Applications of Derivatives - Complete Interactive Lesson

Part 1: Critical Points & Increasing/Decreasing

📈 Applications of Derivatives

Part 1 of 7 — Critical Points & Increasing/Decreasing

Part	Topic
1	Critical Points & Increasing/Decreasing
2	Second Derivative & Concavity
3	Absolute (Global) Extrema
4	Curve Sketching
5	Mean Value Theorem
6	Optimization
7	Review & AP Applications

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

Type	What Happens	Examples
$f'(c) = 0$	Horizontal tangent line	Smooth peaks/valleys
$f'(c)$ undefined	Cusp, corner, or vertical tangent	$
Not a critical point	$f$ cannot have a local extremum	Guaranteed by Fermat's Theorem

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 $x = 1$ .

Interval	Test Value	$f'(x)$	Behavior
$(-\infty, -1)$	$x = -2$	$3(4-1) = 9 > 0$	Increasing
$(-1, 1)$	$x = 0$	$3(0-1) = -3 < 0$	Decreasing
$(1, \infty)$	$x = 2$	$3(4-1) = 9 > 0$	Increasing

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 $c$ :

Sign Change of $f'$	Conclusion	Mnemonic
$+ \to -$	Local maximum	Hill: going up then down
$- \to +$	Local minimum	Valley: going down then up
$+ \to +$ or $- \to -$	Neither	No direction change

Complete Worked Example

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

Interval	$4x^2$	$(x-3)$	$f'(x)$
$x < 0$	$+$	$-$	$-$
$0 < x < 3$	$+$	$-$	$-$
$x > 3$	$+$	$+$	$+$

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

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

Concept	Key Fact
Critical points	Where $f' = 0$ or $f'$ DNE
Increasing	$f' > 0$ on the interval
Decreasing	$f' < 0$ on the interval
Local max	$f'$ changes $+ \to -$
Local min	$f'$ changes $- \to +$
Neither	No sign change

Up Next: Part 2 — Second Derivative & Concavity.

Part 2: Second Derivative & Concavity

📈 Applications of Derivatives

Part 2 of 7 — Second Derivative & Concavity

Concavity

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

$f''(x)$	Concavity	Shape	Tangent Lines
$f''(x) > 0$	Concave up	$\cup$	Lie BELOW the curve
$f''(x) < 0$	Concave down	$\cap$	Lie ABOVE the curve

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 $f'(c) = 0$ :

$f''(c)$	Conclusion	Reason
$f''(c) > 0$	Local minimum	Concave up = valley
$f''(c) < 0$	Local maximum	Concave down = hill
$f''(c) = 0$	Inconclusive	Use First Derivative Test

Worked Example

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

Derivative	Expression	Critical Points
$f'(x)$	$3x^2 - 12x + 9 = 3(x-1)(x-3)$	$x = 1, 3$
$f''(x)$	$6x - 12$	$x = 2$

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

Feature	First Derivative Test	Second Derivative Test
What you check	Sign change of $f'$	Sign of $f''$ at critical point
Requires	Sign chart around $c$	Just $f''(c)$
Always works?	Yes	No ( $f''(c) = 0$ is inconclusive)
Finds inflection points?	No	Yes (as a byproduct)
AP recommendation	Use for justifications	Use when $f''$ is easy to compute

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 $f''$

If $f'$ is...	Then $f$ is...
Positive	Increasing
Negative	Decreasing
Zero (with sign change)	Has local extremum
Increasing	Concave up ( $f'' > 0$ )
Decreasing	Concave down ( $f'' < 0$ )
Has a local extremum	$f$ has an inflection point

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

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

Concept	Key Rule
Concave up	$f'' > 0$ , tangent lines below curve
Concave down	$f'' < 0$ , tangent lines above curve
Inflection point	$f''$ changes sign
2nd Deriv Test (min)	$f'(c) = 0$ and $f''(c) > 0$
2nd Deriv Test (max)	$f'(c) = 0$ and $f''(c) < 0$
Inconclusive	$f'(c) = 0$ and $f''(c) = 0$

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

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.

