Session 05-05 - Graphical Calculus Mastery

Section 05: Differential Calculus

Author

Dr. Nikolai Heinrichs & Dr. Tobias Vlćek

Entry Quiz

Quick Review from Session 05-04

Test your understanding of chain rule and implicit differentiation

  1. Differentiate \(f(x) = (2x^3 - 5)^4\) using the chain rule.

  2. Find \(\frac{dy}{dx}\) if \(x^2 + xy = 10\).

  3. If \(f'(3) = 0\), what does this tell about the graph of \(f(x)\) at \(x = 3\)?

Homework Discussion

Your questions from Session 05-04

What questions do you have regarding the previous session?

Learning Objectives

What You’ll Master Today

  • Sketch \(f'(x)\) from the graph of \(f(x)\) by analyzing slopes
  • Determine properties of \(f(x)\) from the graph of \(f'(x)\)
  • Understand concavity through second derivatives \(f''(x)\)
  • Identify critical points and their classification
  • Find inflection points where concavity changes
  • Master visual analysis - a heavily tested exam skill!
NoteKey Insight

Being able to translate between graphs of \(f\), \(f'\), and \(f''\) is one of the most important visual skills in calculus and frequently tested!

Function to Derivative

Reading the Derivative Graph

The graph of \(f'(x)\) shows the slope of \(f(x)\) at each point.

What to look for in \(f(x)\):

  • Where \(f\) is increasing → \(f'(x) > 0\) (derivative is positive)
  • Where \(f\) is decreasing → \(f'(x) < 0\) (derivative is negative)
  • Where \(f\) has a horizontal tangent → \(f'(x) = 0\) (crosses x-axis)
  • Where \(f\) is steep → \(|f'(x)|\) is large
  • Where \(f\) is flat → \(|f'(x)|\) is small

This shouldn’t be too complicated, right?

Sketching \(f'\) from \(f\)

A parabola and its derivative: \(f(x) = x^2\)

Two panels. Left: the parabola f(x) = x², decreasing left of its minimum at x = 0 and increasing right of it. Right: the derivative f'(x) = 2x, a straight line that is negative for x < 0, crosses zero at x = 0 and is positive for x > 0.

\(f\) has minimum at \(x = 0\) → \(f'\) crosses zero

Steps: Sketching the Derivative

  1. Identify critical points: Where does \(f\) have horizontal tangents? Mark these as zeros of \(f'\)

  2. Determine sign: Where is \(f\) increasing/decreasing? Make \(f'\) positive/negative accordingly

  3. Consider steepness:

    • Where is \(f\) very steep? Make \(|f'|\) large.
    • Where is \(f\) nearly flat? Make \(|f'|\) small.

  4. Check concavity: Is \(f'\) increasing or decreasing? This tells you about the concavity of \(f\)

Complex Example

A cubic function and its derivative.

Two panels. Left: the cubic f(x) = x³ − 3x with a local maximum at (−1, 2) and a local minimum at (1, −2). Right: its derivative f'(x) = 3x² − 3, a parabola with zeros at x = −1 and x = 1, positive outside them and negative in between.

\(f\) increasing, then decreasing, then increasing → \(f'\) crosses zero twice

Critical Points

A point \(x = c\) where either:

  • \(f'(c) = 0\) (horizontal tangent), or
  • \(f'(c)\) does not exist (corner, cusp, vertical tangent)
TipClassification of Critical Points
  • Local maximum: \(f'\) changes from positive to negative
  • Local minimum: \(f'\) changes from negative to positive
  • Saddle point (stationary point without extremum): \(f'\) doesn’t change sign (e.g., \(f(x) = x^3\) at \(x = 0\))

Derivative to Function

Reading Information from \(f'(x)\)

Given the graph of \(f'(x)\), we can determine:

  1. Where \(f\) is increasing/decreasing:

    • \(f'(x) > 0\) → \(f\) is increasing
    • \(f'(x) < 0\) → \(f\) is decreasing

  2. Where \(f\) has local extrema: \(f'\) crosses zero

    • Sign change determines type!

  3. Where \(f\) is steepest: Where \(|f'(x)|\) is largest

  4. Relative heights: Cannot determine absolute \(y\)-values!

    • Can determine relative changes!

Example: Reading from \(f'(x)\)

Two panels. Left, given: the derivative f'(x) = −x² + 4, a downward-opening parabola with zeros at x = −2 and x = 2, positive between them. Right: a possible f that falls to a local minimum at x = −2, rises to a local maximum at x = 2 and falls again.

  • \(f' > 0\) on \((-2; 2)\) → \(f\) increasing, \(f' < 0\) for \(|x| > 2\) → \(f\) decreasing
  • \(f'\) changes at \(x = -2, 2\) → \(f\) has local extrema

Quick Practice

Individual Exercise I

Sketch the derivative of this function!

Given graph of f made of three straight pieces: it rises from (−3, −1) to a marked corner at (−1, 1), runs horizontally to a second marked corner at (1, 1) and then falls to (3, −1).

