W4. Functions and Their Graphs, Elementary Functions and Properties

1. Summary

1.1 Functions, Domain, and Range

A function is a special relationship between two sets of values where each input corresponds to exactly one output. Think of it like a machine: you put something in (the input), and you get exactly one thing out (the output). You cannot put in the same input and get two different outputs.

More formally, a function \(f\) from a set \(D\) to a set \(Y\) is a rule that assigns a single value \(f(x)\) in \(Y\) to each \(x\) in \(D\). This “single value” requirement is crucial—it’s what distinguishes functions from other types of relationships.

The domain of a function is the set of all possible input values that the function can accept. We denote it as \(D(f)\) or \(D_f\). Think of the domain as all the valid things you’re allowed to put into your function machine.

The range of a function is the set of all possible output values that the function can produce as \(x\) varies throughout the domain. The range tells you what values can actually come out of your function.

When we define a function \(f\) with a formula \(y = f(x)\) and the domain is not explicitly stated or restricted by context, we assume the domain is the largest set of real \(x\)-values for which the formula gives real \(y\)-values. This is called the natural domain or implied domain.

1.1.1 Examples of Functions with Their Domains and Ranges

Let’s examine several common functions to understand how domains and ranges work:

• \(y = x^2\): Domain is \((-\infty, \infty)\) (all real numbers), Range is \([0, \infty)\) (non-negative numbers only). No matter what real number you square, you always get a non-negative result.
• \(y = \frac{1}{x}\): Domain is \((-\infty, 0) \cup (0, \infty)\) (all real numbers except zero), Range is \((-\infty, 0) \cup (0, \infty)\). We exclude zero from the domain because division by zero is undefined.
• \(y = \sqrt{x}\): Domain is \([0, \infty)\), Range is \([0, \infty)\). We can only take square roots of non-negative numbers (in the real number system).
• \(y = \sqrt{4-x}\): Domain is \((-\infty, 4]\), Range is \([0, \infty)\). We need \(4-x \geq 0\), which means \(x \leq 4\).
• \(y = \sqrt{4-x^2}\): Domain is \([-2, 2]\). We need \(4-x^2 \geq 0\), which means \(x^2 \leq 4\), so \(-2 \leq x \leq 2\).

1.2 The Graph of a Function

The graph of a function \(f\) with domain \(D\) consists of all points in the Cartesian plane (the standard \(xy\)-plane) whose coordinates are the input-output pairs for \(f\). In mathematical notation, the graph is the set:

\[\{(x, f(x)) \mid x \in D\}\]

In simpler terms, for every input value \(x\) in the domain, we plot a point at coordinates \((x, f(x))\) where \(f(x)\) is the output. When we connect all these points, we get the visual representation of our function.

If a point \((x, y)\) lies on the graph of \(f\), then the value \(y = f(x)\) represents the height of the graph above the point \(x\) on the horizontal axis (if \(f(x)\) is positive) or the depth below the point \(x\) (if \(f(x)\) is negative).

1.3 Floor and Ceiling Functions

1.3.1 The Floor Function

The floor function (also called the greatest integer function) takes any real number and rounds it down to the nearest integer. It is denoted by \(\lfloor x \rfloor\) and gives you the greatest integer less than or equal to \(x\).

Examples:

• \(\lfloor 2.4 \rfloor = 2\) (rounds down)
• \(\lfloor 1.9 \rfloor = 1\) (rounds down)
• \(\lfloor 0 \rfloor = 0\) (already an integer)
• \(\lfloor -1.2 \rfloor = -2\) (rounds down to more negative)
• \(\lfloor 2 \rfloor = 2\) (already an integer)
• \(\lfloor 0.2 \rfloor = 0\) (rounds down)
• \(\lfloor -0.3 \rfloor = -1\) (rounds down to more negative)

1.3.2 The Ceiling Function

The ceiling function (also called the least integer function) takes any real number and rounds it up to the nearest integer. It is denoted by \(\lceil x \rceil\) and gives you the smallest integer greater than or equal to \(x\).

Examples:

• \(\lceil 2.4 \rceil = 3\) (rounds up)
• \(\lceil 1.9 \rceil = 2\) (rounds up)
• \(\lceil 0 \rceil = 0\) (already an integer)
• \(\lceil -1.2 \rceil = -1\) (rounds up to less negative)
• \(\lceil 2 \rceil = 2\) (already an integer)
• \(\lceil 0.2 \rceil = 1\) (rounds up)
• \(\lceil -0.3 \rceil = 0\) (rounds up to less negative)

1.4 Monotone Functions

Monotone functions are functions that consistently move in one direction—they either go up, stay flat, or go down as we move from left to right along the graph. Understanding monotonicity helps us predict function behavior and is crucial for many mathematical proofs.

1.4.1 Types of Monotonicity

Let \(f\) be a function defined on an interval \(I\). We distinguish between several types of monotonic behavior:

(1) Increasing Function: \(f\) is increasing on \(I\) if for all \(x_1, x_2 \in I\) with \(x_1 < x_2\), we have \(f(x_1) \leq f(x_2)\). The function values go up or stay the same as we move right.

(2) Strictly Increasing Function: \(f\) is strictly increasing on \(I\) if for all \(x_1, x_2 \in I\) with \(x_1 < x_2\), we have \(f(x_1) < f(x_2)\). The function values always go up as we move right (no flat parts).

(3) Decreasing Function: \(f\) is decreasing on \(I\) if for all \(x_1, x_2 \in I\) with \(x_1 < x_2\), we have \(f(x_1) \geq f(x_2)\). The function values go down or stay the same as we move right.

(4) Strictly Decreasing Function: \(f\) is strictly decreasing on \(I\) if for all \(x_1, x_2 \in I\) with \(x_1 < x_2\), we have \(f(x_1) > f(x_2)\). The function values always go down as we move right (no flat parts).

(5) Monotone Function: The function \(f\) is (strictly) monotone on \(I\) if it is either (strictly) increasing or (strictly) decreasing on \(I\).