Hypothesis Fails	Example	What Goes Wrong
Not continuous	$f(x) = \frac{1}{x}$ on $[-1,1]$	Blows up at $x = 0$
Not closed	$f(x) = x$ on $(0,1)$	Can get arbitrarily close to 0 and 1 but never reach them
Both fail	$f(x) = \tan(x)$ on $(-\frac{\pi}{2}, \frac{\pi}{2})$	Unbounded, open interval

Candidates Test (Closed Interval Method)

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

Step	Action	Detail
1	Find $f'(x)$	Differentiate
2	Find critical points	Where $f'(x) = 0$ or $f'(x)$ is undefined
3	Filter	Keep only critical points in $(a,b)$
4	Evaluate	Compute $f$ at each critical point and at $a$ and $b$
5	Compare	Largest = absolute max, Smallest = absolute min

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).

$x$	$f(x)$	Candidate Type
$-2$	$-8+6+1 = -1$	Endpoint
$-1$	$-1+3+1 = 3$	Critical point
$1$	$1-3+1 = -1$	Critical point
$2$	$8-6+1 = 3$	Endpoint

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

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 $(-\infty, \infty)$ :

Interval Type	Strategy
Open $(a,b)$	Find critical points, check limits as $x \to a^+$ and $x \to b^-$
Half-open $[a, b)$	Include endpoint $a$ , check limit as $x \to b^-$
$(-\infty, \infty)$	Only one critical point? Use 2nd Derivative Test

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

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

Concept	Key Rule
EVT	Continuous on $[a,b]$ → abs max and min exist
Candidates Test	Evaluate at CPs + endpoints; compare
Open intervals	No Candidates Test; use limits and single-CP shortcuts
One critical point	If only local min $\Rightarrow$ absolute min (and vice versa)
AP justification	Must list ALL candidates and state why largest/smallest

Up Next: Part 4 — Curve Sketching.

Part 4: Curve Sketching

📈 Applications of Derivatives

Part 4 of 7 — Curve Sketching

The 7-Step Procedure

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

Step	What to Find	How
1	Domain	Where is $f$ defined?
2	Intercepts	$y$ -int: set $x=0$ . $x$ -int: set $f(x)=0$
3	Symmetry	Even: $f(-x)=f(x)$ . Odd: $f(-x)=-f(x)$
4	$f'$ analysis	Critical points, inc/dec, local extrema
5	$f''$ analysis	Concavity, inflection points
6	End behavior	$\lim_{x \to \pm\infty} f(x)$ or asymptotes
7	Sketch	Combine all info into a graph

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

Interval	Sign of $f'$	$f$ behavior
$(-\infty, 0)$	$(+)(-)= -$	Decreasing
$(0, 3)$	$(+)(-) = -$	Decreasing
$(3, \infty)$	$(+)(+) = +$	Increasing

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

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

Interval	Sign of $f''$	Concavity
$(-\infty, 0)$	$(-)(-) = +$	Up
$(0, 2)$	$(+)(-) = -$	Down
$(2, \infty)$	$(+)(+) = +$	Up

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 $f(x)$ :

Feature of $f'$ graph	Conclusion about $f$
$f'$ crosses $x$ -axis ( $+ \to -$ )	$f$ has local MAX
$f'$ crosses $x$ -axis ( $- \to +$ )	$f$ has local MIN
$f'$ touches $x$ -axis (no sign change)	No extremum (flat spot)
$f' > 0$	$f$ is increasing
$f' < 0$	$f$ is decreasing
$f'$ is increasing	$f$ is concave UP
$f'$ is decreasing	$f$ is concave DOWN
$f'$ has a local extremum	$f$ has an inflection point

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

Degree	End Behavior	Max Turning Points	Max Inflection Points
2 (quadratic)	Same direction both ends	1	0
3 (cubic)	Opposite directions	2	1
4 (quartic)	Same direction both ends	3	2
$n$	Depends on leading coeff.	$n-1$	$n-2$

Multiplicity and Behavior at Roots of $f'$

Multiplicity of root in $f'$	Sign change?	Feature of $f$
Odd (1, 3, 5, ...)	Yes	Local extremum
Even (2, 4, 6, ...)	No	Flat spot (no extremum)

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

Concept	Key Insight
7-step procedure	Systematic: domain → intercepts → symmetry → $f'$ → $f''$ → ends → sketch
Reading $f'$ graphs	$f'$ sign → inc/dec; $f'$ zero with sign change → extremum
$f'$ increasing/decreasing	Tells you concavity of $f$
Local extrema of $f'$	= inflection points of $f$
Even multiplicity in $f'$	Flat spot, no extremum

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

Up Next: Part 5 — Mean Value Theorem.

