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Higher Mathematics

Functions

This chapter is the starting point of higher mathematics. We will systematically study the three core concepts: functions, limits, and continuity. They are not only the foundation for learning calculus but also key tools for describing and solving practical problems using mathematical language.

Functions

A function is a fundamental mathematical model that describes the dependency relationship between variables.

Concept and Representation of Functions

Definition: Given two variables x and y, if for every value of x in a certain range D, there is a unique value of y corresponding to it, then y is called a function of x, denoted as y=f(x)y = f(x). Here, x is called the independent variable, y is the dependent variable, and D is the domain of the function.

Representation Methods:

  1. Analytical (Formula) Method: Use a mathematical expression to represent the function relationship, such as y=x2+1y = x^2 + 1. This is the most common method.
  2. Tabular Method: List the independent variable and the corresponding function values in a table.
  3. Graphical Method: Use a curve on the coordinate plane to represent the function relationship.

Properties of Functions

Boundedness

  • Definition: If there exists a constant M such that for any xDx \in D, f(x)M|f(x)| \leq M, then f(x)f(x) is said to be bounded on D.

Monotonicity

  • Definition: Let f(x)f(x) be defined on interval I.
    • If for any x1,x2Ix_1, x_2 \in I, when x1<x2x_1 < x_2, f(x1)<f(x2)f(x_1) < f(x_2) always holds, then f(x)f(x) is said to be increasing on I.
    • If for any x1,x2Ix_1, x_2 \in I, when x1<x2x_1 < x_2, f(x1)>f(x2)f(x_1) > f(x_2) always holds, then f(x)f(x) is said to be decreasing on I.

Periodicity

  • Definition: If there exists a nonzero constant T such that for any x in the domain, f(x+T)=f(x)f(x + T) = f(x) always holds, then f(x)f(x) is called a periodic function, and T is called its period. Usually, we refer to the smallest positive period.

Parity

  • Prerequisite: The domain of the function is symmetric about the origin.
    • Even Function: If f(x)=f(x)f(-x) = f(x), then f(x)f(x) is an even function, and its graph is symmetric about the y-axis.
    • Odd Function: If f(x)=f(x)f(-x) = -f(x), then f(x)f(x) is an odd function, and its graph is symmetric about the origin.

Special Types of Functions

  1. Composite Function: If y=f(u)y = f(u) and u=g(x)u = g(x), then y=f[g(x)]y = f[g(x)] is called the composite function of ff and gg.
  2. Inverse Function: If y=f(x)y = f(x) is monotonic, then it has an inverse function, denoted as x=f1(y)x = f^{-1}(y) or y=f1(x)y = f^{-1}(x). The graphs of the original and inverse functions are symmetric about the line y=xy = x.
  3. Piecewise Function: A function represented by different expressions in different parts of its domain.
  4. Implicit Function: A function relationship determined by the equation F(x,y)=0F(x, y) = 0.

Elementary Functions

  • Basic Elementary Functions: Include power functions, exponential functions, logarithmic functions, trigonometric functions, and inverse trigonometric functions.
  • Elementary Functions: Functions obtained by a finite number of algebraic operations and compositions of basic elementary functions.

Exercises

  1. Find the domain of the function f(x)=14x2+ln(x)f(x) = \frac{1}{\sqrt{4 - x^2}} + \ln(x).
  2. Determine the parity (odd/even) of the function f(x)=ln(1x1+x)f(x) = \ln(\frac{1 - x}{1 + x}).
  3. Find the inverse function of y=e2x+1y = e^{2x} + 1.
Reference Answers

1. Find the domain of f(x)=14x2+ln(x)f(x) = \frac{1}{\sqrt{4 - x^2}} + \ln(x)

Solution: The domain is the set of x values for which all parts of the function are defined. Consider each part separately:

  • For 14x2\frac{1}{\sqrt{4 - x^2}}, the denominator cannot be zero and the expression under the square root must be positive: 4x2>0    2<x<24 - x^2 > 0 \implies -2 < x < 2.
  • For ln(x)\ln(x), x>0x > 0.
  • The intersection is 0<x<20 < x < 2.

Answer: The domain is (0,2)(0, 2).


2. Determine the parity of f(x)=ln(1x1+x)f(x) = \ln(\frac{1 - x}{1 + x})

Solution:

  • The domain is (1,1)(-1, 1), which is symmetric about the origin.
  • f(x)=ln(1+x1x)=ln(1x1+x)=f(x)f(-x) = \ln(\frac{1 + x}{1 - x}) = -\ln(\frac{1 - x}{1 + x}) = -f(x)

Answer: The function is odd.


3. Find the inverse function of y=e2x+1y = e^{2x} + 1

Solution:

  • y=e2x+1    y1=e2x    ln(y1)=2x    x=12ln(y1)y = e^{2x} + 1 \implies y - 1 = e^{2x} \implies \ln(y - 1) = 2x \implies x = \frac{1}{2}\ln(y - 1)
  • Swap x and y: y=12ln(x1)y = \frac{1}{2}\ln(x - 1)
  • The domain of the inverse function is (1,)(1, \infty)

Answer: y=12ln(x1)y = \frac{1}{2}\ln(x - 1), domain (1,)(1, \infty).