1.4.2 Examples of Monotone Functions

Example 1: Linear Functions

Let \(a, b \in \mathbb{R}\), and \(f(x) = ax + b\). The monotonicity depends entirely on the coefficient \(a\):

• If \(a > 0\): \(f\) is strictly increasing on \(\mathbb{R}\) (the line slopes upward)
• If \(a = 0\): \(f\) is constant on \(\mathbb{R}\) (horizontal line, hence both increasing and decreasing)
• If \(a < 0\): \(f\) is strictly decreasing on \(\mathbb{R}\) (the line slopes downward)

Example 2: Quadratic Function \(f(x) = x^2\)

The function \(f: \mathbb{R} \to \mathbb{R}\), \(f(x) = x^2\):

• Is strictly increasing on \(I = [0, \infty)\) (right side of the parabola)
• Is strictly decreasing on \(I = (-\infty, 0]\) (left side of the parabola)

All functions \(f(x) = x^n\) with \(n \geq 4\) even have the same monotonic behavior as \(f(x) = x^2\).

Example 3: Cubic Function \(f(x) = x^3\)

The function \(f: \mathbb{R} \to \mathbb{R}\), \(f(x) = x^3\) is strictly increasing on all of \(\mathbb{R}\). All functions like \(f(x) = x^n\) with \(n\) odd have analogous monotonic behavior.

1.5 Combining Functions: Sums, Differences, Products, and Quotients

Just like numbers, functions can be combined through arithmetic operations to create new functions. This allows us to build complex functions from simpler building blocks.

If \(f\) and \(g\) are functions, then for every \(x \in D(f) \cap D(g)\) (meaning \(x\) must be in both domains):

(1) Sum: \((f + g)(x) = f(x) + g(x)\)

(2) Difference: \((f - g)(x) = f(x) - g(x)\)

(3) Product: \((fg)(x) = f(x)g(x)\)

(4) Quotient: At any point \(x \in (D(f) \cap D(g)) \setminus \{x \mid g(x) = 0\}\) (points in both domains where \(g(x) \neq 0\)), we define:

\[(f/g)(x) = \frac{f(x)}{g(x)}, \quad g(x) \neq 0\]

(5) Scalar Multiplication: Functions can also be multiplied by constants. If \(c\) is a real number, then the function \(cf\) is defined for all \(x\) in the domain of \(f\) by:

\[(cf)(x) = cf(x)\]

The domain of the combined functions \(f + g\), \(f - g\), and \(fg\) is \(D(f) \cap D(g)\). The domain of \(f/g\) is \(D(f) \cap D(g)\) with all points where \(g(x) = 0\) removed.

1.6 Composition of Functions

Composition is a way of combining two functions where the output of one function becomes the input of another. This is like a two-stage process: first apply one function, then apply another function to the result.

The composition of functions \(f\) and \(g\), denoted \(f \circ g\) (read as “\(f\) composed with \(g\)”), is defined by:

\[(f \circ g)(x) = f(g(x))\]

To evaluate \(f \circ g\): first find \(g(x)\), then apply \(f\) to that result.

The domain of \(f \circ g\) consists of all numbers \(x\) in the domain of \(g\) for which \(g(x)\) lies in the domain of \(f\). In other words, we need: 1. \(x\) must be a valid input for \(g\) 2. The output \(g(x)\) must be a valid input for \(f\)

Similarly, to evaluate \(g \circ f\) (when defined): first find \(f(x)\), then find \(g(f(x))\). The domain of \(g \circ f\) is the set of numbers \(x\) in the domain of \(f\) such that \(f(x)\) lies in the domain of \(g\).

Important: The functions \(f \circ g\) and \(g \circ f\) are usually quite different—order matters in composition!

1.6.1 Examples of Function Composition

Let \(f(x) = \sqrt{x}\) and \(g(x) = x + 1\). Let’s explore various compositions:

(1) \((f \circ g)(x) = f(g(x)) = \sqrt{g(x)} = \sqrt{x+1}\)

Domain: \(D(f \circ g) = [-1, \infty)\) (we need \(x + 1 \geq 0\))

(2) \((g \circ f)(x) = g(f(x)) = f(x) + 1 = \sqrt{x} + 1\)

Domain: \(D(g \circ f) = [0, \infty)\) (we need \(x \geq 0\) for the square root)

(3) \((f \circ f)(x) = f(f(x)) = \sqrt{f(x)} = \sqrt{\sqrt{x}} = x^{1/4}\)

Domain: \(D(f \circ f) = [0, \infty)\)

(4) \((g \circ g)(x) = g(g(x)) = g(x) + 1 = (x + 1) + 1 = x + 2\)

Domain: \(D(g \circ g) = (-\infty, \infty)\)

Important Note: If \(f(x) = x^2\) and \(g(x) = \sqrt{x}\), then \((f \circ g)(x) = (\sqrt{x})^2 = x\). However, the domain of \(f \circ g\) is \([0, \infty)\), not \((-\infty, \infty)\), because \(\sqrt{x}\) requires \(x \geq 0\).

1.7 One-to-One Functions

In many mathematical problems, we want to reverse a function—that is, given an output value, find the unique input that produced it. This is called the inverse problem. However, not all functions can be reversed uniquely.

Consider the function \(f(x) = x^2\) with domain all real numbers. If we’re told that \(f(x) = 16\) and asked to find \(x\), we have a problem: both \(f(4) = 16\) and \(f(-4) = 16\). With the information given, we cannot determine a unique value of \(x\) such that \(f(x) = 16\). This means the inverse problem does not have a unique solution for this function.

We can avoid this difficulty by restricting the domain or by working with functions that naturally avoid this issue. This leads us to the concept of one-to-one functions.

A function \(f(x)\) is one-to-one on a domain \(D\) if \(f(x_1) \neq f(x_2)\) whenever \(x_1 \neq x_2\) in \(D\). In other words, different inputs always produce different outputs. This property is also called being injective.

1.7.1 Examples of One-to-One Functions

(1) \(f(x) = \sqrt{x}\) is one-to-one on any domain of nonnegative numbers because \(\sqrt{x_1} \neq \sqrt{x_2}\) whenever \(x_1 \neq x_2\).

