💡 Key Idea: If the graph is curving upward at a critical point, it's a minimum. If curving downward, it's a maximum!

The Test (Formal Statement)

Let $c$ be a critical point where $f'(c) = 0$.

Evaluate the second derivative at $c$:

Case 1: Local Minimum

If $f''(c) > 0$ (concave up):

• Then $f(c)$ is a local minimum ∪

Case 2: Local Maximum

If $f''(c) < 0$ (concave down):

• Then $f(c)$ is a local maximum ∩

Case 3: Inconclusive

If $f''(c) = 0$:

• The test fails - use the First Derivative Test instead
• Could be a max, min, or neither

Why It Works

Think about concavity:

Concave Up ($f'' > 0$): Graph curves like ∪

• If you also have $f'(c) = 0$ (horizontal tangent)
• The point must be at the bottom → minimum

Concave Down ($f'' < 0$): Graph curves like ∩

• If you also have $f'(c) = 0$ (horizontal tangent)
• The point must be at the top → maximum

💡 Memory Trick: "Concave up = cup = holds minimum" and "Concave down = frown = maximum"

Step-by-Step Process

Step 1: Find $f'(x)$

Step 2: Find critical points by solving $f'(x) = 0$

Step 3: Find $f''(x)$

Step 4: Evaluate $f''(c)$ at each critical point $c$

Step 5: Apply the test:

• $f''(c) > 0$ → local min
• $f''(c) < 0$ → local max
• $f''(c) = 0$ → use First Derivative Test

Step 6: Calculate $f(c)$ to get the actual min/max value

Example 1: Basic Application

Classify the critical points of $f(x) = x^3 - 6x^2 + 9x + 1$

Step 1: Find first derivative

$f'(x) = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x-1)(x-3)$

Step 2: Find critical points

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

Critical points: $x = 1$ and $x = 3$

Step 3: Find second derivative

$f''(x) = 6x - 12 = 6(x - 2)$

Step 4: Evaluate at critical points

At $x = 1$: $f''(1) = 6(1-2) = -6 < 0$ → LOCAL MAX ∩

At $x = 3$: $f''(3) = 6(3-2) = 6 > 0$ → LOCAL MIN ∪

Step 5: Find the values

$f(1) = 1 - 6 + 9 + 1 = 5$

$f(3) = 27 - 54 + 27 + 1 = 1$

Answer: Local maximum of 5 at $x = 1$, local minimum of 1 at $x = 3$

When the Test Fails

If $f''(c) = 0$, the Second Derivative Test gives no information.

Example: $f(x) = x^4$

$f'(x) = 4x^3$, so critical point at $x = 0$

$f''(x) = 12x^2$, so $f''(0) = 0$

The test is inconclusive!

But using the First Derivative Test:

• $f'(x) < 0$ for $x < 0$
• $f'(x) > 0$ for $x > 0$
• Changes from − to + → local minimum at $x = 0$ ✓

Example: $f(x) = x^3$

$f'(x) = 3x^2$, so critical point at $x = 0$

$f''(x) = 6x$, so $f''(0) = 0$

The test fails again!

Using First Derivative Test:

• $f'(x) > 0$ for all $x \neq 0$
• No sign change → neither max nor min (inflection point) ✓

Comparing the Two Tests

First Derivative Test

✅ Always works (when derivative exists nearby) ✅ Gives increasing/decreasing information ✅ Can handle points where $f'$ is undefined ❌ Requires checking intervals on both sides ❌ More work (sign chart needed)

Second Derivative Test

✅ Faster - only evaluate at one point ✅ Simpler - just check one sign ❌ Fails when $f''(c) = 0$ ❌ Doesn't give increasing/decreasing info ❌ Requires finding second derivative

💡 Strategy: Try Second Derivative Test first (it's faster). If it fails ($f'' = 0$), use First Derivative Test.