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):

$f$ is continuous on $[a,b]$
$f$ 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

Theorem	Extra Condition	Conclusion
MVT	None	$f'(c) = \frac{f(b)-f(a)}{b-a}$
Rolle's	$f(a) = f(b)$	$f'(c) = 0$

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)

$x$	$0$	$2$	$5$	$9$	$12$
$f(x)$	$1$	$4$	$7$	$13$	$19$

$f$ 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

Consequence	Statement	Why It Matters
Speed analogy	If avg speed was 70 mph, you were going EXACTLY 70 at some moment	Real-world MVT interpretation
Bounding derivatives	If $	f'(x)
Zero derivative	If $f'(x) = 0$ for all $x$ , then $f$ is constant	Proves constant functions
Equal derivatives	If $f'(x) = g'(x)$ for all $x$ , then $f = g + C$	Functions with same derivative differ by constant

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

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

Concept	Key Rule
MVT hypotheses	Continuous on $[a,b]$ , differentiable on $(a,b)$
MVT conclusion	$f'(c) =$ average rate for some $c \in (a,b)$
Rolle's Theorem	MVT when $f(a) = f(b)$ : then $f'(c) = 0$
AP justification	MUST state both hypotheses before applying
Table data MVT	Compute average rate between table values

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

Part 6: Related Rates (Mini-Review)

📈 Applications of Derivatives

Part 6 of 7 — Optimization

The Optimization Framework

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

Step	Action	Detail
1	Identify	What quantity to maximize/minimize?
2	Write objective	Express the quantity as a function
3	Find constraint	A second equation relating the variables
4	Eliminate	Use constraint to get one-variable function
5	Differentiate	Set $f'(x) = 0$ , find critical points
6	Verify	Confirm it's actually a max/min (not saddle)
7	Answer	State the answer in context with units

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.

Variable	Meaning
$x$	Side of square base
$h$	Height

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

$x$	$D = d^2$	$d$
$0$	$1$	$1$
$\pm\frac{1}{\sqrt{2}}$	$\frac{1}{4} - \frac{1}{2} + 1 = \frac{3}{4}$	$\frac{\sqrt{3}}{2}$

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

Common Optimization Setups

Problem Type	Objective	Typical Constraint
Fencing	Maximize area	Fixed perimeter
Box/can	Minimize surface area	Fixed volume
Distance	Minimize $d^2$	Point on curve
Revenue	Maximize $R = p \cdot q$	Demand equation
Travel time	Minimize $t = \frac{d}{v}$	Different speeds on different terrain

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

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

Concept	Key Rule
Setup	Identify what to maximize/minimize and the constraint
Eliminate	Use constraint to reduce to one variable
Solve	$f'(x) = 0$ to find critical points
Justify	Use FDT, SDT, or Candidates Test to verify max/min
Distance trick	Minimize $d^2$ instead of $d$
Domain	Always determine the feasible domain

Up Next: Part 7 — Comprehensive Review & Assessment.

Part 7: Comprehensive Assessment

📈 Applications of Derivatives — Review

Part 7 of 7 — Comprehensive Assessment

Complete Topic Summary

Part	Topic	Key Tool
1	Critical Points & First Derivative Test	Sign chart of $f'$
2	Second Derivative & Concavity	Sign of $f''$
3	Absolute Extrema	Candidates Test: CPs + endpoints
4	Curve Sketching	7-step procedure, reading $f'$ graphs
5	Mean Value Theorem	$f'(c) = \frac{f(b)-f(a)}{b-a}$
6	Optimization	Objective + constraint → single variable

Quick Reference

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

Decision Guide: Which Test to Use?