(2) \(g(x) = \sin x\) is not one-to-one on the interval \([0, \pi]\) because \(\sin(\pi/6) = \sin(5\pi/6) = 1/2\), but \(\pi/6 \neq 5\pi/6\).

Important Proposition: If \(f\) is strictly monotone on its domain, then \(f\) is one-to-one.

Proof idea: If \(f\) is strictly increasing and \(x_1 \neq x_2\), then either \(x_1 < x_2\) (giving \(f(x_1) < f(x_2)\)) or \(x_2 < x_1\) (giving \(f(x_2) < f(x_1)\)). Either way, \(f(x_1) \neq f(x_2)\). The same logic applies for strictly decreasing functions.

Important Remark: The reverse of this proposition is not always true. A function \(f\) may be one-to-one without being strictly monotone on its domain. For example:

\[f(x) = \begin{cases} \frac{1}{x} & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases}\]

This function is one-to-one on \((-\infty, \infty)\) (its graph intersects each horizontal line at most once), but it is neither increasing nor decreasing on its entire domain.

1.8 Inverse Functions

Once we have a one-to-one function, we can define its inverse—a function that “undoes” what the original function does.

Suppose that \(f\) is a one-to-one function on a domain \(D\) with range \(R\). The inverse function \(f^{-1}\) is defined by:

\[f^{-1}(b) = a \quad \text{if and only if} \quad f(a) = b\]

The domain of \(f^{-1}\) is \(R\) (the range of \(f\)), and the range of \(f^{-1}\) is \(D\) (the domain of \(f\)). The inverse function essentially reverses the input-output relationship: if \(f\) takes \(a\) to \(b\), then \(f^{-1}\) takes \(b\) back to \(a\).

1.8.1 Properties of Inverse Functions

Composing a function and its inverse:

(1) \((f^{-1} \circ f)(x) = x\) for all \(x\) in the domain of \(f\)

(2) \((f \circ f^{-1})(y) = y\) for all \(y\) in the domain of \(f^{-1}\) (or equivalently, the range of \(f\))

These properties confirm that \(f\) and \(f^{-1}\) truly “undo” each other.

When does a function have an inverse?

Suppose \(f\) is a function whose domain is an interval:

• If \(f\) is increasing, then it satisfies \(f(x_1) > f(x_2)\) when \(x_1 > x_2\), so it is one-to-one and has an inverse.
• If \(f\) is decreasing, then it also has an inverse.
• Functions that are neither increasing nor decreasing may still be one-to-one and have an inverse, as shown in the example above.

Graphical Property: The graph of \(y = f^{-1}(x)\) is obtained by reflecting the graph of \(y = f(x)\) about the line \(y = x\). This is because if \((a, b)\) is on the graph of \(f\), then \((b, a)\) is on the graph of \(f^{-1}\).

1.8.2 Example: Finding an Inverse Function

Let \(f(x) = \frac{4x+5}{2x+3}\). To find \(f^{-1}(x)\):

Set \(y = \frac{4x+5}{2x+3}\) and solve for \(x\) in terms of \(y\):

\[y(2x+3) = 4x+5\] \[2xy + 3y = 4x + 5\] \[2xy - 4x = 5 - 3y\] \[x(2y - 4) = 5 - 3y\] \[x = \frac{5-3y}{2y-4}\]

Therefore, \(f^{-1}(y) = \frac{5-3y}{2y-4}\), or replacing \(y\) with \(x\): \(f^{-1}(x) = \frac{5-3x}{2x-4}\).

1.9 Even and Odd Functions

Even and odd functions have special symmetry properties that make them easier to analyze and work with.

Let \(f: D \subset \mathbb{R} \to \mathbb{R}\) be a function with a domain that is symmetric with respect to the origin (meaning if \(x \in D\), then \(-x \in D\) as well).

• The function \(f\) is even if \(f(-x) = f(x)\) for all \(x \in D\). The graph of an even function is symmetric with respect to the \(y\)-axis (vertical axis). Examples: \(f(x) = x^2\), \(f(x) = \cos x\), \(f(x) = |x|\).
• The function \(f\) is odd if \(f(-x) = -f(x)\) for all \(x \in D\). The graph of an odd function is symmetric with respect to the origin (if you rotate the graph 180° about the origin, it looks the same). Examples: \(f(x) = x^3\), \(f(x) = \sin x\), \(f(x) = x\).

1.9.1 Properties of Even and Odd Functions

(1) The sum (or difference) of two even functions is even.

(2) The sum (or difference) of two odd functions is odd.

(3) The sum of an even and odd function is neither even nor odd, unless one of the functions is zero.

(4) The product and quotient of two even functions is even.

(5) The product and quotient of two odd functions is even.

(6) The product and quotient of an even function and an odd function is odd.

(7) The composition of two even functions is even.

(8) The composition of two odd functions is odd.

(9) The composition of an even function and an odd function is even.

(10) If an odd function is invertible, then its inverse is also odd.

1.9.2 Decomposition into Even and Odd Parts

Important Fact: Any function with a domain that is symmetric with respect to the origin can be uniquely decomposed as the sum of an even function and an odd function. These are called the even part (or even component) and the odd part (or odd component) of the function:

\[f_{\text{even}}(x) = \frac{f(x) + f(-x)}{2}\]

\[f_{\text{odd}}(x) = \frac{f(x) - f(-x)}{2}\]

We can verify: \(f(x) = f_{\text{even}}(x) + f_{\text{odd}}(x)\).

1.10 Power Functions

Power functions are among the most fundamental functions in mathematics. To understand them fully, we need to carefully define what we mean by powers of real numbers.

1.10.1 Powers of a Real Number

Let \(x \in \mathbb{R}\). We define powers progressively:

(1) Natural Number Exponents: If \(m \in \mathbb{N}\) (positive integer), then:

\[x^m := \underbrace{x \cdot x \cdots x}_{m \text{ times}}\]

(2) Zero Exponent: If \(x \neq 0\), then \(x^0 := 1\). Note that \(0^0\) is undefined.

