The Second Derivative Test

∫ Concavity & Inflection Points

Part 1 of 7 — Understanding Concavity

🔑 Key Concept: The second derivative tells us about the shape of a curve — whether it bends upward (concave up) or downward (concave down).

📖 What Is Concavity?

$f''(x) > 0 \implies \text{concave up (cup shape, } \cup \text{)}$ $f''(x) < 0 \implies \text{concave down (cap shape, } \cap \text{)}$

Interpretations

Visual Test

• Concave up: Tangent lines lie BELOW the curve
• Concave down: Tangent lines lie ABOVE the curve
• If you drive along the curve, you'd turn your steering wheel left (concave up) or right (concave down)

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

$f''(x) = 6x$

• $x < 0$: $f'' < 0$ → concave down
• $x > 0$: $f'' > 0$ → concave up
• $x = 0$: inflection point (concavity changes!)

Finding Intervals of Concavity

Step-by-Step Process

1. Find $f''(x)$
2. Set $f''(x) = 0$ and find where $f''$ is undefined → candidate inflection points
3. Test intervals: create a sign chart for $f''$
4. $f'' > 0$ → concave up; $f'' < 0$ → concave down

Example: $f(x) = x^4 - 6x^2 + 1$

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

$f''(x) = 0$: $x = -1, 1$

Sign chart for $f''$:

Concavity Basics 🎯

📌 Inflection Points

An inflection point is where concavity changes.

Requirements

1. $f''(c) = 0$ or $f''(c)$ is undefined
2. $f''$ changes sign at $c$
3. $f(c)$ exists (point must be on the curve)

⚠️ $f''(c) = 0$ Does NOT Guarantee an Inflection Point!

$f(x) = x^4$: $f''(x) = 12x^2$, $f''(0) = 0$

But $f''$ does NOT change sign at $0$ (it's positive on both sides).

So $x = 0$ is NOT an inflection point.

Shortcut: If $f''$ changes sign → inflection point. If $f''$ doesn't change sign → not an inflection point.

AP Tip: When asked to "justify" an inflection point, you MUST show the sign change in $f''$. Simply stating $f''(c) = 0$ is insufficient.

Find Inflection Points 🧮

1) $f(x) = x^3 - 3x$. Inflection point x-coordinate: (integer)

2) $f(x) = x^4 - 4x^3$. Number of inflection points: (integer)

Concavity Concepts 🔽

Exit Quiz ✅

∫ Concavity & Inflection Points — Part 2

Part 2 of 7 — The Second Derivative Test for Extrema

📖 Second Derivative Test

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

$\boxed{f''(c) > 0 \implies \text{local minimum at } c}$ $\boxed{f''(c) < 0 \implies \text{local maximum at } c}$ $f''(c) = 0 \implies \text{test is inconclusive}$

Why It Works

• $f'(c) = 0$ and $f''(c) > 0$: curve is concave up at $c$ → $\cup$ shape → minimum
• $f'(c) = 0$ and $f''(c) < 0$: curve is concave down at $c$ → $\cap$ shape → maximum

Example: $f(x) = x^3 - 12x + 1$

$f'(x) = 3x^2 - 12 = 3(x^2 - 4) = 0$ → $x = \pm 2$

$f''(x) = 6x$

• $f''(2) = 12 > 0$ → local min at $x = 2$; $f(2) = 8 - 24 + 1 = -15$
• $f''(-2) = -12 < 0$ → local max at $x = -2$; $f(-2) = -8 + 24 + 1 = 17$

Second Derivative Test vs. First Derivative Test

When Second DT Fails

$f(x) = x^4$: $f'(0) = 0$, $f''(0) = 0$ → inconclusive!

Use First Derivative Test: $f' = 4x^3$ changes from $-$ to $+$ → minimum.

AP Tip: If the second derivative test is inconclusive, explicitly state so and switch to the first derivative test.

Second Derivative Test 🎯

📌 Comprehensive Example

$f(x) = 3x^4 - 4x^3 - 12x^2 + 5$

$f'(x) = 12x^3 - 12x^2 - 24x = 12x(x^2 - x - 2) = 12x(x-2)(x+1)$

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

$f''(x) = 36x^2 - 24x - 24$

Local max: $(0, 5)$. Local mins: $(-1, 0)$ and $(2, -27)$.

