Functions Useful in Economics and Management

NoteLearning objectives

• Identify a single-variable function and determine its domain of definition.
• Study the sign, the sense of variation, and the parity of a function.
• Recognise the usual functions (affine, quadratic, square root, logarithm, exponential).
• Solve simple and general second-degree equations.

1 Diagnostic Activity

Answer these questions before reading the module. They take about five minutes and will show you where to focus your attention.

1. What type of function is \(f(x) = 3x + 2\): affine, quadratic, exponential, or logarithmic?
2. Compute the discriminant of \(x^2 - 5x + 6 = 0\).
3. Is \(\ln(x)\) defined for \(x = -1\)? Why?

NoteCheck your answers

1. Affine — it has the form \(ax + b\).
2. \(\Delta = (-5)^2 - 4(1)(6) = 25 - 24 = 1\).
3. No — the logarithm is only defined for \(x > 0\). \(x = -1\) is a forbidden value.

Use your results to identify topics needing attention.

2 Opening Problem

A firm’s sales data shows rapid growth in the first year, followed by a gradual slowdown as the market matures. A consultant proposes modelling monthly revenue with one of four function types: affine, quadratic, exponential, or logarithmic.

Which function type best captures rapid initial growth that later flattens off? Write down your reasoning before continuing.

3 Core Concepts

Single-Variable Functions

NoteDefinition

A function is a process that associates to a number a unique number. We write \(f : x \mapsto f(x)\); \(x\) is the variable and \(f(x)\) is the image of \(x\).

Functions are very useful: they let us model a situation, produce a graphical representation, or solve problems.

Domain of definition and forbidden values. The domain \(D_f\) is the set of values of \(x\) for which \(f(x)\) can be computed. Some values are forbidden, in particular:

• values that make a denominator zero (division by zero is undefined);
• negative values under a square root;
• non-positive values inside a logarithm.

Sign. Studying the sign of \(f\) means determining for which \(x\) we have \(f(x) > 0\), \(f(x) = 0\), or \(f(x) < 0\) — for example, the values where a profit function is positive.

Sense of variation. A function is increasing on an interval if \(x\) rising implies \(f(x)\) rising, and decreasing if \(x\) rising implies \(f(x)\) falling.

Parity. A function is even if \(f(-x) = f(x)\) (graph symmetric about the \(y\)-axis) and odd if \(f(-x) = -f(x)\) (graph symmetric about the origin).

The Usual Functions

TipReference functions

• Affine: \(f(x) = a x + b\). A straight line; slope \(a\), intercept \(b\). Increasing if \(a > 0\), decreasing if \(a < 0\). Domain \(\mathbb{R}\).
• Degree-2 polynomial: \(f(x) = a x^{2} + b x + c\) with \(a \neq 0\). A parabola, opening upward if \(a > 0\), downward if \(a < 0\). Domain \(\mathbb{R}\).
• Square root: \(f(x) = \sqrt{x}\). Increasing; domain \([0, +\infty[\).
• Logarithm: \(f(x) = \ln x\). Increasing; domain \(]0, +\infty[\); turns products into sums: \(\ln(ab) = \ln a + \ln b\).
• Exponential: \(f(x) = e^{x}\). Increasing and always positive; domain \(\mathbb{R}\); turns sums into products: \(e^{a+b} = e^{a} e^{b}\).

TipInverse relationship between \(\ln\) and \(\exp\)

\(\ln(e^x) = x\) and \(e^{\ln x} = x\) for \(x > 0\). These inverse functions are central to finance: the exponential models compound growth; the logarithm measures the time needed to reach a target.

Second-Degree Equations

A simple second-degree equation such as \(x^{2} = 2\) is solved directly: \(x = \sqrt{2}\) or \(x = -\sqrt{2}\). Note that \(x^2 = k\) has no real solution when \(k < 0\).

For the general equation \(a x^{2} + b x + c = 0\) (with \(a \neq 0\)), we use the discriminant \(\Delta = b^{2} - 4ac\).

TipRoots of a quadratic

• If \(\Delta > 0\): two distinct real roots \(x = \dfrac{-b \pm \sqrt{\Delta}}{2a}\).
• If \(\Delta = 0\): one double root \(x = \dfrac{-b}{2a}\).
• If \(\Delta < 0\): no real root.

