📐 The Second Derivative Test

What is the Second Derivative Test?

The Second Derivative Test is a quick way to classify critical points using the second derivative $f''(x)$ instead of checking sign changes in $f'(x)$.

💡 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

| $f''(c)$ | Shape | Result | |----------|-------|--------| | $> 0$ | ∪ (concave up) | Local MIN | | $< 0$ | ∩ (concave down) | Local MAX | | $= 0$ | ? | INCONCLUSIVE |

---

📝 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