Session 05-07 - Function Determination & Funktionsscharen

Section 05: Differential Calculus

Author

Dr. Nikolai Heinrichs & Dr. Tobias Vlćek

Entry Quiz

Quick Review from Session 05-06

Test your understanding of optimization and curve sketching

  1. For \(f(x) = x^3 - 6x^2 + 9x\), find all critical points and classify them.

  2. What is the difference between a local maximum and an absolute maximum?

  3. Find the absolute extrema of \(g(x) = x^2 - 4x + 1\) on \([0; 3]\).

  4. A profit function is \(P(x) = -2x^2 + 40x - 100\). What production level maximizes profit?

Homework Discussion

Your questions from Session 05-06

What questions do you have regarding the previous session?

Learning Objectives

What You’ll Master Today

  • Set up systems of equations from function conditions
  • Apply point conditions to determine unknown coefficients
  • Use tangent/slope conditions with derivatives
  • Incorporate extrema conditions (f’(a) = 0 and f(a) = b)
  • Master Funktionsscharen (function families with parameters)
  • Analyze how parameters affect zeros, extrema, inflection points
  • Apply systematic problem-solving to business scenarios

Funktionsscharen are heavily tested on exams! Both function determination and Funktionsscharen share a key skill: setting up equations from conditions systematically.

Point Conditions

The General Approach

Strategy for Finding Unknown Functions:

  1. Choose the function form (polynomial degree, rational, etc.)
  2. Count unknowns (how many coefficients?)
  3. Identify conditions (points, slopes, extrema, etc.)
  4. Set up equations from conditions
  5. Solve the system
  6. Verify the solution

Number of conditions = Number of unknowns

For a function with \(n\) unknowns, you need exactly \(n\) independent conditions!

Example: Quadratic Three Points

Problem:

  • Find the quadratic function \(f(x) = ax^2 + bx + c\) passing through \((1, 4)\), \((2, 3)\), and \((3, 4)\).

Solution:

  • Three unknowns \(a, b, c\) - so we need three point conditions
    • \(f(1) = 4: \quad a(1)^2 + b(1) + c = 4 \quad \implies \quad a + b + c = 4\)
    • \(f(2) = 3: \quad a(4) + b(2) + c = 3 \quad \implies \quad 4a + 2b + c = 3\)
    • \(f(3) = 4: \quad a(9) + b(3) + c = 4 \quad \implies \quad 9a + 3b + c = 4\)

Solving the System

System of equations:

\(\begin{cases} a + b + c = 4 \\ 4a + 2b + c = 3 \\ 9a + 3b + c = 4 \end{cases}\)

Answer: \(f(x) = x^2 - 4x + 7\)

Check: Does \(f(x) = x^2 - 4x + 7\) pass through all three points?

  • \(f(1) = 1 - 4 + 7 = 4\) ✓
  • \(f(2) = 4 - 8 + 7 = 3\) ✓
  • \(f(3) = 9 - 12 + 7 = 4\) ✓

Visualizing the Solution

Parabola f(x) = x² − 4x + 7 running exactly through the three given points (1, 4), (2, 3) and (3, 4); the middle point (2, 3) is its vertex.

Quadratic through three points

Derivative Conditions

Using Derivative Conditions

Key Idea: Derivative conditions give us additional equations!

Common derivative conditions:

  1. Slope at a point: \(f'(a) = m\)

    • “The tangent line at \(x = a\) has slope \(m\)”

  2. Horizontal tangent: \(f'(a) = 0\)

    • “The function has a horizontal tangent at \(x = a\)”

  3. Parallel tangents: \(f'(a) = f'(b)\)

    • “Slopes at two points are equal”

Remember: Each derivative condition counts as one equation!

Example 2: Quadratic Point & Slope

Problem:

  • Find \(f(x) = ax^2 + bx + c\) such that:
    • Passes through \((1, 3)\)
    • Has slope \(2\) at \(x = 1\)
    • Passes through \((2, 5)\)

Solution:

Step 1: \(f(x) = ax^2 + bx + c\) and \(f'(x) = 2ax + b\) (3 unknowns)

Solving with Derivative Conditions

Step 2: Set up equations from conditions:

  1. Point \((1, 3)\): \(a + b + c = 3\)
  2. Slope at \(x=1\): \(f'(1) = 2a + b = 2\)
  3. Point \((2, 5)\): \(4a + 2b + c = 5\)

System: \(\begin{cases} a + b + c = 3 \\ 2a + b = 2 \\ 4a + 2b + c = 5 \end{cases}\)

Answer: \(f(x) = 2x + 1\) (actually linear, not quadratic!)