(3) Negative Integer Exponents: If \(x \neq 0\) and \(m \in \mathbb{N}\), then:

\[x^{-m} := \frac{1}{x^m}\]

(4) Roots (Fractional Exponents with Numerator 1): If \(m\) is a positive integer and \(x\) is a real number, then \(x^{1/m}\) (also denoted \(\sqrt[m]{x}\), the \(m\)th root of \(x\)) is defined to be the real number satisfying:

\[(x^{1/m})^m = x\]

subject to the following conditions:

• If \(x < 0\) and \(m\) is even, then \(x^{1/m}\) is undefined (no real even root of negative numbers).
• If \(x > 0\) and \(m\) is even, then \(x^{1/m}\) is chosen to be the positive number satisfying the equation above (principal root).
• If \(m\) is odd, then \(x^{1/m}\) is defined for all real \(x\) and has the same sign as \(x\).

(5) Rational Exponents: If \(\frac{n}{m}\) is a fraction in reduced form, where \(n\) and \(m\) are integers and \(m > 0\), then \(x^{n/m}\) is defined by:

\[x^{n/m} = (x^{1/m})^n\]

whenever this makes sense (respecting the conditions for roots above).

(6) Real Exponents: If \(x \geq 0\) and \(\alpha \in \mathbb{R}^+ \setminus \mathbb{Q}\) (positive irrational number), we can define the power \(x^\alpha\) by approximating \(\alpha\) with rational numbers:

• If \(\alpha \geq 1\), then: \[x^\alpha := \sup \{x^{n/m} \mid n/m \leq \alpha, \; n/m \in \mathbb{Q}\}\]

• If \(0 < \alpha < 1\), then: \[x^\alpha := \inf \{x^{n/m} \mid n/m \geq \alpha, \; n/m \in \mathbb{Q}\}\]

• If \(x > 0\) and \(\alpha < 0\) (negative real), then: \[x^\alpha := \frac{1}{x^{-\alpha}}\] where \(-\alpha > 0\).

1.10.2 Example: Computing an Irrational Power

For \(2^{\sqrt{3}}\), we use the decimal expansion \(\sqrt{3} = 1.732050808...\). We consider powers of 2 with increasingly accurate approximations:

\[2^1, \; 2^{1.7}, \; 2^{1.73}, \; 2^{1.732}, \; 2^{1.7320}, \; 2^{1.73205}, \; ...\]

Which gives approximately:

\[2.000000000, \; 3.249009585, \; 3.317278183, \; 3.321880096, \; ...\]

Taking better approximations to \(\sqrt{3}\) gives better approximations to \(2^{\sqrt{3}} \approx 3.321997086\).

1.11 Polynomials and Rational Functions

Polynomials and rational functions are among the most important and frequently used functions in mathematics, appearing in virtually every area of applied mathematics and science.

1.11.1 Polynomials

A function \(p\) is a polynomial if it can be written in the form:

\[p(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0\]

where \(n\) is a non-negative integer and the numbers \(a_0, a_1, a_2, \ldots, a_n\) are real constants called the coefficients of the polynomial.

All polynomials have domain \((-\infty, \infty)\) (all real numbers).

If the leading coefficient \(a_n \neq 0\), then \(n\) is called the degree of the polynomial.

Special types of polynomials:

• Linear functions are polynomials of degree 1: \(f(x) = a_1 x + a_0\) (with \(a_1 \neq 0\)). These represent straight lines.
• Quadratic functions are polynomials of degree 2: \(p(x) = ax^2 + bx + c\) (with \(a \neq 0\)). Their graphs are parabolas.
• Cubic functions are polynomials of degree 3: \(p(x) = ax^3 + bx^2 + cx + d\) (with \(a \neq 0\)).

1.11.2 Rational Functions

A rational function is a quotient or ratio of two polynomials:

\[f(x) = \frac{p(x)}{q(x)}\]

where \(p\) and \(q\) are polynomials. The name “rational” comes from “ratio.”

The domain of a rational function is the set of all real \(x\) for which \(q(x) \neq 0\). We must exclude any values of \(x\) that make the denominator zero, as division by zero is undefined.

Examples:

• \(f(x) = \frac{1}{x}\) is a rational function with domain \((-\infty, 0) \cup (0, \infty)\).
• \(g(x) = \frac{x^2 - 1}{x^2 + 1}\) is a rational function with domain \((-\infty, \infty)\) (the denominator is never zero).
• \(h(x) = \frac{x + 2}{x^2 - 4} = \frac{x+2}{(x-2)(x+2)}\) has domain \((-\infty, -2) \cup (-2, 2) \cup (2, \infty)\).

1.12 Exponential Functions

Exponential functions model growth and decay processes and are fundamental in describing natural phenomena like population growth, radioactive decay, compound interest, and many other applications.

1.12.1 Definition and Basic Properties

Let \(a > 0\) be a positive real number. A function of the form \(f(x) = a^x\) is called an exponential function with base \(a\).

Properties of Exponential Functions:

(1) All exponential functions have domain \((-\infty, \infty)\) and range \((0, \infty)\). This means an exponential function never assumes the value 0—it can get arbitrarily close to zero but never reaches it.

(2) If \(a > 1\), the exponential function is strictly increasing. The function grows rapidly as \(x\) increases.

(3) If \(a = 1\), then we get the constant function \(f(x) = 1\) for all \(x \in \mathbb{R}\).

(4) If \(0 < a < 1\), the function is strictly decreasing. The function decays as \(x\) increases.

The most important exponential function has base \(e \approx 2.718281828...\), known as Euler’s number. The function \(f(x) = e^x\) is called the natural exponential function and appears throughout mathematics and science.

1.13 Logarithmic Functions

Logarithmic functions are the inverses of exponential functions. They answer the question: “To what power must we raise the base to get a given number?”

1.13.1 Definition of Logarithm

The logarithm function with base \(a\), written \(y = \log_a x\), is the inverse of the base \(a\) exponential function \(y = a^x\) (with \(a > 0\) and \(a \neq 1\)). Thus:

\[y = \log_a x \quad \Longleftrightarrow \quad x = a^y\]

This means: “\(\log_a x\) is the exponent to which we must raise \(a\) to get \(x\).”

