Applications of Derivatives

📈 Applications of Derivatives

Part 1 of 7 — Critical Points & Increasing/Decreasing

Critical Points

$\boxed{\text{A critical point of } f \text{ occurs where } f'(c) = 0 \text{ or } f'(c) \text{ is undefined (but } f(c) \text{ exists)}}$

Key Fact: Critical points are the ONLY candidates for local extrema. If $f$ has a local max or min at $x = c$, then $c$ must be a critical point.

Why Critical Points Matter

First Derivative Test for Increasing/Decreasing

$\boxed{f'(x) > 0 \Rightarrow f \text{ increasing} \qquad f'(x) < 0 \Rightarrow f \text{ decreasing}}$

Worked Example

Find where $f(x) = x^3 - 3x + 1$ is increasing and decreasing.

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

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

AP Tip: Always use a sign chart or number line to organize your analysis. Pick a test value in each interval — don't just guess the sign.

Critical Points 🎯

First Derivative Test for Local Extrema

At a critical point $c$:

Complete Worked Example

For $f(x) = x^4 - 4x^3$: $f'(x) = 4x^2(x-3)$

• At $x = 0$: $f'$ stays negative ($- \to -$) → Neither max nor min
• At $x = 3$: $f'$ changes $- \to +$ → Local minimum at $f(3) = 81 - 108 = -27$

Key Concept: A critical point where $f' = 0$ does NOT guarantee a local extremum. You must verify with a sign change analysis.

Classify Critical Points 🎯

Sign chart analysis 🔍

For $f(x) = x^3 - 12x$, $f'(x) = 3(x-2)(x+2)$. Classify each critical point.

Find the critical points. ✍️

Key Takeaways — Part 1

$\boxed{f'(c) = 0 \text{ or } f'(c) \text{ undefined} \Rightarrow c \text{ is a critical point}}$

Up Next: Part 2 — Second Derivative & Concavity.

📈 Applications of Derivatives

Part 2 of 7 — Second Derivative & Concavity

Concavity

$\boxed{f''(x) > 0 \Rightarrow \text{Concave UP (cup)} \qquad f''(x) < 0 \Rightarrow \text{Concave DOWN (cap)}}$

Key Concept: Concavity tells you how the SLOPE is changing. Concave up means the slope is increasing (even if the function is decreasing). Concave down means the slope is decreasing.

Inflection Points

An inflection point is where concavity changes.

$\boxed{\text{Inflection point at } x = c \iff f'' \text{ changes sign at } x = c}$

Warning: $f''(c) = 0$ does NOT guarantee an inflection point! You must verify the sign change. Example: $f(x) = x^4$ has $f''(0) = 0$ but NO inflection point (concave up on both sides).

Second Derivative Test for Extrema

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

Worked Example

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

Classify using Second Derivative Test:

• $f''(1) = 6(1) - 12 = -6 < 0$ → Local max at $x = 1$
• $f''(3) = 6(3) - 12 = 6 > 0$ → Local min at $x = 3$
• $f''(2) = 0$ and sign changes ($- \to +$) → Inflection point at $x = 2$

Second Derivative Analysis 🎯

First vs Second Derivative Test

AP Tip: On free-response questions, use the First Derivative Test for justifications — it ALWAYS gives a definitive answer. The Second Derivative Test is faster for multiple-choice when $f''$ is easy to compute.

Connecting $f$, $f'$, and $f''$

Concavity & Inflection 🎯

Analyze $f(x) = x^3 - 3x^2 + 2$ 🔍

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

Find the inflection point. ✍️

Key Takeaways — Part 2

$\boxed{f'' > 0: \text{ concave up} \qquad f'' < 0: \text{ concave down}}$

Up Next: Part 3 — Absolute (Global) Extrema.

📈 Applications of Derivatives

Part 3 of 7 — Absolute (Global) Extrema

Extreme Value Theorem (EVT)

$\boxed{\text{If } f \text{ is continuous on } [a,b], \text{ then } f \text{ attains an absolute max and absolute min on } [a,b].}$

Key Fact: The EVT has TWO hypotheses: (1) $f$ is continuous, (2) the interval is CLOSED $[a,b]$. If either fails, the conclusion is NOT guaranteed.

Candidates Test (Closed Interval Method)

$\boxed{\text{Abs. extrema on } [a,b]: \text{ compare } f \text{ at all critical points AND endpoints}}$

Worked Example

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

$f'(x) = 3x^2 - 3 = 3(x-1)(x+1) = 0$ → $x = \pm 1$ (both in the interval).

$\boxed{\text{Absolute max} = 3 \text{ at } x = -1 \text{ and } x = 2}$ $\boxed{\text{Absolute min} = -1 \text{ at } x = -2 \text{ and } x = 1}$

AP Tip: On free-response, you MUST evaluate at EVERY critical point AND both endpoints. Missing even one candidate can cost you points.

Absolute Extrema 🎯

Absolute Extrema on Open or Infinite Intervals

The Candidates Test only works on closed intervals. For open intervals or $(-\infty, \infty)$:

Key Concept: If $f$ has exactly ONE critical point on $(-\infty,\infty)$ and it's a local min, then it's also the absolute min. Same for max. This is extremely useful for optimization!