Extrema Conditions

Extrema Give TWO Conditions

Important: When a function has an extremum (max or min) at \((a, b)\):

You get TWO conditions:

  1. Point condition: \(f(a) = b\) (the function passes through the point)
  2. Horizontal tangent: \(f'(a) = 0\) (slope is zero at extremum)

Total: 2 equations from one extremum condition!

Don’t forget the \(f'(a) = 0\) condition!

An extremum at \((a, b)\) gives you both the point and the derivative condition.

Example 3: Quadratic & Maximum

Problem: Find the quadratic function with a maximum at \((2, 5)\) that passes through \((0, 1)\).

Solution:

\(f(x) = ax^2 + bx + c\) (3 unknowns)

Conditions:

  1. Point \((0, 1)\): \(f(0) = 1\)
  2. Maximum at \((2, 5)\): \(f(2) = 5\)
  3. Horizontal tangent at max: \(f'(2) = 0\)

Solving the Extremum System

Important: \(f'(x) = 2ax + b\)

Equations:

\(\begin{cases} c = 1 & \text{from } f(0) = 1 \\ 4a + 2b + c = 5 & \text{from } f(2) = 5 \\ 4a + b = 0 & \text{from } f'(2) = 0 \end{cases}\)

Answer: \(f(x) = -x^2 + 4x + 1\)

Verification: \(f''(x) = -2 < 0\) confirms maximum ✓

Visualizing the Maximum

Downward-opening parabola f(x) = −x² + 4x + 1 that passes through (0, 1) and peaks at (2, 5), where a dashed horizontal tangent touches it: the extremum supplies both a point and a zero-slope condition.

Quadratic with maximum at (2, 5)

Example 4: Vertex Form Parabola

Problem: Find the parabola with vertex at \((3, -2)\) through \((1, 6)\).

Two approaches:

  • Approach 1 - Standard form: \(f(x) = ax^2 + bx + c\) (harder)
  • Approach 2 - Vertex: \(f(x) = a(x - h)^2 + k\) where \((h, k)\) vertex

Now it’s easy!

  • Using vertex form with \((h, k) = (3, -2)\):
  • \(f(x) = a(x - 3)^2 - 2\)
  • Only 1 unknown (\(a\)) now!

Vertex Simplification

Condition: Passes through \((1, 6)\):

  • \(f(1) = a(1 - 3)^2 - 2 = 6\)
  • \(4a - 2 = 6\)
  • \(a = 2\)

Answer:

  • \(f(x) = 2(x - 3)^2 - 2\)
  • Expanded: \(f(x) = 2x^2 - 12x + 16\)

Break

Mixed Conditions

Multiple Conditions I

Strategy for Complex Problems:

  1. List all unknowns clearly

  2. Identify each condition type:

    • Point: \(f(a) = b\)
    • Slope: \(f'(a) = m\)
    • Extremum: \(f(a) = b\) AND \(f'(a) = 0\)
    • Inflection: \(f''(a) = 0\) (and possibly \(f(a) = b\))

Remember that the number of unknowns must match the number of conditions!

Multiple Conditions II

Strategy for Complex Problems:

  1. Write derivative(s) before setting up equations
  2. Organize your system (group similar types)
  3. Solve systematically (substitution or elimination)
  4. Verify your answer

Not too complicated, right?

Example 5: Cubic Mixed Conditions

Problem: Find \(f(x) = ax^3 + bx^2 + cx + d\) such that:

  • Has a local maximum at \((0, 4)\)
  • Has an inflection point at \((1, 2)\)

Analysis: 4 unknowns, need 4 equations

  • Derivatives:
    • \(f'(x) = 3ax^2 + 2bx + c\)
    • \(f''(x) = 6ax + 2b\)

The Conditions

Now, remember the conditions we have:

  1. \(f(0) = 4\): point condition
  2. \(f'(0) = 0\): horizontal tangent at max
  3. \(f(1) = 2\): point condition at inflection
  4. \(f''(1) = 0\): inflection point condition

You will need to know these by heart in the exam!