Second DT Practice 🧮

1) $f(x) = x^3 - 3x$. $f''(-1) =$ (integer, determines max/min)

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

Test Selection 🔽

Exit Quiz ✅

∫ Concavity & Inflection Points — Part 3

Part 3 of 7 — Complete Curve Sketching Framework

📖 The Complete Curve Sketching Checklist

1. Domain and key features ($x$-intercepts, $y$-intercept)
2. Symmetry (even, odd, periodic)
3. First derivative → critical points, increasing/decreasing
4. Second derivative → inflection points, concavity
5. End behavior → limits as $x \to \pm\infty$
6. Asymptotes (vertical, horizontal)
7. Plot key points and connect

Example: $f(x) = x^3 - 3x^2 + 4$

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

Critical points: $x = 0$ (max, $f = 4$), $x = 2$ (min, $f = 0$)

$f''(x) = 6x - 6 = 6(x - 1)$

Inflection point: $x = 1$ ($f = 2$)

The Four Shape Combinations

Transitions

• $f' = 0$: changes from increasing ↔ decreasing (potential extremum)
• $f'' = 0$ with sign change: changes concavity (inflection point)
• Both change: the four shapes cycle through combinations

Shape Analysis 🎯

📌 AP-Style Complete Analysis

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

Domain: All real numbers

Symmetry: Odd ($f(-x) = -f(x)$)

$f'(x)$: Using quotient rule: $f'(x) = \frac{(x^2+1)(1) - x(2x)}{(x^2+1)^2} = \frac{1-x^2}{(x^2+1)^2}$

Critical: $x = \pm 1$. $f(-1) = -1/2$ (min), $f(1) = 1/2$ (max)

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

Inflection: $x = 0, \pm\sqrt{3}$

End behavior: $\lim_{x \to \pm\infty} f(x) = 0$ (HA: $y = 0$)

This is a classic AP exam function.

Curve Analysis 🧮

1) $f(x) = x^4 - 4x^3$. Number of inflection points: (integer)

2) $f(x) = x^3 - 12x$. Local min value: (integer)

Curve Sketching Steps 🔽

Exit Quiz ✅

∫ Concavity & Inflection Points — Part 4

Part 4 of 7 — Reading Derivative Graphs

📖 From $f'$ Graph to $f$ Information

The graph of $f'$ tells us the shape of $f$ (but not its values, which need an initial condition such as $f(0)$):

Critical Insight: The concavity of $f$ comes from the slope of $f'$, not its sign!

From $f''$ Graph to $f$ Information

Example Reading

If $f''$ graph is a line crossing zero at $x = 3$:

• $f$ has an inflection point at $x = 3$
• Concavity changes at $x = 3$

If $f''$ graph is a parabola touching zero at $x = 2$:

• NOT an inflection point at $x = 2$ (no sign change)

Derivative Graph Reading 🎯

📌 Reconstructing $f$ from $f'$ or $f''$

Key AP Skill

Given a graph of $f'$, sketch $f$:

1. Find zeros of $f'$ → critical points of $f$
2. Where $f' > 0$: $f$ goes up; where $f' < 0$: $f$ goes down
3. Where $f'$ increases: $f$ is concave up
4. Where $f'$ decreases: $f$ is concave down
5. Use $f'$ values for steepness info

Common AP Setup

Given the graph of $f'$ and $f(0) = 2$, find $f(3)$.

$f(3) = f(0) + \int_0^3 f'(x) dx = 2 + \text{(area under f')}$

This connects derivatives to integrals via the FTC!

Graph Reading 🧮

1) If $f'$ has 3 local extrema, $f$ has ___ inflection points: (integer)

2) If $f'(x) = 0$ at $x = 1, 4$ and $f' > 0$ on $(1,4)$, $f$ is ___ on $(1,4)$: (enter "increasing" or "decreasing")

Graph Feature Mapping 🔽

Exit Quiz ✅

∫ Concavity & Inflection Points — Part 5

Part 5 of 7 — Position, Velocity, Acceleration & Concavity

📖 Concavity in Motion

For position $s(t)$:

• $s'(t) = v(t)$ = velocity
• $s''(t) = a(t)$ = acceleration

Concavity ↔ Acceleration

⚠️ Speeding Up vs. Slowing Down

An object speeds up when $v$ and $a$ have the same sign:

• $v > 0, a > 0$ → speeding up forward
• $v < 0, a < 0$ → speeding up backward

An object slows down when $v$ and $a$ have opposite signs.

