Mathematics

Functions: Overview

Concept of a function, domain and range, symmetries, periodicity, rate of change, and composition of functions.

A function is a relation between xx and yy that assigns exactly one value of y=f(x)y = f(x) to each value of xx.
It can be defined by an algebraic expression, a table of values, or a graph.
y=f(x)=x2−1y = f(x) = x^2 - 1
x2+y2=1x^2 + y^2 = 1
Graph of a parabola, indicating it is a function
Graph of a circle, indicating it is not a function
Is a function
Not a function
Domain: The set of all possible input values (xx-values) for which the function is defined.
Dom(f)={x∈R;∃y=f(x)∈R}Dom(f) = \{x \in \mathbb{R} ; \exists y = f(x) \in \mathbb{R}\}
Example: Dom(f)=(−∞,1)∪[2,∞)Dom(f) = (-\infty, 1) \cup [2, \infty)
Graph showing the domain of a function on the X-axis
Range (or Image): The set of all possible output values (yy-values) produced by the function.
Example: Range(f)=(−∞,0)∪[1,∞)\text{Range}(f) = (-\infty, 0) \cup [1, \infty)
Graph showing the range of a function on the Y-axis
Symmetries:
Even Symmetry
Satisfies f(x)=f(−x)f(x) = f(-x)
Odd Symmetry
Satisfies f(x)=−f(−x)f(x) = -f(-x)
Graph with even symmetry with respect to the Y-axis
Graph with odd symmetry with respect to the origin
Periodicity: If a function is periodic, the period is the xx-interval after which the output values repeat.
Graph of a periodic function showing its period
Other important characteristics:
  • Intercepts:
    • X-intercepts: set y=0y=0 and solve the equation (there can be multiple).
    • Y-intercept: set x=0x=0 and evaluate the function (there can be at most one).
  • Sign of the function: the xx-intervals where the function takes positive or negative values.
  • Relative (Local) Extrema: points where the function reaches a local maximum (peak) or local minimum (valley).
  • Monotonicity: the xx-intervals where the function is increasing or decreasing.
  • Absolute (Global) Extrema: the highest (absolute maximum) or lowest (absolute minimum) point over the entire domain.
  • Boundedness: a function is bounded if all its values (its range) fall between two real numbers; that is, it does not tend to infinity or negative infinity.
Note: The "Function Analysis" section covers the study of these characteristics in greater detail.
Graph of a piecewise function to analyze its characteristics
  • Domain: Dom(f)=(−∞,2]∪(4,∞)Dom(f) = (-\infty, 2] \cup (4, \infty)
  • Range: Range(f)=[−1,∞)\text{Range}(f) = [-1, \infty)
  • Symmetries: none
  • Periodicity: none
  • Intercepts:
    • Y-axis: (0,0)(0, 0)
    • X-axis: (0,0),(10,0)(0, 0), (10, 0)
  • Sign intervals:
    • Positive: (−∞,0)∪(0,2]∪(4,8]∪(10,∞)(-\infty, 0) \cup (0, 2] \cup (4, 8] \cup (10, \infty)
    • Negative: (8,10)(8, 10)
  • Relative extrema:
    • Relative maximum: (6,3)(6, 3)
    • Relative minimums: (0,0),(9,−1)(0, 0), (9, -1)
  • Monotonicity:
    • Increasing: (0,2)∪(4,6)∪(9,∞)(0, 2) \cup (4, 6) \cup (9, \infty)
    • Decreasing: (−∞,0)∪(6,9)(-\infty, 0) \cup (6, 9)
  • Absolute extrema: Absolute minimum at (9,−1)(9, -1). No absolute maximum.
  • Boundedness: Not bounded.
Function composition occurs when one function is applied to the result of another. It is denoted by a small circle and read as "gg composed with ff" or "gg of ff":
g(f(x))=(g∘f)(x)g(f(x)) = (g \circ f)(x)
To calculate gg composed with ff, we substitute the expression of function ff into every instance of the variable xx in function gg.
Example:
g(x)=x2+2xg(x) = x^2 + 2x

f(x)=xf(x) = \sqrt{x}
(g∘f)(x)=(x)2+2(x)=x+2x(g \circ f)(x) = (\sqrt{x})^2 + 2(\sqrt{x}) = x + 2\sqrt{x}
Attention: Function composition is not commutative.
(g∘f)(x)≠(f∘g)(x)(g \circ f)(x) \neq (f \circ g)(x)
The inverse of a function (denoted by f−1f^{-1}), if it exists, is calculated using the following steps:
  1. Swap xx and yy in the algebraic expression.
  2. Solve for yy, which will be the inverse function.
  3. Verify if the inverse is correct. It must satisfy:
(f∘f−1)(x)=(f−1∘f)(x)=x(f \circ f^{-1})(x) = (f^{-1} \circ f)(x) = x
Example: Inverse of y=f(x)=2x+1y = f(x) = 2x + 1
x=2y+1  ⟹  y=x−12=f−1(x)x = 2y + 1 \implies y = \frac{x - 1}{2} = f^{-1}(x)
Verification:
(f−1∘f)(x)=(2x+1)−12=x(f^{-1} \circ f)(x) = \frac{(2x + 1) - 1}{2} = x