Properties:

• The domain of \(\log_a x\) is \((0, \infty)\) (same as the range of \(a^x\)). We can only take logarithms of positive numbers.
• The range of \(y = \log_a x\) is \((-\infty, \infty)\) (same as the domain of \(a^x\)).

Special Logarithms:

Logarithms with base \(e \approx 2.718\) and base 10 are so important that they have special notation:

• \(\log_e x\) is written as \(\ln x\) and called the natural logarithm function.
• \(\log_{10} x\) is written as \(\log x\) and called the common logarithm function.

1.13.2 Algebraic Properties of the Natural Logarithm

For any numbers \(x_1 > 0\) and \(x_2 > 0\), the natural logarithm satisfies the following fundamental rules:

(1) Product Rule: \(\ln(x_1 x_2) = \ln(x_1) + \ln(x_2)\)

The logarithm of a product is the sum of the logarithms. By induction:

\[\ln\left(\prod_{i=1}^n x_i\right) = \sum_{i=1}^n \ln(x_i) \quad \text{for all } x_i > 0\]

(2) Quotient Rule: \(\ln\left(\frac{x_1}{x_2}\right) = \ln(x_1) - \ln(x_2)\)

The logarithm of a quotient is the difference of the logarithms.

(3) Reciprocal Rule: \(\ln\left(\frac{1}{x_1}\right) = -\ln(x_1)\)

This follows from the quotient rule with \(x_1 = 1\) in the numerator.

(4) Power Rule: \(\ln(x_1^\alpha) = \alpha \ln(x_1)\) for all \(\alpha \in \mathbb{R}\)

The logarithm of a power brings the exponent down as a multiplier.

1.13.3 Additional Properties of Logarithms

Because \(a^x\) and \(\log_a x\) are inverses, composing them in either order gives the identity function:

(1) Base \(a\) (where \(a > 0\), \(a \neq 1\)):

• \(a^{\log_a x} = x\) for all \(x > 0\)
• \(\log_a(a^x) = x\) for all \(x \in \mathbb{R}\)

(2) Base \(e\):

• \(e^{\ln x} = x\) for all \(x > 0\)
• \(\ln(e^x) = x\) for all \(x \in \mathbb{R}\)

Expressing exponentials as powers of \(e\):

Substituting \(a^x\) for \(x\) in the equation \(x = e^{\ln x}\) enables us to rewrite \(a^x\) as a power of \(e\):

\[a^x = e^{\ln(a^x)} = e^{x \ln(a)} = e^{(\ln a) x}\]

This shows that any exponential function can be written in terms of the natural exponential.

Change-of-Base Formula:

Every logarithmic function is a constant multiple of the natural logarithm:

\[\log_a x = \frac{\ln(x)}{\ln(a)} \quad \text{for all } x > 0, \; a > 0, \; a \neq 1\]

Proof: For any \(x > 0\), we have:

\[x = a^{\log_a x} \implies \ln(x) = \ln(a^{\log_a x}) \implies \ln(x) = \log_a(x) \cdot \ln(a) \implies \log_a x = \frac{\ln(x)}{\ln(a)}\]

This formula allows us to compute logarithms in any base using only natural logarithms.

1.14 The Six Basic Trigonometric Functions

Trigonometric functions are fundamental in mathematics, physics, and engineering. They originally arose from the study of triangles and circles, but their applications extend far beyond geometry.

1.14.1 Definitions

We can define the trigonometric functions in terms of the coordinates of the point \(P(x, y)\) where an angle’s terminal ray intersects a circle with radius \(r > 0\) centered at the origin.

For an angle \(\theta\):

• sine: \(\sin \theta = \frac{y}{r}\)
• cosine: \(\cos \theta = \frac{x}{r}\)
• tangent: \(\tan \theta = \frac{y}{x}\) (defined when \(x \neq 0\))
• cosecant: \(\csc \theta = \frac{r}{y} = \frac{1}{\sin \theta}\) (defined when \(y \neq 0\))
• secant: \(\sec \theta = \frac{r}{x} = \frac{1}{\cos \theta}\) (defined when \(x \neq 0\))
• cotangent: \(\cot \theta = \frac{x}{y} = \frac{1}{\tan \theta}\) (defined when \(y \neq 0\))

1.14.2 Periodic Functions

A function \(f(x)\) is periodic if there is a positive number \(p\) such that \(f(x + p) = f(x)\) for every value of \(x\). The smallest such value of \(p\) is called the period of \(f\).

• The functions \(\sin x\), \(\cos x\), \(\sec x\), and \(\csc x\) are periodic with period \(2\pi\). We have \(\sin(x + 2\pi) = \sin x\) for all \(x\).
• The functions \(\tan x\) and \(\cot x\) are periodic with period \(\pi\).

1.14.3 Domains, Ranges, and Periods

\(y = \cos x\)

• Domain: \(-\infty < x < \infty\)
• Range: \(-1 \leq y \leq 1\)
• Period: \(2\pi\)

\(y = \sin x\)

• Domain: \(-\infty < x < \infty\)
• Range: \(-1 \leq y \leq 1\)
• Period: \(2\pi\)

\(y = \tan x\)

• Domain: \(x \neq \pm\frac{\pi}{2}, \pm\frac{3\pi}{2}, \ldots\) (all \(x \neq \frac{\pi}{2} + n\pi\) for \(n \in \mathbb{Z}\))
• Range: \(-\infty < y < \infty\)
• Period: \(\pi\)

\(y = \sec x\)

• Domain: \(x \neq \pm\frac{\pi}{2}, \pm\frac{3\pi}{2}, \ldots\)
• Range: \(y \leq -1\) or \(y \geq 1\)
• Period: \(2\pi\)

\(y = \csc x\)

• Domain: \(x \neq 0, \pm\pi, \pm 2\pi, \ldots\) (all \(x \neq n\pi\) for \(n \in \mathbb{Z}\))
• Range: \(y \leq -1\) or \(y \geq 1\)
• Period: \(2\pi\)