Solving the Equations

Problem: Find \(f(x) = ax^3 + bx^2 + cx + d\) such that:

  • Has a local maximum at \((0, 4)\)
  • Has an inflection point at \((1, 2)\)

Equations:

  1. \(f(0) = 4\): \(d = 4\)
  2. \(f'(0) = 0\): \(c = 0\)
  3. \(f(1) = 2\): \(a + b + c + d = 2\), so \(a + b = -2\)
  4. \(f''(1) = 0\): \(6a + 2b = 0\), so \(3a + b = 0\)

Verification

Answer: \(f(x) = x^3 - 3x^2 + 4\)

Check all conditions for \(f(x) = x^3 - 3x^2 + 4\):

  • \(f(0) = 4\) ✓
  • \(f'(x) = 3x^2 - 6x\), so \(f'(0) = 0\) ✓
  • \(f(1) = 1 - 3 + 4 = 2\) ✓
  • \(f''(x) = 6x - 6\), so \(f''(1) = 0\) ✓

Additional check: \(f''(0) = -6 < 0\), confirming local maximum ✓

Visualizing Mixed Conditions

Graph of f(x) = x³ − 3x² + 4 with a dashed horizontal tangent at its local maximum (0, 4) and a marked inflection point at (1, 2), where the falling curve switches from concave down to concave up.

Cubic with maximum and inflection point

Funktionsscharen

Introduction to Funktionsscharen

Definition: A Funktionsschar is a family of functions that depend on a parameter we can vary (usually \(t\), \(a\), or \(k\)).

  • Notation: \(f_t(x)\) or \(f(x, t)\)
  • Example: \(f_t(x) = x^2 - tx + 1\)
    • When \(t = 0\): \(f_0(x) = x^2 + 1\)
    • When \(t = 2\): \(f_2(x) = x^2 - 2x + 1 = (x-1)^2\)
    • When \(t = 4\): \(f_4(x) = x^2 - 4x + 1\)
  • Each value of \(t\) gives a different parabola!

Visualizing a Function Family

Four parabolas of the family f_t(x) = x² − tx + 1 for t = 0, 2, 4 and 6. All of them pass through (0, 1); as t grows, the vertex moves to the right and down, from (0, 1) to (1, 0), (2, −3) and (3, −8).

A collection of functions that are related to each other by a parameter.

Why Study Funktionsscharen?

Heavily tested on exams! Common question types:

  • For which \(t\) does \(f_t(x)\) have exactly 2 zeros?
  • For which \(t\) does \(f_t(x)\) have a local maximum at \(x = 3\)?
  • For which \(t\) is \(f_t(2) = 5\)?
  • Find the parameter \(t\) such that the inflection point is at \(x = 1\).

Practice these on your own as well, as we are coming to the end of this section!

Example

Problem: For which \(t\) does \(f_t(x) = x^2 - tx + t\) have exactly one zero?

  • Solution: Exactly one zero when the discriminant equals zero.
  • For \(x^2 - tx + t = 0\): \(\Delta = b^2 - 4ac = (-t)^2 - 4(1)(t) = t^2 - 4t\)
  • Set \(\Delta = 0\):
    • \(t^2 - 4t = 0\)
    • \(t(t - 4) = 0\)
    • \(t = 0 \text{ or } t = 4\)
  • Answer: For \(t = 0\) or \(t = 4\), the function has exactly one zero.

General Strategy: Funktionsscharen

  1. Identify condition: zeros, extrema, inflection, function values

  2. Set up the equation:

    • Zeros: Use discriminant or factor
    • Extrema: \(f_t'(x) = 0\)
    • Inflection: \(f_t''(x) = 0\)
    • Function value: \(f_t(a) = c\)

  3. Substitute the given point (if specified)

  4. Solve for the parameter \(t\)

  5. Verify your answer makes sense

Guided Practice

Set A - Work in Pairs

Complete these problems and then we discuss

  1. Find the quadratic passing through \((1, 0)\), \((2, 3)\), and \((3, 8)\).
  2. Find the cubic \(f(x) = ax^3 + bx^2 + cx + d\) with \(f(0) = 2\), \(f(1) = 3\), \(f'(0) = 1\), and \(f'(1) = 4\).
  3. Find the parabola with vertex at \((2, -1)\) passing through \((0, 7)\).
  4. For \(h_t(x) = x^2 - tx + 3\), find \(t\) such that \(h_t(2) = 5\).

