$ at each critical point

Step 3: Evaluate $f$ at the endpoints $a$ and $b$

Step 4: Compare all values:

• Largest value = absolute maximum
• Smallest value = absolute minimum

Example: Closed Interval Method

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

Step 1: 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 2: Evaluate at critical points

$f(-1) = (-1)^3 - 3(-1) + 1 = -1 + 3 + 1 = 3$

$f(1) = 1^3 - 3(1) + 1 = 1 - 3 + 1 = -1$

Step 3: Evaluate at endpoints

$f(-2) = (-2)^3 - 3(-2) + 1 = -8 + 6 + 1 = -1$

$f(2) = 2^3 - 3(2) + 1 = 8 - 6 + 1 = 3$

Step 4: Compare

Values: $f(-2) = -1$, $f(-1) = 3$, $f(1) = -1$, $f(2) = 3$

Answer:

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

Common Types of Critical Points

Type 1: Horizontal Tangent ($f'(c) = 0$)

The most common type - the derivative equals zero.

Examples:

• Tops of hills (local max)
• Bottoms of valleys (local min)
• Inflection points with horizontal tangent

Type 2: Sharp Corner

The function is continuous but not differentiable.

Example: $f(x) = |x|$ at $x = 0$

• $f$ is continuous at $x = 0$
• $f'(0)$ does not exist (sharp turn)
• $x = 0$ is a critical point (absolute minimum!)

Type 3: Vertical Tangent

Example: $f(x) = x^{1/3}$ at $x = 0$

• $f'(x) = \frac{1}{3}x^{-2/3} = \frac{1}{3x^{2/3}}$
• $f'(0)$ is undefined (vertical tangent)
• $x = 0$ is a critical point (inflection point)

Type 4: Discontinuity

If $f$ is discontinuous at $x = c$, it's NOT a critical point (must be in domain).

⚠️ Common Mistakes

Mistake 1: Assuming All Critical Points Are Extrema

❌ Critical points might be inflection points or saddle points ✅ Always verify using a test (First Derivative Test, Second Derivative Test)

Mistake 2: Forgetting Endpoints

When finding absolute extrema on $[a, b]$, always check the endpoints!

Mistake 3: Domain Errors

$x = c$ is only a critical point if it's in the domain of $f$.

Mistake 4: Undefined vs. Zero

$f'(c) = 0$ and $f'(c)$ undefined are both critical points, but they're different!

Quick Reference

Critical Point Checklist

• Find $f'(x)$
• Solve $f'(x) = 0$
• Find where $f'(x)$ is undefined
• Verify points are in domain of $f$

Absolute Extrema Checklist (on $[a, b]$)

• Find all critical points in $(a, b)$
• Evaluate $f$ at critical points
• Evaluate $f$ at endpoints $a$ and $b$
• Largest value = absolute max
• Smallest value = absolute min

📝 Practice Tips

1. Always find the derivative first before looking for critical points
2. Set derivative equal to zero and solve carefully
3. Check for undefined points in the derivative
4. Verify domain - make sure critical points are where $f$ exists
5. For closed intervals, don't forget the endpoints!
6. To classify critical points, use First or Second Derivative Test (next lessons!)

Find the derivative:

$f'(x) = 3x^2 - 12x + 9$

Set equal to zero:

$3x^2 - 12x + 9 = 0$

$3(x^2 - 4x + 3) = 0$

$3(x - 1)(x - 3) = 0$

$x = 1$ or $x = 3$

Find corresponding $y$-values:

$f(1) = 1 - 6 + 9 + 1 = 5$ $f(3) = 27 - 54 + 27 + 1 = 1$

Critical points: $(1, 5)$ and $(3, 1)$

Step 2: Find critical points

$6x^2 + 6x - 12 = 0$

$6(x^2 + x - 2) = 0$

$6(x + 2)(x - 1) = 0$

So $x = -2$ or $x = 1$

Both are in $[-3, 2]$ ✓

Step 3: Evaluate at critical points

$g(-2) = 2(-2)^3 + 3(-2)^2 - 12(-2) + 5$ $= 2(-8) + 3(4) + 24 + 5$ $= -16 + 12 + 24 + 5 = 25$

$g(1) = 2(1)^3 + 3(1)^2 - 12(1) + 5$ $= 2 + 3 - 12 + 5 = -2$

Step 4: Evaluate at endpoints

$g(-3) = 2(-3)^3 + 3(-3)^2 - 12(-3) + 5$ $= 2(-27) + 3(9) + 36 + 5$ $= -54 + 27 + 36 + 5 = 14$

$g(2) = 2(2)^3 + 3(2)^2 - 12(2) + 5$ $= 2(8) + 3(4) - 24 + 5$ $= 16 + 12 - 24 + 5 = 9$

Step 5: Compare all values

$g(-3) = 14$, $g(-2) = 25$, $g(1) = -2$, $g(2) = 9$

Answer:

• Absolute maximum: $25$ at $x = -2$
• Absolute minimum: $-2$ at $x = 1$

Critical points: $x = 3$ and $x = -1$ (both in the interval)

Step 2: Evaluate $f$ at critical points and endpoints.

$f(-2) = (-2)^3 - 3(-2)^2 - 9(-2) + 5 = -8 - 12 + 18 + 5 = 3$

$f(-1) = (-1)^3 - 3(-1)^2 - 9(-1) + 5 = -1 - 3 + 9 + 5 = 10$

$f(3) = (3)^3 - 3(3)^2 - 9(3) + 5 = 27 - 27 - 27 + 5 = -22$

$f(4) = (4)^3 - 3(4)^2 - 9(4) + 5 = 64 - 48 - 36 + 5 = -15$

Step 3: Compare values.

Absolute maximum: $f(-1) = 10$ Absolute minimum: $f(3) = -22$

Step 2: Find the derivative

$h'(x) = \frac{5}{3}x^{2/3} - 5 \cdot \frac{2}{3}x^{-1/3}$

$h'(x) = \frac{5}{3}x^{2/3} - \frac{10}{3}x^{-1/3}$

Step 3: Factor using common denominator

$h'(x) = \frac{5x^{2/3}}{3} - \frac{10}{3x^{1/3}}$

$h'(x) = \frac{5x^{2/3} \cdot x^{1/3} - 10}{3x^{1/3}}$

$h'(x) = \frac{5x - 10}{3x^{1/3}}$

$h'(x) = \frac{5(x - 2)}{3x^{1/3}}$

Step 4: Find where $h'(x) = 0$

Numerator = 0: $5(x - 2) = 0$

So $x = 2$ ✓

Step 5: Find where $h'(x)$ is undefined

Denominator = 0: $3x^{1/3} = 0$

So $x = 0$ ✓

$h(0) = 0^{2/3}(0 - 5) = 0$ (in domain) ✓

Answer: Critical points at $x = 0$ and $x = 2$

Note: At $x = 0$, there's a vertical tangent (cusp).