Inflection Point of $s(t)$

$s''(t_0) = 0$ with sign change → acceleration changes direction.

This is when velocity reaches a local maximum or minimum. Speed $|v|$ may have a local extreme there too, but check the sign of $v$ and the endpoints before calling it the greatest speed.

Example: Complete Motion Analysis

$s(t) = t^3 - 6t^2 + 9t + 1$ for $t \geq 0$

$v(t) = 3t^2 - 12t + 9 = 3(t-1)(t-3)$ $a(t) = 6t - 12 = 6(t-2)$

At $t = 2$: inflection point of $s(t)$, maximum leftward velocity.

AP Pattern: The particle changes direction at $t = 1$ and $t = 3$ (where $v$ changes sign). On $(1, 3)$ it moves left, and its speed is greatest on that interval at $t = 2$, where $a = 0$ (speed $|v(2)| = 3$). That is not the greatest speed for all $t \geq 0$: $|v(0)| = 9$, and the speed grows without bound as $t \to \infty$.

Motion Analysis 🎯

📌 Free Fall & Projectile Motion

$s(t) = -\frac{1}{2}gt^2 + v_0 t + s_0$ (with $g = 32$ $ft/s^{2}$ or $9.8$ $m/s^{2}$)

$v(t) = -gt + v_0$ $a(t) = -g$ (constant, always concave down)

Since $a < 0$ always:

• While going up ($v > 0$): slowing down
• At the top ($v = 0$): instantaneously stopped
• While falling ($v < 0$): speeding up (downward)

Maximum Height

$v(t) = 0$: $t = v_0/g$

$s_{max} = s_0 + \frac{v_0^2}{2g}$

Motion Practice 🧮

1) $s(t) = t^3 - 3t$. When does $a = 0$? $t =$ (integer)

2) At that time, $v =$ (integer)

Motion Concepts 🔽

Exit Quiz ✅

∫ Concavity & Inflection Points — Part 6

Part 6 of 7 — Connecting Concavity to Optimization

📖 Second Derivative Test in Optimization

In optimization problems, after finding critical points, the second derivative test confirms max/min:

Example: Maximize Area

A farmer has 400 ft of fencing for a rectangular pen against a barn. Maximize area.

$A = x(400 - 2x) = 400x - 2x^2$

$A' = 400 - 4x = 0 \implies x = 100$

$A'' = -4 < 0$ → maximum confirmed!

$A_{max} = 100(200) = 20,000$ $ft^{2}$

When the Domain is Restricted

If $f''< 0$ everywhere on an interval, then:

• Only ONE critical point → it MUST be the absolute max
• No need for endpoint checking if domain is all reals

If $f'' > 0$ everywhere:

• The single critical point is the absolute min

Economic Applications

Marginal Analysis

$C(x)$ = cost, $R(x)$ = revenue, $P(x) = R(x) - C(x)$ = profit

$P'(x) = R'(x) - C'(x) = 0 \implies R'(x) = C'(x)$

Profit is maximized when marginal revenue = marginal cost.

$P''(x) = R''(x) - C''(x)$

If $R''(x) < 0$ (diminishing returns) and $C''(x) > 0$ (increasing marginal cost), then $P'' < 0$ → confirmed maximum.

Point of Diminishing Returns

The inflection point of $R(x)$: where revenue growth starts slowing.

$R''(x) = 0$ with sign change from + to - → this is where each additional unit adds less revenue than the previous one.

Optimization & Concavity 🎯

📌 Global vs. Local Extrema

Closed Interval Method (Absolute Extrema)

On a closed interval $[a, b]$:

1. Find all critical points in $(a, b)$
2. Evaluate $f$ at critical points AND endpoints
3. Largest value = absolute max, smallest = absolute min

When Second Derivative Helps

If there's only one critical point $c$ on the domain:

• $f''(c) > 0$ → absolute minimum
• $f''(c) < 0$ → absolute maximum

Example: Minimize Surface Area

Cylinder with volume $1000$ $cm^{3}$. Minimize $SA = 2\pi r^2 + 2\pi r h$.

Constraint: $\pi r^2 h = 1000 \implies h = 1000/(\pi r^2)$

$SA = 2\pi r^2 + 2000/r$

$SA' = 4\pi r - 2000/r^2 = 0 \implies r^3 = 500/\pi \implies r = (500/\pi)^{1/3}$

$SA'' = 4\pi + 4000/r^3 > 0$ always → confirmed minimum!