Using Both Derivatives Together

The most complete analysis uses both:

Complete Analysis Template

For critical point $c$:

1. Location: $x = c$
2. First derivative: $f'(c) = 0$ (confirms it's critical)
3. Second derivative:
   • $f''(c) > 0$ → local min
   • $f''(c) < 0$ → local max
   • $f''(c) = 0$ → inconclusive
4. Value: $f(c) = ?$

Relationship to Concavity

The Second Derivative Test is really a concavity test:

Remember:

• $f''(x) > 0$ → graph is concave up ∪
• $f''(x) < 0$ → graph is concave down ∩

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

• Concave up + flat tangent = bottom of curve = minimum
• Concave down + flat tangent = top of curve = maximum

⚠️ Common Mistakes

Mistake 1: Using Wrong Derivative

❌ Evaluating $f'(c)$ instead of $f''(c)$ ✅ Second Derivative Test uses $f''(c)$

Mistake 2: Forgetting f'(c) = 0

The test only works at critical points where $f'(c) = 0$, NOT where $f'$ is undefined!

Mistake 3: Thinking 0 Means Maximum or Minimum

If $f''(c) = 0$, the test is inconclusive, not "neither"!

Mistake 4: Wrong Sign Interpretation

• $f''(c) > 0$ is POSITIVE → minimum (concave up ∪)
• $f''(c) < 0$ is NEGATIVE → maximum (concave down ∩)

Don't mix these up!

Special Cases

Case 1: Multiple Critical Points

Test each one separately.

Example: Critical points at $x = 1, 3, 5$

• Check $f''(1)$, $f''(3)$, $f''(5)$ individually

Case 2: Second Derivative is Constant

Example: $f(x) = x^2$ → $f''(x) = 2 > 0$ everywhere

The second derivative is always positive, so any critical point is a minimum.

Case 3: Optimization Problems

The Second Derivative Test is especially useful in optimization to verify that a critical point is indeed a max or min!

Quick Reference

📝 Practice Tips

1. Find both derivatives first: $f'(x)$ and $f''(x)$
2. Critical points come from $f'(x) = 0$
3. Evaluate $f''$ at each critical point (just plug in!)
4. Check the sign: positive = min, negative = max, zero = use other test
5. Calculate $f(c)$ to get the actual extremum value
6. Remember: This test ONLY works where $f'(c) = 0$, not where $f'$ is undefined

Step 2: Find critical points

$4x(x-1)(x-2) = 0$

Critical points: $x = 0, 1, 2$

Step 3: Find second derivative

$g''(x) = 12x^2 - 24x + 8$

Step 4: Test at $x = 0$

$g''(0) = 12(0)^2 - 24(0) + 8 = 8 > 0$

Positive → LOCAL MINIMUM ∪

Step 5: Test at $x = 1$

$g''(1) = 12(1)^2 - 24(1) + 8 = 12 - 24 + 8 = -4 < 0$

Negative → LOCAL MAXIMUM ∩

Step 6: Test at $x = 2$

$g''(2) = 12(2)^2 - 24(2) + 8 = 48 - 48 + 8 = 8 > 0$

Positive → LOCAL MINIMUM ∪

Step 7: Calculate values

$g(0) = 0$

$g(1) = 1 - 4 + 4 = 1$

$g(2) = 16 - 32 + 16 = 0$

Answer:

• Local minimum of $0$ at $x = 0$
• Local maximum of $1$ at $x = 1$
• Local minimum of $0$ at $x = 2$

Step 2: Find critical points

$5(x-1)(x+1)(x^2+1) = 0$

Since $x^2 + 1 > 0$ for all real $x$:

Critical points: $x = -1$ and $x = 1$

Step 3: Find second derivative

$h''(x) = 20x^3$

Step 4: Test at $x = -1$

$h''(-1) = 20(-1)^3 = -20 < 0$

Negative → LOCAL MAXIMUM ∩

Step 5: Test at $x = 1$

$h''(1) = 20(1)^3 = 20 > 0$

Positive → LOCAL MINIMUM ∪

Step 6: Calculate values

$h(-1) = (-1)^5 - 5(-1) = -1 + 5 = 4$

$h(1) = 1^5 - 5(1) = 1 - 5 = -4$

Note: The Second Derivative Test worked at both points! If we had gotten $h''(c) = 0$ anywhere, we would have needed the First Derivative Test.

Answer:

• Local maximum of $4$ at $x = -1$
• Local minimum of $-4$ at $x = 1$