Scenario	Best Approach
Classify a critical point	First or Second Derivative Test
Find abs extrema on $[a,b]$	Candidates Test
$f''(c) = 0$ at critical point	Use First Derivative Test (SDT inconclusive)
Reading a graph of $f'$	$f'$ zeros = potential extrema; $f'$ extrema = inflection
Justify on free-response	State hypotheses, then conclusion
Optimization	Write objective, use constraint, differentiate
MVT application	Verify continuous on $[a,b]$ , differentiable on $(a,b)$

Common AP Mistakes to Avoid

Mistake	Correction
$f'(c) = 0$ → extremum	Need sign change (or use SDT)
$f''(c) = 0$ → inflection	Need sign change in $f''$
Forgetting endpoints in Candidates Test	ALWAYS check $f(a)$ and $f(b)$
Not justifying optimization answer	Must verify max/min (FDT, SDT, or Candidates)
Using SDT when $f''(c) = 0$	Switch to FDT — SDT is inconclusive
MVT without stating hypotheses	Must say "continuous" and "differentiable"

Comprehensive Assessment 🎯

Mixed Applications 🎯

AP-Style Table Problem 🔍

$x$	$1$	$3$	$5$	$7$
$f(x)$	$2$	$10$	$6$	$14$

$f$ 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

Claim	Required Justification
$f$ has a local max at $c$	$f'$ changes $+ \to -$ at $c$
$f$ has a local min at $c$	$f'$ changes $- \to +$ at $c$
$f$ has inflection at $c$	$f''$ changes sign at $c$
$f$ has abs max/min	Candidates Test: list ALL values
MVT applies	State: continuous on $[a,b]$ , differentiable on $(a,b)$
Optimization answer is a max/min	FDT, SDT, or only critical point argument

Applications of Derivatives

Applications of Derivatives - Complete Interactive Lesson

Part 1: Critical Points & Increasing/Decreasing

📈 Applications of Derivatives

Critical Points

Why Critical Points Matter

First Derivative Test for Increasing/Decreasing

Worked Example

First Derivative Test for Local Extrema

Complete Worked Example

Key Takeaways — Part 1

Part 2: Second Derivative & Concavity

📈 Applications of Derivatives

Concavity

Inflection Points

Second Derivative Test for Extrema

Worked Example

First vs Second Derivative Test

Connecting fff, f′f'f′, and f′′f''f′′

Key Takeaways — Part 2

Part 3: Absolute (Global) Extrema

📈 Applications of Derivatives

Extreme Value Theorem (EVT)

Candidates Test (Closed Interval Method)

Worked Example

Absolute Extrema on Open or Infinite Intervals

Example: One Critical Point

Key Takeaways — Part 3

Part 4: Curve Sketching

📈 Applications of Derivatives

The 7-Step Procedure

Complete Worked Example

Reading f′f'f′ Graphs — AP Essential Skill

Polynomial Curve Sketching Shortcuts

Multiplicity and Behavior at Roots of f′f'f′

Key Takeaways — Part 4

Part 5: Mean Value Theorem

📈 Applications of Derivatives

Statement (MVT)

MVT vs Rolle's Theorem

Worked Example 1

Worked Example 2 (Table Data — AP Style)

Important Consequences of MVT

Common MVT Justification Template (AP Free-Response)

Key Takeaways — Part 5

Part 6: Related Rates (Mini-Review)

📈 Applications of Derivatives

The Optimization Framework

Classic Example 1: Maximize Area

Classic Example 2: Minimize Material

Minimizing Distance

Common Optimization Setups

Key Takeaways — Part 6

Part 7: Comprehensive Assessment

📈 Applications of Derivatives — Review

Complete Topic Summary

Quick Reference

Decision Guide: Which Test to Use?

Common AP Mistakes to Avoid

Applications of Derivatives — Complete! ✅

AP Free-Response Justification Checklist

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

Reading $f'$ Graphs — AP Essential Skill

Multiplicity and Behavior at Roots of $f'$