Optimization Check 🧮

1) $A = 200x - x^2$. $A'' =$ (integer, confirms max/min?)

2) $A$ is maximized at $x =$ (integer)

Optimization Strategies 🔽

Exit Quiz ✅

∫ Concavity & Inflection Points — Part 7

Part 7 of 7 — AP Free Response Justifications

📖 AP Justification Standards

The AP exam requires specific language when justifying conclusions about concavity and inflection points.

Justifying Local Extrema

Using First Derivative Test: "Since $f'$ changes from positive to negative at $x = c$, $f$ has a local maximum at $x = c$." ✅

Using Second Derivative Test: "Since $f'(c) = 0$ and $f''(c) < 0$, $f$ has a local maximum at $x = c$." ✅

Justifying Inflection Points

"Since $f''$ changes sign at $x = c$ (from negative to positive), $f$ has an inflection point at $x = c$." ✅

"$f''(c) = 0$, so $c$ is an inflection point." ❌ (Insufficient! Need sign change!)

AP Free Response Walkthrough

Given: $f'(x) = (x-1)(x-3)^2(x-5)$

(a) Find where $f$ has local extrema.

$f'(x) = 0$ at $x = 1, 3, 5$.

Sign analysis of $f'$:

• $f' > 0$ on $(-\infty, 1)$, $f' < 0$ on $(1, 3)$
• $f' < 0$ on $(3, 5)$ ($(x-3)^2 \geq 0$, doesn't change sign!)
• $f' > 0$ on $(5, \infty)$

Local max at $x = 1$ (+ to -). Local min at $x = 5$ (- to +). $x = 3$ is NOT an extremum (no sign change).

(b) Find inflection points.

Inflection points of $f$ are where $f''$ changes sign, which is where $f'$ switches between increasing and decreasing.

Let $u = x - 3$. Then $(x-1)(x-5) = (u+2)(u-2) = u^2 - 4$, so $f'(x) = u^2(u^2 - 4) = u^4 - 4u^2$.

$f''(x) = 4u^3 - 8u = 4u(u^2 - 2) = 4(x-3)\left(x - 3 - \sqrt{2}\right)\left(x - 3 + \sqrt{2}\right)$

$f'' = 0$ at $x = 3 - \sqrt{2}$, $x = 3$, and $x = 3 + \sqrt{2}$.

Sign analysis of $f''$:

• $f'' < 0$ on $(-\infty, 3 - \sqrt{2})$, $f'' > 0$ on $(3 - \sqrt{2}, 3)$
• $f'' < 0$ on $(3, 3 + \sqrt{2})$, $f'' > 0$ on $(3 + \sqrt{2}, \infty)$

$f''$ changes sign at all three values, so $f$ has inflection points at $x = 3 - \sqrt{2}$, $x = 3$, and $x = 3 + \sqrt{2}$.

Check with $f'$: $f' \le 0$ on $[1, 5]$ with $f'(3) = 0$, so $f'$ has a local max at $x = 3$ and a local min in each of $(1, 3)$ and $(3, 5)$. Those three extrema of $f'$ are exactly the three inflection points. Note that $x = 3$ is an inflection point even though it is not an extremum of $f$.

AP Justifications 🎯

📌 Common AP Setups

Type 1: Given $f'$ as a formula

Find critical points, use $f''$ or sign chart.

Type 2: Given $f'$ as a graph

Read increasing/decreasing, extrema, concavity from the graph.

Type 3: Given a table of $f$, $f'$, $f''$ values

Use values to determine properties at specific points.

From this table:

• At $x = 1$: $f'= 0$ and $f'' < 0$ → local maximum
• At $x = 0$: $f$ is decreasing, concave up
• At $x = 2$: $f''(2) = 0$, but $f''(1) = -1$ and $f''(3) = -3$ are both negative. The table shows no sign change around $x = 2$, so it gives no evidence of an inflection point there. A zero of $f''$ is only a candidate.
• Between $x = 0$ and $x = 1$: $f''$ goes from $4$ to $-1$. If $f''$ is continuous, it changes sign somewhere in $(0, 1)$, so $f$ has an inflection point in that interval.

AP Practice 🧮

1) $f(x) = x^4/4 - 2x^2$. Number of inflection points: (integer)

2) Same $f$: $x$-value of local maximum: (integer)

Final Concepts 🔽

Final Exit Quiz ✅