\(y = \cot x\)

• Domain: \(x \neq 0, \pm\pi, \pm 2\pi, \ldots\)
• Range: \(-\infty < y < \infty\)
• Period: \(\pi\)

1.14.4 Trigonometric Identities

The most fundamental trigonometric identities are:

\[\sin^2 x + \cos^2 x = 1\] \[1 + \tan^2 x = \sec^2 x\] \[1 + \cot^2 x = \csc^2 x\]

Addition Formulas:

• \(\sin(x \pm y) = \sin x \cos y \pm \cos x \sin y\)
• \(\cos(x \pm y) = \cos x \cos y \mp \sin x \sin y\)
• \(\tan(x \pm y) = \frac{\tan x \pm \tan y}{1 \mp \tan x \tan y}\)

Double-Angle Formulas (derived from addition formulas):

• \(\sin(2x) = 2 \sin x \cos x\)
• \(\cos(2x) = \cos^2 x - \sin^2 x\)
• \(\tan(2x) = \frac{2 \tan x}{1 - \tan^2 x}\)

Half-Angle Formulas (derived from \(\cos^2 x + \sin^2 x = 1\) and \(\cos(2x)\)):

• \(\cos^2 x = \frac{1 + \cos(2x)}{2}\)
• \(\sin^2 x = \frac{1 - \cos(2x)}{2}\)

Product-to-Sum Identities:

• \(\cos x \cos y = \frac{1}{2}[\cos(x-y) + \cos(x+y)]\)
• \(\sin x \sin y = \frac{1}{2}[\cos(x-y) - \cos(x+y)]\)
• \(\sin x \cos y = \frac{1}{2}[\sin(x+y) + \sin(x-y)]\)
• \(\cos x \sin y = \frac{1}{2}[\sin(x+y) - \sin(x-y)]\)

1.15 Inverse Trigonometric Functions

The six basic trigonometric functions are not one-to-one over their natural domains because their values repeat periodically. However, we can restrict their domains to intervals on which they are one-to-one, allowing us to define inverse functions.

1.15.1 Restricted Domains for One-to-One Behavior

To make the trigonometric functions one-to-one, we restrict their domains as follows:

\(y = \sin x\)

• Restricted domain: \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\)
• Range: \([-1, 1]\)

\(y = \cos x\)

• Restricted domain: \([0, \pi]\)
• Range: \([-1, 1]\)

\(y = \tan x\)

• Restricted domain: \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\)
• Range: \((-\infty, \infty)\)

\(y = \cot x\)

• Restricted domain: \((0, \pi)\)
• Range: \((-\infty, \infty)\)

\(y = \sec x\)

• Restricted domain: \(\left[0, \frac{\pi}{2}\right) \cup \left(\frac{\pi}{2}, \pi\right]\)
• Range: \((-\infty, -1] \cup [1, \infty)\)

\(y = \csc x\)

• Restricted domain: \(\left[-\frac{\pi}{2}, 0\right) \cup \left(0, \frac{\pi}{2}\right]\)
• Range: \((-\infty, -1] \cup [1, \infty)\)

1.15.2 Definitions of Inverse Trigonometric Functions

Since these restricted functions are now one-to-one, they have inverses, which we denote using two equivalent notations:

• \(y = \sin^{-1} x\) or \(y = \arcsin x\)
• \(y = \cos^{-1} x\) or \(y = \arccos x\)
• \(y = \tan^{-1} x\) or \(y = \arctan x\)
• \(y = \cot^{-1} x\) or \(y = \text{arccot} \, x\)
• \(y = \sec^{-1} x\) or \(y = \text{arcsec} \, x\)
• \(y = \csc^{-1} x\) or \(y = \text{arccsc} \, x\)

Important Caution: \(\sin^{-1} x\) means the inverse function of sine, not \(\frac{1}{\sin x}\). The reciprocal of \(\sin x\) is \((\sin x)^{-1} = \frac{1}{\sin x} = \csc x\).

Formal Definitions:

(1) \(y = \arcsin x\) is the number in \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) for which \(\sin y = x\).

(2) \(y = \arccos x\) is the number in \([0, \pi]\) for which \(\cos y = x\).

(3) \(y = \arctan x\) is the number in \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\) for which \(\tan y = x\).

(4) \(y = \text{arccot} \, x\) is the number in \((0, \pi)\) for which \(\cot y = x\).

(5) \(y = \text{arcsec} \, x\) is the number in \(\left[0, \frac{\pi}{2}\right) \cup \left(\frac{\pi}{2}, \pi\right]\) for which \(\sec y = x\).

(6) \(y = \text{arccsc} \, x\) is the number in \(\left[-\frac{\pi}{2}, 0\right) \cup \left(0, \frac{\pi}{2}\right]\) for which \(\csc y = x\).

1.15.3 Domains and Ranges of Inverse Trigonometric Functions

\(y = \arcsin x\)

• Domain: \(-1 \leq x \leq 1\)
• Range: \(-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}\)

\(y = \arccos x\)

• Domain: \(-1 \leq x \leq 1\)
• Range: \(0 \leq y \leq \pi\)

\(y = \arctan x\)

• Domain: \(-\infty < x < \infty\)
• Range: \(-\frac{\pi}{2} < y < \frac{\pi}{2}\)

\(y = \text{arccot} \, x\)

• Domain: \(-\infty < x < \infty\)
• Range: \(0 < y < \pi\)

\(y = \text{arcsec} \, x\)

• Domain: \(x \leq -1\) or \(x \geq 1\)
• Range: \(0 \leq y \leq \pi\), \(y \neq \frac{\pi}{2}\)

\(y = \text{arccsc} \, x\)

• Domain: \(x \leq -1\) or \(x \geq 1\)
• Range: \(-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}\), \(y \neq 0\)

1.15.4 Relationships Among Inverse Trigonometric Functions

Complementary Angles:

• \(\arcsin x + \arccos x = \frac{\pi}{2}\) for all \(x \in [-1, 1]\)
• \(\arctan x + \text{arccot} \, x = \frac{\pi}{2}\) for all \(x \in \mathbb{R}\)
• \(\text{arcsec} \, x + \text{arccsc} \, x = \frac{\pi}{2}\) for all \(|x| \geq 1\)

Negative Arguments:

• \(\arcsin(-x) = -\arcsin x\)
• \(\text{arccsc}(-x) = -\text{arccsc} \, x\)
• \(\arccos(-x) = \pi - \arccos x\)
• \(\text{arcsec}(-x) = \pi - \text{arcsec} \, x\)
• \(\arctan(-x) = -\arctan x\)
• \(\text{arccot}(-x) = \pi - \text{arccot} \, x\)

Reciprocal Arguments:

• \(\arcsin\left(\frac{1}{x}\right) = \text{arccsc} \, x\) for \(|x| \geq 1\)
• \(\text{arccsc}\left(\frac{1}{x}\right) = \arcsin x\) for \(|x| \leq 1\)
• \(\arccos\left(\frac{1}{x}\right) = \text{arcsec} \, x\) for \(|x| \geq 1\)
• \(\text{arcsec}\left(\frac{1}{x}\right) = \arccos x\) for \(|x| \leq 1\)
• \(\arctan\left(\frac{1}{x}\right) = \frac{\pi}{2} - \arctan x\) for all \(x > 0\)
• \(\arctan\left(\frac{1}{x}\right) = -\frac{\pi}{2} - \arctan x\) for all \(x < 0\)
• \(\text{arccot}\left(\frac{1}{x}\right) = \frac{\pi}{2} - \text{arccot} \, x\) for all \(x > 0\)
• \(\text{arccot}\left(\frac{1}{x}\right) = \frac{3\pi}{2} - \text{arccot} \, x\) for all \(x < 0\)

1.15.5 Example Proofs of Inverse Trigonometric Identities

Example 1: Let us prove \(\arcsin x + \arccos x = \frac{\pi}{2}\) for all \(x \in [-1, 1]\).

Let \(\theta := \arcsin x\). By definition of arcsin, we have \(\sin \theta = x\) and \(\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).

We know that \(\cos\left(\frac{\pi}{2} - \theta\right) = \sin \theta = x\).

Since \(\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\), we have \(\frac{\pi}{2} - \theta \in [0, \pi]\).

Therefore, by definition of arccos: \(\arccos x = \frac{\pi}{2} - \theta\).

Thus, \(\arcsin x + \arccos x = \theta + \left(\frac{\pi}{2} - \theta\right) = \frac{\pi}{2}\). ∎

Example 2: Let us prove \(\arcsin(-x) = -\arcsin x\) for all \(x \in [-1, 1]\).

Let \(\theta := \arcsin x\). Then \(\sin \theta = x\) and \(\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).

We know that \(\sin(-\theta) = -\sin \theta = -x\).

Since \(\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\), we have \(-\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).

Therefore, by definition of arcsin: \(\arcsin(-x) = -\theta = -\arcsin x\). ∎

Example 3: Let us prove \(\arcsin\left(\frac{1}{x}\right) = \text{arccsc} \, x\) for all \(x \in (-\infty, -1] \cup [1, \infty)\).

Let \(\theta := \arcsin\left(\frac{1}{x}\right)\). Because \(x \in (-\infty, -1] \cup [1, \infty)\), we have \(\frac{1}{x} \in [-1, 0) \cup (0, 1]\), so \(\theta \in \left[-\frac{\pi}{2}, 0\right) \cup \left(0, \frac{\pi}{2}\right]\).

By definition of arcsin: \(\frac{1}{x} = \sin \theta\), which means \(x = \frac{1}{\sin \theta} = \csc \theta\).

Because \(\theta \in \left[-\frac{\pi}{2}, 0\right) \cup \left(0, \frac{\pi}{2}\right]\), by definition of arccsc, we have \(\theta = \text{arccsc} \, x\).

Therefore, \(\arcsin\left(\frac{1}{x}\right) = \text{arccsc} \, x\). ∎

1.16 The Six Basic Hyperbolic Functions

Hyperbolic functions are formed by taking combinations of the two exponential functions \(e^x\) and \(e^{-x}\). Despite their name suggesting a connection to hyperbolas (just as trigonometric functions relate to circles), hyperbolic functions have their own unique properties and applications. They simplify many mathematical expressions and occur frequently in mathematical and engineering applications, including modeling hanging cables (catenary curves), describing relativistic velocity addition, and solving certain differential equations.

1.16.1 Definitions

(1) Hyperbolic sine: \(\sinh x = \frac{e^x - e^{-x}}{2}\)

(2) Hyperbolic cosine: \(\cosh x = \frac{e^x + e^{-x}}{2}\)

(3) Hyperbolic tangent: \(\tanh x = \frac{\sinh x}{\cosh x} = \frac{e^x - e^{-x}}{e^x + e^{-x}}\)

(4) Hyperbolic cotangent: \(\coth x = \frac{1}{\tanh x} = \frac{\cosh x}{\sinh x} = \frac{e^x + e^{-x}}{e^x - e^{-x}}\) (defined for \(x \neq 0\))

(5) Hyperbolic secant: \(\text{sech} \, x = \frac{1}{\cosh x} = \frac{2}{e^x + e^{-x}}\)

(6) Hyperbolic cosecant: \(\text{csch} \, x = \frac{1}{\sinh x} = \frac{2}{e^x - e^{-x}}\) (defined for \(x \neq 0\))

Important Symmetry Properties:

Note that:

• \(\sinh(-x) = -\sinh(x)\) (odd function)
• \(\cosh(-x) = \cosh x\) (even function)

Thus, the functions \(\cosh x\) and \(\text{sech} \, x\) are even, while \(\sinh x\), \(\text{csch} \, x\), \(\coth x\), and \(\tanh x\) are odd.

1.16.2 Identities for Hyperbolic Functions

The hyperbolic functions satisfy identities analogous to (but slightly different from) trigonometric identities.

Fundamental Identity:

\[\cosh^2 x - \sinh^2 x = 1\]

This is the hyperbolic analogue of \(\cos^2 x + \sin^2 x = 1\) (note the minus sign!).