Example: One Critical Point

$f(x) = e^x + e^{-x}$ on $(-\infty, \infty)$.

$f'(x) = e^x - e^{-x} = 0$ → $e^{2x} = 1$ → $x = 0$.

$f''(0) = e^0 + e^0 = 2 > 0$: local min. Only one critical point → absolute min $f(0) = 2$.

EVT & Open Intervals 🎯

Classify the extrema of $f(x) = 2x^3 - 9x^2 + 12x$ on $[0, 4]$ 🔍

$f'(x) = 6x^2 - 18x + 12 = 6(x-1)(x-2)$

Candidates: $f(0) = 0$, $f(1) = 5$, $f(2) = 4$, $f(4) = 128 - 144 + 48 = 32$

Find the absolute minimum. ✍️

Key Takeaways — Part 3

$\boxed{\text{Candidates Test: compare } f \text{ at critical points + endpoints}}$

Up Next: Part 4 — Curve Sketching.

📈 Applications of Derivatives

Part 4 of 7 — Curve Sketching

The 7-Step Procedure

$\boxed{\text{Domain} \to \text{Intercepts} \to \text{Symmetry} \to f' \to f'' \to \text{End behavior} \to \text{Sketch}}$

AP Tip: On the AP exam, you rarely sketch from scratch. Instead, you're given a graph of $f'$ and must deduce properties of $f$. Master reading $f'$ graphs!

Complete Worked Example

$f(x) = x^4 - 4x^3$

Step 1 — Domain: All real numbers.

Step 2 — Intercepts: $f(0) = 0$, $f(x) = x^3(x-4) = 0$ at $x = 0, 4$.

Step 3 — Symmetry: Neither even nor odd.

Step 4 — First Derivative: $f'(x) = 4x^3 - 12x^2 = 4x^2(x-3)$

Local min at $x = 3$: $f(3) = 81 - 108 = -27$. No extremum at $x = 0$ (no sign change!).

Step 5 — Second Derivative: $f''(x) = 12x^2 - 24x = 12x(x-2)$

Inflection points at $x = 0$ and $x = 2$.

Step 6 — End behavior: $\lim_{x \to \pm\infty} f(x) = +\infty$.

Key Concept: $x = 0$ gives $f'(0) = 0$ but NO extremum — it's a "flat spot" where $f$ still decreases. This happens when a factor in $f'$ has EVEN multiplicity.

Reading $f'$ Graphs — AP Essential Skill

Given a GRAPH of $f'(x)$, determine features of $f(x)$:

Key Fact: A local max of $f'$ corresponds to an inflection point of $f$ where concavity changes from UP to DOWN. A local min of $f'$ corresponds to an inflection point where concavity changes from DOWN to UP.

Curve Sketching from Derivatives 🎯

Given $f'(x) = (x-1)^2(x-4)$:

Polynomial Curve Sketching Shortcuts

Multiplicity and Behavior at Roots of $f'$

Reading Derivative Graphs 🎯

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

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

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

Find the $y$-coordinate of the inflection point. ✍️

Key Takeaways — Part 4

$\boxed{f' > 0: \text{ inc} \quad f' < 0: \text{ dec} \quad f'' > 0: \text{ CU} \quad f'' < 0: \text{ CD}}$

Up Next: Part 5 — Mean Value Theorem.

📈 Applications of Derivatives

Part 5 of 7 — Mean Value Theorem

Statement (MVT)

$\boxed{f'(c) = \frac{f(b) - f(a)}{b - a} \quad \text{for some } c \in (a,b)}$

Hypotheses (BOTH required):

1. $f$ is continuous on $[a,b]$
2. $f$ is differentiable on $(a,b)$

Geometric meaning: There is a point where the tangent line is parallel to the secant line through $(a, f(a))$ and $(b, f(b))$.

Key Fact: MVT says instantaneous rate of change EQUALS average rate of change somewhere in the interval. This is one of the most tested theorems on the AP exam.

MVT vs Rolle's Theorem

Rolle's is just MVT when the secant line is horizontal!

Worked Example 1

$f(x) = x^3$ on $[0, 2]$.

Step 1: Average rate = $\frac{f(2)-f(0)}{2-0} = \frac{8-0}{2} = 4$.

Step 2: Set $f'(c) = 4$: $3c^2 = 4$ → $c = \frac{2\sqrt{3}}{3} \approx 1.155$.

Since $c \in (0, 2)$: MVT confirmed. ✓

Worked Example 2 (Table Data — AP Style)

$f$ is continuous and differentiable on $[0, 12]$.

Average rate on $[0, 12]$: $\frac{19-1}{12-0} = \frac{18}{12} = \frac{3}{2}$.

MVT guarantees $f'(c) = \frac{3}{2}$ for some $c \in (0, 12)$.