Individual Exercise II

The Questions:

  1. Where is \(f'(x) > 0\), \(f'(x) < 0\), and \(f'(x) = 0\)?

  2. Sketch the graph of \(f'(x)\).

  3. At what points does \(f'(x)\) not exist?

Break

Concavity

Understanding Concavity

Describes the curving behavior of a function.

  • Concave up (⌣): Curves upward like a smile (“holds water”)
    • \(f''(x) > 0\)
    • The slope \(f'(x)\) is increasing
  • Concave down (⌢): Curves downward like a frown (“spills water”)
    • \(f''(x) < 0\)
    • The slope \(f'(x)\) is decreasing

Visualizing Concavity

Concave up vs. concave down

Two panels with short dashed tangent segments drawn along each curve. Left: f(x) = x² is concave up, its tangent slopes increase from left to right. Right: f(x) = −x² + 4 is concave down, its tangent slopes decrease from left to right.

\(f''(x)\) tells us how \(f'(x)\) is changing, which determines the shape of \(f(x)\).

Inflection Points

A point where the curvature (concavity) changes sign!

  • \(f''(x) = 0\) or \(f''(x)\) does not exist
  • Concavity changes (from ⌣ to ⌢ or vice versa)

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

  • \(f'(x) = 3x^2\)
  • \(f''(x) = 6x\)
  • \(f''(0) = 0\) and concavity changes at \(x = 0\)
  • So \((0, 0)\) is an inflection point

Relationships Between \(f\), \(f'\), and \(f''\)

Three panels for f(x) = x³ − 3x. f has a local maximum at (−1, 2), a local minimum at (1, −2) and an inflection point at the origin. f'(x) = 3x² − 3 is zero at x = −1 and x = 1 and has its minimum at x = 0. f''(x) = 6x is negative left of 0 and positive right of it, crossing zero at the inflection point.

  • Where \(f'' < 0\): \(f\) concave down, \(f'\) decreasing, where \(f'' > 0\): \(f\) concave up, \(f'\) increasing
  • Where \(f'' = 0\): Inflection point in \(f\), extremum in \(f'\)

Complete Analysis

The Complete Picture

Given a function \(f(x)\), complete analysis involves:

  1. Critical points: Solve \(f'(x) = 0\)
  2. First derivative test: Check sign changes of \(f'\) to classify extrema
  3. Inflection points: Solve \(f''(x) = 0\) and check for concavity change
  4. Intervals: Determine where \(f\) is increasing/decreasing and where concave up/down
  5. Key points: Evaluate \(f\) at critical points and inflection points

Example: Complete Analysis

Analyze \(f(x) = x^4 - 4x^3\)

First derivative:

\[f'(x) = 4x^3 - 12x^2 = 4x^2(x - 3)\]

Critical points: \(x = 0, 3\)

Second derivative:

\[f''(x) = 12x^2 - 24x = 12x(x - 2)\]

Inflection points: \(x = 0, 2\)

Analysis Visualization

Three stacked graphs and a text summary. f has a flat critical point at x = 0 that is no extremum and a local minimum at (3, −27); its inflection points are at x = 0 and x = 2. f'(x) = 4x²(x − 3) touches zero at x = 0 and turns from negative to positive at x = 3. f''(x) = 12x(x − 2) is negative only between 0 and 2, where f is concave down.

Complete analysis of f(x) = x⁴ - 4x³

Guided Practice

Practice Set A: Derivative Sketches

For each function graphed below, sketch \(f'(x)\) and identify:

  • Where \(f'(x) > 0, < 0, = 0\) and any points where \(f'\) does not exist

Problem 1: a full sine wave from 0 to 2π. Problem 2: a V with its corner at the origin. Problem 3: a W shape with a local maximum at (0, 1) and minima at height −1 near x = ±1.4. Problem 4: constant −1 for x < −1, a parabola arc from 0 down to −1 and back to 0 between x = −1 and 1, constant 1 for x > 1, with a vertical step at each end of the arc.

Practice B: From Derivative I

Continue working for 10 minutes

Given the graph of \(f'(x)\), answer:

  1. Where is \(f(x)\) increasing/decreasing?
  2. Where does \(f(x)\) have local extrema? Classify them.
  3. Sketch a possible graph of \(f(x)\).

Practice B: From Derivative II

Given graph of the derivative f'(x) = x² − 4: an upward-opening parabola with its lowest point at (0, −4) and marked zeros at x = −2 and x = 2. It lies above the x-axis outside the zeros and below it between them.

Quick Practice: Derivative Sketch

Work individually for 5 minutes

For each function described, sketch both \(f(x)\) and \(f'(x)\):

  1. \(f(x) = x^3 - 3x\) (cubic with local max and min)

  2. \(f(x) = |x - 2|\) (V-shape shifted right)

  3. \(f(x)\) is constant for \(x < 0\), then increases linearly for \(x \geq 0\)

  4. \(f(x)\) has \(f'(x) > 0\) everywhere but \(f'(x) \to 0\) as \(x \to \infty\)

Coffee Break

Business Applications

Profit Function Analysis

Business Context: A company’s monthly profit (in thousands EUR) is modeled by: \[P(t) = -t^3 + 12t^2 - 36t + 50\] where \(t\) is months since product launch.

