6️⃣ Functions Useful in Economics and Management

Published

September 17, 2025

Abstract

Present single-variable functions and their properties (domain, forbidden values, sign, sense of variation, parity), the usual functions (affine, degree-2 polynomial, square root, logarithm, exponential), and the solving of second-degree equations.

Learning 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.

Single-Variable Functions

Definition

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

Reference 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: \(e^{a+b} = e^{a} e^{b}\).

The logarithm and the exponential are inverse of each other: \(\ln(e^{x}) = x\) and \(e^{\ln x} = x\) for \(x > 0\). They are central to finance (continuous growth, doubling times) and to modelling elasticities.

Worked example — 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 \([1, 3[\,\cup\,]3, +\infty[\).

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 \(\Delta = b^{2} - 4ac\).

Roots 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.

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

  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 \(x \in \,]2, 8[\).
Important

TODO: interactive function/parabola plotter (vary \(a\), \(b\), \(c\)) deferred — see plan Complexity Tracking.