CautionDerivation — the quadratic formula (completing the square, click)

Start from \(a x^{2} + b x + c = 0\) and divide by \(a \neq 0\): \(x^{2} + \dfrac{b}{a}x + \dfrac{c}{a} = 0\). Complete the square:

\[ \left(x + \frac{b}{2a}\right)^{2} - \frac{b^{2}}{4a^{2}} + \frac{c}{a} = 0 \;\Longrightarrow\; \left(x + \frac{b}{2a}\right)^{2} = \frac{b^{2} - 4ac}{4a^{2}}. \]

Writing \(\Delta = b^{2} - 4ac\), if \(\Delta \geq 0\) we take square roots: \(x + \dfrac{b}{2a} = \pm \dfrac{\sqrt{\Delta}}{2a}\), hence

\[ x = \frac{-b \pm \sqrt{\Delta}}{2a}. \]

If \(\Delta < 0\) the right-hand side is negative and no real square root exists, so there is no real solution.

Worked example — solving a quadratic

Solve \(x^{2} - 5x + 6 = 0\). Here \(\Delta = 25 - 24 = 1 > 0\), so \(x = \dfrac{5 \pm 1}{2}\), giving \(x = 3\) or \(x = 2\). (Consistent with the factored form \((x - 2)(x - 3) = 0\).)

Practice exercises

CautionExercises (click for solutions)

1. Domain of \(f(x) = \ln(x - 2)\) → \(]2, +\infty[\).
2. Solve \(2x^{2} + 3x - 2 = 0\). \(\Delta = 9 + 16 = 25\), so \(x = \dfrac{-3 \pm 5}{4}\), i.e. \(x = \tfrac12\) or \(x = -2\).
3. Is \(f(x) = x^{2}\) even or odd? \(f(-x) = (-x)^2 = x^2 = f(x)\), so even.
4. A profit function is \(P(x) = -x^{2} + 10x - 16\). For which \(x\) is the firm profitable (\(P(x) > 0\))? Roots: \(\Delta = 100 - 64 = 36\), \(x = 2\) or \(x = 8\); since the parabola opens downward, \(P(x) > 0\) for \(x \in \,]2, 8[\).

4 Worked Example

Worked example — identifying and using the right function

Part A — Domain of definition

Find the domain of \(f(x) = \dfrac{\sqrt{x-1}}{x-3}\).

We need \(x - 1 \geq 0\) (square root) and \(x - 3 \neq 0\) (denominator), so \(x \geq 1\) and \(x \neq 3\). The domain is \([1, 3[\,\cup\,]3, +\infty[\).

Part B — Revenue model

A firm’s monthly revenue (thousands of euros) is modelled by \(R(t) = -2t^2 + 20t + 5\), where \(t\) is the number of months since launch.

1. Month of maximum revenue. \(R'(t) = -4t + 20 = 0 \Rightarrow t = 5\). Maximum revenue: \(R(5) = -2(25) + 20(5) + 5 = -50 + 100 + 5 = 55\) thousand €.

2. Months when revenue is positive. We solve \(R(t) = 0\): \(\Delta = 20^2 - 4(-2)(5) = 400 + 40 = 440\). Since \(\sqrt{440} \approx 20.98\), the roots are \(t = \dfrac{-20 \pm \sqrt{440}}{2(-2)} = \dfrac{20 \mp \sqrt{440}}{4}\), giving \(t \approx -0.25\) and \(t \approx 10.25\). Because the parabola opens downward (\(a = -2 < 0\)), revenue is positive for \(t \in \,]0, 10.25[\), i.e., during approximately the first 10 months of activity.

5 Guided Practice

A new subscription service models its cumulative subscriber count after \(t\) weeks as \(N(t) = 500 \times e^{0.12t}\).

1. How many subscribers does the service have at launch (\(t = 0\))?
2. How many subscribers does it have after 10 weeks?
3. After how many weeks does the count reach 2,000 subscribers?

TipHint

For Q3, solve \(500 \times e^{0.12t} = 2000\), i.e., \(e^{0.12t} = 4\). Apply \(\ln\) to both sides: \(0.12t = \ln(4)\).

CautionShow the solution

• At launch: \(N(0) = 500 \times e^0 = 500\) subscribers.
• After 10 weeks: \(N(10) = 500 \times e^{1.2} \approx 500 \times 3.320 \approx 1{,}660\) subscribers.
• Reach 2,000: \(0.12t = \ln(4) \approx 1.386 \Rightarrow t \approx 11.55\) weeks, i.e., approximately 12 weeks.