Questions:

  1. When is profit increasing/decreasing?
  2. When does profit reach local extrema?
  3. When is the rate of profit change accelerating/decelerating?

Profit Analysis Solution

First derivative (profit rate of change): \[P'(t) = -3t^2 + 24t - 36 = -3(t^2 - 8t + 12) = -3(t - 2)(t - 6)\]

Critical points: \(t = 2, 6\) months

Second derivative (acceleration of profit change): \[P''(t) = -6t + 24 = -6(t - 4)\]

Inflection point: \(t = 4\) months

Profit Visualization

Three panels over months 0 to 10. Profit P(t) falls to a local minimum at month 2, rises through an inflection point at month 4 to a local maximum at month 6, then falls. The profit rate P'(t) is positive only between months 2 and 6. The acceleration P''(t) is a falling line that crosses zero at month 4.

Question: How would you describe the behaviour?

Profit Interpretation

TipReading Business Graphs: Quick Rules
  • \(P'(t) > 0\) → Profit is growing (good news!)
  • \(P'(t) < 0\) → Profit is shrinking (warning sign)
  • \(P'(t) = 0\) → Profit has reached a turning point (decision time)

The sign of the derivative tells you the direction of change!

NoteInflection Points in Business

The inflection point (\(P''(t) = 0\) at month 4) marks where:

  • Before: Profit is accelerating (growth speeding up)
  • After: Profit is decelerating (growth slowing down)

Business insight: Month 4 is when the company should start planning for the eventual peak. Growth is still positive but momentum is fading!

Team Problem-Solving

Challenge: Full Function Analysis

Scenario: A manufacturing company models their weekly profit (in thousands EUR) as a function of production quantity:

\[\Pi(q) = -0.02q^3 + 1.5q^2 - 20q + 50\]

where \(q\) is quantity produced (in hundreds of units), with \(q \in [0; 50]\).

Tasks

Work in groups of 3-4

  1. Find \(\Pi'(q)\) and determine where profit is increasing vs. decreasing. Interpret economically.

  2. Find \(\Pi''(q)\) and locate any inflection points - what do they say about profit growth?

  3. Sketch the graphs of \(\Pi(q)\), \(\Pi'(q)\), and \(\Pi''(q)\) on \(q \in [0; 50]\).

  4. Find all critical points and classify them as local maxima or minima using the sign change of \(\Pi'(q)\).

  5. Business question: Current production is 1,500 units (\(q = 15\)) - raise or lower it, and by how much? Justify with calculus.

Think-Pair-Share

Discussion Question

Think individually, then discuss with class

Question: Consider these scenarios:

  • Company A: Stock price \(S_A(t)\) is \(S'_A > 0\) but \(S''_A < 0\)
  • Company B: Stock price \(S_B(t)\) is \(S'_B < 0\) but \(S''_B > 0\)
  1. Which company is in a better position right now?
  2. Which company shows more promising momentum for the future?
  3. In financial terms, what do \(S'\) and \(S''\) represent?
  4. Can you think of real-world examples of each scenario?

Wrap-Up & Takeaways

The Derivative Overview

If you know… You can determine…
\(f' > 0\) \(f\) is increasing
\(f' < 0\) \(f\) is decreasing
\(f' = 0\) Possible local extremum
\(f'' > 0\) \(f\) concave up, \(f'\) increasing
\(f'' < 0\) \(f\) concave down, \(f'\) decreasing
\(f'' = 0\) Possible inflection point

These are heavily Tested Skills in FSP!!!

Final Assessment

Quick Check I

Consider this function:

Given graph of f: a curve that rises steeply from the lower left, levels off to a momentarily flat stretch at the marked point (1, 0) and then rises ever more steeply to the upper right. It never falls.

Quick Check II

Work individually, then we compare

  1. Where is \(f'(x) > 0\)?

  2. At \(x = 1\), is this a local max, local min, or neither?

  3. Where is \(f(x)\) concave up?

  4. True or False: If \(f'(c) = 0\), then \(f\) must have a local extremum at \(x = c\).

Next Session Preview

Session 05-06

Optimization & Curve Sketching

  • First and second derivative tests for extrema classification
  • Global maxima/minima on closed intervals
  • Complete curve sketching algorithm (6 steps)
  • Business optimization: profit maximization, cost minimization
  • Interpreting results in real-world context

Complete Tasks 05-05!

Appendix

Three panels over quantity q from 0 to 50. Profit falls to a local minimum near q = 7.9, rises through the inflection point at q = 25 to a local maximum near q = 42.1, then falls. Marginal profit is a downward-opening parabola, positive between the two critical points. The second derivative is a falling line that crosses zero at q = 25.

Profit function and its derivatives