AP Tip: On free-response with table data, you can also apply MVT to sub-intervals. On $[2, 5]$: $\frac{7-4}{5-2} = 1$. On $[5, 9]$: $\frac{13-7}{9-5} = \frac{3}{2}$. Since $f'$ takes values $1$ and $\frac{3}{2}$ at different points, by IVT applied to $f'$, $f'$ takes every value between.

Mean Value Theorem 🎯

Important Consequences of MVT

Common MVT Justification Template (AP Free-Response)

"Since $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, by the Mean Value Theorem, there exists $c \in (a,b)$ such that $f'(c) = \frac{f(b)-f(a)}{b-a} = \text{[value]}$."

Key Concept: Always STATE the hypotheses (continuous, differentiable) before applying MVT. This is required for full credit on free-response.

MVT Applications 🎯

Verify MVT for $f(x) = x^2 - 4x + 1$ on $[1, 5]$ 🔍

Apply MVT. ✍️

Key Takeaways — Part 5

$\boxed{f'(c) = \frac{f(b)-f(a)}{b-a} \quad \text{(instantaneous = average somewhere)}}$

Up Next: Part 6 — L'Hôpital's Rule & Optimization.

📈 Applications of Derivatives

Part 6 of 7 — Optimization

The Optimization Framework

$\boxed{\text{Optimize} \to \text{Constraint} \to \text{Single Variable} \to \text{Calculus} \to \text{Verify}}$

AP Tip: On free-response optimization problems, you MUST justify why your critical point is a maximum or minimum. Use the First or Second Derivative Test, or the Candidates Test on a closed interval.

Classic Example 1: Maximize Area

A farmer has 200 m of fence. Maximize the area of a rectangular pen against a barn (3 sides needed).

Objective: $A = xy$ (maximize)

Constraint: $x + 2y = 200$ → $x = 200 - 2y$

Substitute: $A(y) = (200-2y)y = 200y - 2y^2$

Differentiate: $A'(y) = 200 - 4y = 0$ → $y = 50$

Then $x = 200 - 100 = 100$.

$\boxed{A_{\max} = 100 \times 50 = 5000 \text{ m}^2}$

Verify: $A''(y) = -4 < 0$: concave down → confirmed maximum.

Classic Example 2: Minimize Material

Make an open-top box with volume $32$ cm$^3$ from a square base. Minimize surface area.

Objective: $S = x^2 + 4xh$ (minimize, no top)

Constraint: $V = x^2 h = 32$ → $h = \frac{32}{x^2}$

Substitute: $S(x) = x^2 + 4x \cdot \frac{32}{x^2} = x^2 + \frac{128}{x}$

Differentiate: $S'(x) = 2x - \frac{128}{x^2} = 0$ → $2x^3 = 128$ → $x = 4$

Then $h = \frac{32}{16} = 2$.

$\boxed{S_{\min} = 16 + 32 = 48 \text{ cm}^2}$

Optimization Basics 🎯

Minimizing Distance

Find the point on $y = x^2$ closest to $(0, 1)$.

Objective: Minimize $d = \sqrt{x^2 + (x^2-1)^2}$

Key Concept: Minimize $d^2$ instead! Same critical points, avoids the square root.

$D = d^2 = x^2 + (x^2-1)^2 = x^2 + x^4 - 2x^2 + 1 = x^4 - x^2 + 1$

$D' = 4x^3 - 2x = 2x(2x^2 - 1) = 0$

$x = 0$ or $x = \pm\frac{1}{\sqrt{2}}$

$\boxed{\text{Closest points: } \left(\pm\frac{1}{\sqrt{2}},\, \frac{1}{2}\right) \text{ at distance } \frac{\sqrt{3}}{2}}$

Common Optimization Setups

Advanced Optimization 🎯

Optimization Setup: An open box is made by cutting squares of side $x$ from each corner of a $12 \times 12$ sheet and folding up. 🔍

Dimensions: $(12-2x) \times (12-2x) \times x$. Volume: $V(x) = x(12-2x)^2$.

Solve this optimization problem. ✍️

Key Takeaways — Part 6

$\boxed{\text{Objective + Constraint} \to \text{Single variable} \to \text{Calculus}}$

Up Next: Part 7 — Comprehensive Review & Assessment.

📈 Applications of Derivatives — Review

Part 7 of 7 — Comprehensive Assessment

Complete Topic Summary

Quick Reference

$\boxed{f' > 0: \uparrow \quad f' < 0: \downarrow \quad f'' > 0: \cup \quad f'' < 0: \cap}$

Decision Guide: Which Test to Use?

Common AP Mistakes to Avoid

Comprehensive Assessment 🎯

Mixed Applications 🎯

AP-Style Table Problem 🔍

$f$ is continuous and differentiable on $[1, 7]$.

Final Optimization Problem ✍️

Applications of Derivatives — Complete! ✅

You have mastered:

• ✅ Critical points and the First Derivative Test
• ✅ Concavity, inflection points, and the Second Derivative Test
• ✅ Absolute extrema via EVT and Candidates Test
• ✅ Curve sketching and reading $f'$ graphs
• ✅ Mean Value Theorem and Rolle's Theorem
• ✅ Optimization: objective + constraint approach

AP Free-Response Justification Checklist