Set B - Funktionsscharen Practice

Work individually for 10 minutes, then compare
For each Funktionsschar, solve the given problem:

  1. For \(f_t(x) = x^2 - 2tx + 3\), find all \(t\) such that \(f_t\) has exactly two zeros.

  2. For \(g_t(x) = tx^2 - 4x + t\), find \(t\) such that \(g_t\) has a zero at \(x = 2\).

  3. For \(h_t(x) = x^3 - tx^2 + 5x\), find \(t\) such that \(h_t\) has a local extremum at \(x = 1\).

  4. For \(p_t(x) = x^2 + tx - 2t\), find \(t\) such that \(p_t(3) = 10\).

  5. For \(q_t(x) = tx^2 - 6x + 9\), for which \(t \neq 0\) does \(q_t\) have exactly one zero?

Coffee Break

Business Applications

Cost Function from Marginal Cost

Scenario: A company knows its marginal cost is:

\[MC(x) = C'(x) = 3x^2 - 12x + 20\]

The fixed cost (when \(x = 0\)) is 500 EUR.

Question: Find the total cost function \(C(x)\).

Solution: We need to find \(C(x)\) such that \(C'(x) = 3x^2 - 12x + 20\).

Integration

Integration (reverse of differentiation) undoes the power rule term by term: raise each exponent by 1, divide by the new exponent.

\[C(x) = x^3 - 6x^2 + 20x + k\]

where \(k\) is a constant. Check: \(C'(x) = 3x^2 - 12x + 20\) ✓

Use the fixed cost condition \(C(0) = 500\):

\[C(0) = 0 - 0 + 0 + k = 500\]

\[k = 500\]

Answer: \(C(x) = x^3 - 6x^2 + 20x + 500\)

Cost Function from Four Conditions

Scenario: A company describes its cubic cost function \(C(x) = ax^3 + bx^2 + cx + d\) (in EUR, \(x\) in units) as follows:

  • Fixed costs are 200 EUR
  • Producing 10 units costs 600 EUR in total
  • At 2 units, the marginal cost is 24 EUR per unit
  • The lowest increase in costs occurs at \(x = 4\)

Question: Determine \(C(x)\).

Translating the Conditions

4 unknowns, 4 conditions - write the derivatives first:

\(C'(x) = 3ax^2 + 2bx + c, \qquad C''(x) = 6ax + 2b\)

Condition Equation
Fixed costs 200 EUR \(C(0) = 200\)
10 units cost 600 EUR \(C(10) = 600\)
Marginal cost at 2 units is 24 EUR \(C'(2) = 24\)
Lowest cost increase at \(x = 4\) \(C''(4) = 0\)

“Lowest increase in costs” = minimum of the marginal cost \(C'\), so \(C''(4) = 0\). This is the inflection point of \(C\)!

Setting up the System

  1. \(C(0) = 200\): \(d = 200\)
  2. \(C(10) = 600\): \(1000a + 100b + 10c + d = 600\)
  3. \(C'(2) = 24\): \(12a + 4b + c = 24\)
  4. \(C''(4) = 0\): \(24a + 2b = 0\)

Solving the Cost System

Solve by substitution:

  • From (4): \(b = -12a\)
  • Into (3): \(12a - 48a + c = 24 \implies c = 24 + 36a\)
  • Into (2): \(1000a - 1200a + 240 + 360a + 200 = 600 \implies 160a = 160\)

The Cost Function

Answer: \(a = 1\), \(b = -12\), \(c = 60\), \(d = 200\)

\[C(x) = x^3 - 12x^2 + 60x + 200\]

Check:

  • \(C(10) = 1000 - 1200 + 600 + 200 = 600\) ✓
  • \(C'(x) = 3x^2 - 24x + 60\), so \(C'(2) = 12 - 48 + 60 = 24\) ✓
  • \(C''(x) = 6x - 24\), so \(C''(4) = 0\) ✓ and \(C'(4) = 12 > 0\): costs still rise, but at the lowest rate

Quick Practice: Cost Function

Work in pairs

Determine the cubic cost function \(C(x)\) with:

  • Fixed costs of 100 EUR
  • Total cost of 316 EUR at 6 units
  • Marginal cost of 42 EUR at 2 units
  • The lowest increase in costs at \(x = 5\)

Demand Function Determination

Scenario: A product’s demand function is quadratic: \(D(p) = ap^2 + bp + c\) where \(p\) is price.

Known information:

  • At price 10 EUR, demand is 100 units: \(D(10) = 100\)
  • At price 20 EUR, demand is 60 units: \(D(20) = 60\)
  • The rate of change of demand at \(p = 10\) is \(-6\) units/euro
    • \(D'(10) = -6\)

Question: Find the demand function.

Solving the Demand System

\(D'(p) = 2ap + b\)

System: \[\begin{cases} 100a + 10b + c = 100 \\ 400a + 20b + c = 60 \\ 20a + b = -6 \end{cases}\]

Answer: \(D(p) = 0.2p^2 - 10p + 180\)

Visualizing the Demand Function

Demand curve falling from 180 units at price 0 through the two given points (10, 100) and (20, 60) to a minimum of 55 units at price 25, then turning up slightly. A short dashed tangent at price 10 shows the required slope of −6.

Demand function D(p) = 0.2p² - 10p + 180

Team Problem-Solving

Group Challenge I

Scenario:

An engineer is designing a roller coaster section modeled by a cubic function \(h(x) = ax^3 + bx^2 + cx + d\), where \(h\) is height (meters) and \(x\) is horizontal distance (meters).

Design requirements:

  • Starts at ground level: \(h(0) = 0\)
  • Reaches a peak (maximum) of 20 meters at \(x = 4\)
  • Returns to ground level at \(x = 10\): \(h(10) = 0\)

Group Challenge II

Work in groups of 3-4 students

  1. How many unknowns and conditions do you have?
  2. Set up the complete system of equations.
  3. Solve for the coefficients \(a, b, c, d\).
  4. Verify that \(x = 4\) is indeed a maximum (not minimum).
  5. Find the height at \(x = 2\) and \(x = 8\).
  6. Sketch the roller coaster section.

Wrap-Up & Takeaways

Summary of Session 05-07

Function Determination - Systematic Approach:

  1. Count unknowns = count conditions needed
  2. Write derivatives before setting up equations
  3. Organize conditions by type: point, slope, extremum, inflection

Funktionsscharen - Parameter Analysis:

  1. Identify the parameter (usually \(t\), \(a\), or \(k\))
  2. Set up condition equations involving the parameter
  3. Solve for parameter values meeting given criteria

Final Assessment

Quick Check

Complete individually and then discuss

  1. How many conditions do you need to find a cubic function? How many does an extremum at \((a, b)\) provide?
  2. Find the parabola with vertex at \((1, 4)\) passing through \((0, 2)\).
  3. For \(f_t(x) = x^2 - 2tx + t\), find \(t\) such that \(f_t(x)\) has exactly one zero.
  4. Find the quadratic \(g(x) = ax^2 + bx + c\) with \(g(0) = 3\), \(g(1) = 2\), and \(g'(1) = -4\).

Next Session Preview

Next Session: Mock Exam 06

Session 05-08 is exam only - no lecture

  • 180 minutes working time plus 5 minutes reading time, 100 points
  • Covers all differential calculus topics in three full-length problems - your first complete exam simulation
  • Allowed: non-programmable calculator, drawing instruments, monolingual dictionary
  • Not allowed: notes, formula sheets, any electronic devices
  • Counts toward the 60% of test points needed for exam admission

Preparing for Mock Exam 06

Complete Tasks 05-07, then prepare for Mock Exam 06 with FSP Training 05 (sidebar, Section 05 worksheets). Get a good night’s sleep and bring your calculator!

Thank You!

See you next session!