6 Interactive Exploration

Use the controls below to explore how each function family behaves as you vary its parameters.

viewof funcType = Inputs.select(
  ["affine", "quadratic", "exponential", "logarithmic"],
  {value: "affine", label: "Function type"}
)

viewof paramA = Inputs.range([-5, 5], {
  value: 1,
  step: 0.5,
  label: "Parameter a"
})

viewof paramB = Inputs.range([-10, 10], {
  value: 0,
  step: 0.5,
  label: "Parameter b"
})

xDomain = Array.from({length: 200}, (_, i) => -5 + i * 0.05)

funcData = xDomain.flatMap(x => {
  let y
  if (funcType === "affine") y = paramA * x + paramB
  else if (funcType === "quadratic") y = paramA * x * x + paramB
  else if (funcType === "exponential") y = Math.exp(paramA * x) + paramB
  else if (funcType === "logarithmic") {
    if (x <= 0) return []
    y = paramA * Math.log(x) + paramB
  }
  if (y === undefined || !isFinite(y) || Math.abs(y) > 30) return []
  return [{x, y}]
})

funcChart = Plot.plot({
  marks: [
    Plot.line(funcData, {x: "x", y: "y", stroke: "steelblue"}),
    Plot.ruleY([0], {stroke: "#aaa", strokeDash: [2,2]}),
    Plot.ruleX([0], {stroke: "#aaa", strokeDash: [2,2]})
  ],
  x: {domain: [-5, 5], label: "x"},
  y: {domain: [-15, 15], label: "f(x)"},
  width: 500,
  height: 320
})

domainText = funcType === "logarithmic"
  ? "Domain: ]0, +∞[ (x must be strictly positive)"
  : "Domain: ℝ (all real numbers)"

html`
<div style="margin-bottom:1rem">${funcChart}</div>
<div class="metric-grid">
  <div class="metric-card">
    <span class="metric-label">Function type</span>
    <strong>${funcType}</strong>
  </div>
  <div class="metric-card">
    <span class="metric-label">Domain</span>
    <strong>${domainText}</strong>
  </div>
</div>
`

Investigation

1. Select logarithmic. What happens when you try to evaluate at \(x = 0\) or \(x < 0\)? Why does the graph start at \(x > 0\)?
2. Switch to quadratic and set \(a > 0\), then \(a < 0\). How does the sign of \(a\) affect the graph shape and the location of the maximum or minimum?
3. Which function type best models the “rapid growth then plateau” pattern from the Opening Problem? Set the parameters to demonstrate your answer.

7 Applied Problem

A firm’s profit function (in thousands of euros) is \(P(x) = -x^2 + 10x - 16\), where \(x\) is the number of units produced (in hundreds).

1. For what production levels is the firm profitable (i.e., \(P(x) > 0\))?
2. Find the production level that maximises profit.
3. What is the maximum profit?
4. Below what production level does the firm begin losing money?

8 Common Errors

WarningCommon error — using ln on a non-positive argument

A student writes \(\ln(-3) = -\ln(3)\) by analogy with the rule \(\ln(1/x) = -\ln(x)\).

This is incorrect. The natural logarithm \(\ln(x)\) is only defined for \(x > 0\). \(\ln(-3)\) does not exist in the real numbers.

The rule \(\ln(1/x) = -\ln(x)\) applies only when \(x > 0\), so that both \(x\) and \(1/x\) are positive.

9 Check Your Understanding

1. Find the domain of \(g(x) = \sqrt{2x - 6}\).
2. Is \(f(x) = x^3\) an even function, an odd function, or neither?
3. A demand function is \(D(p) = -3p + 90\). For what price range is demand positive?
4. Solve \(2x^2 - 7x + 3 = 0\).
5. A student says the exponential \(e^x\) can be negative for some values of \(x\). Is this correct? Why?

10 Summary

1. A function assigns a unique output to each input; the domain is the set of valid inputs — forbidden values arise from division by zero, negative arguments under a square root, and non-positive arguments in a logarithm.
2. The four main function families (affine, quadratic, exponential, logarithmic) model different economic relationships: linear proportionality, parabolic revenue/cost, exponential growth/decay, and logarithmic saturation.
3. The discriminant \(\Delta = b^2 - 4ac\) determines whether a quadratic equation has two real roots, one double root, or no real root — and identifies the break-even or profit boundaries of a quadratic profit model.

11 Independent Work

• Find the domain for functions 1–8 in the exercise bank.
• Solve the quadratic equations in problems 1–6.
• Identify the function type in the applied-context problems.
• Record errors in your error log.
• Review glossary entries for: domain of definition, discriminant, exponential, logarithm, sense of variation.

12 Continue

Continue to Module 7 →