Double-Angle Formulas:

• \(\sinh(2x) = 2 \sinh x \cosh x\)
• \(\cosh(2x) = \cosh^2 x + \sinh^2 x\)
• \(\tanh(2x) = \frac{2 \tanh x}{1 + \tanh^2 x}\)

Half-Angle-Type Formulas:

• \(\cosh^2 x = \frac{\cosh(2x) + 1}{2}\)
• \(\sinh^2 x = \frac{\cosh(2x) - 1}{2}\)

Pythagorean-Type Identities:

• \(\tanh^2 x = 1 - \text{sech}^2 x\)
• \(\coth^2 x = 1 + \text{csch}^2 x\)

Addition Formulas:

• \(\sinh(x \pm y) = \sinh x \cosh y \pm \cosh x \sinh y\)
• \(\cosh(x \pm y) = \cosh x \cosh y \pm \sinh x \sinh y\)
• \(\tanh(x \pm y) = \frac{\tanh x \pm \tanh y}{1 \pm \tanh x \tanh y}\)

Product-to-Sum Identities:

• \(\sinh(mx) \cosh(nx) = \frac{1}{2}[\sinh((m+n)x) + \sinh((m-n)x)]\)
• \(\cosh(mx) \cosh(nx) = \frac{1}{2}[\cosh((m+n)x) + \cosh((m-n)x)]\)
• \(\sinh(mx) \sinh(nx) = \frac{1}{2}[\cosh((m+n)x) - \cosh((m-n)x)]\)

1.17 Inverse Hyperbolic Functions

Just as we defined inverse trigonometric functions, we can define inverses of hyperbolic functions. However, the situation is simpler here because several hyperbolic functions are naturally one-to-one.

1.17.1 Existence of Inverses

• The hyperbolic sine (\(\sinh x\)) is a strictly increasing function on all of \(\mathbb{R}\). We denote its inverse by \(y = \sinh^{-1} x\) (also written as \(\text{arcsinh} \, x\), \(\text{arsinh} \, x\), or \(\text{argsinh} \, x\)).
• The function \(y = \cosh x\) is not one-to-one over all real numbers. However, the restricted function \(y = \cosh x\) for \(x \geq 0\) is one-to-one and therefore has an inverse, denoted by \(y = \cosh^{-1} x\).
• The function \(y = \text{sech} \, x\) is not one-to-one on its natural domain. But its restriction to nonnegative values of \(x\) (i.e., \(x \geq 0\)) does have an inverse, denoted by \(y = \text{sech}^{-1} x\).
• The hyperbolic tangent (\(\tanh x\)), cotangent (\(\coth x\)), and cosecant (\(\text{csch} \, x\)) are one-to-one on their domains and therefore have inverses, denoted by:

\[y = \tanh^{-1} x, \quad y = \coth^{-1} x, \quad y = \text{csch}^{-1} x\]

1.17.2 Logarithmic Formulas for Inverse Hyperbolic Functions

Since the hyperbolic functions can be expressed in terms of exponential functions, it is possible to express the inverse hyperbolic functions in terms of logarithms:

(1) \(\sinh^{-1} x = \ln(x + \sqrt{x^2 + 1})\) for \(-\infty < x < \infty\)

(2) \(\cosh^{-1} x = \ln(x + \sqrt{x^2 - 1})\) for \(x \geq 1\)

(3) \(\tanh^{-1} x = \frac{1}{2} \ln\left(\frac{1+x}{1-x}\right)\) for \(|x| < 1\)

(4) \(\text{sech}^{-1} x = \ln\left(\frac{1 + \sqrt{1-x^2}}{x}\right)\) for \(0 < x \leq 1\)

(5) \(\text{csch}^{-1} x = \ln\left(\frac{1}{x} + \frac{\sqrt{1+x^2}}{|x|}\right)\) for \(x \neq 0\)

(6) \(\coth^{-1} x = \frac{1}{2} \ln\left(\frac{x+1}{x-1}\right)\) for \(|x| > 1\)

1.17.3 Example Proofs of Logarithmic Formulas

Example 1: Let us prove \(\sinh^{-1} x = \ln(x + \sqrt{x^2 + 1})\) for \(-\infty < x < \infty\).

Let \(y = \sinh^{-1} x\). Then, by definition:

\[x = \sinh y = \frac{e^y - e^{-y}}{2}\]

Thus:

\[2x = e^y - e^{-y}\]

Multiplying both sides by \(e^y\):

\[2xe^y = e^{2y} - 1\]

Rearranging:

\[e^{2y} - 2xe^y - 1 = 0\]

This is a quadratic equation in \(e^y\). Let \(z = e^y\). Then:

\[z^2 - 2xz - 1 = 0\]

Using the quadratic formula:

\[z = \frac{2x \pm \sqrt{4x^2 + 4}}{2} = \frac{2x \pm 2\sqrt{x^2 + 1}}{2} = x \pm \sqrt{x^2 + 1}\]

Since \(z = e^y > 0\), and \(x - \sqrt{x^2 + 1} < 0\) (because \(\sqrt{x^2 + 1} > |x| \geq x\)), we must take the positive root:

\[e^y = x + \sqrt{x^2 + 1}\]

Taking the natural logarithm of both sides:

\[y = \sinh^{-1} x = \ln(x + \sqrt{x^2 + 1})\]

∎

Example 2: Let us prove \(\tanh^{-1} x = \frac{1}{2} \ln\left(\frac{1+x}{1-x}\right)\) for \(|x| < 1\).

Let \(y = \tanh^{-1} x\). Then, by definition:

\[x = \tanh y = \frac{\sinh y}{\cosh y} = \frac{e^y - e^{-y}}{e^y + e^{-y}}\]

Multiplying both sides by \(e^y + e^{-y}\):

\[x(e^y + e^{-y}) = e^y - e^{-y}\]

\[xe^y + xe^{-y} = e^y - e^{-y}\]

\[(x-1)e^y + (x+1)e^{-y} = 0\]

Multiplying both sides by \(e^y\):

\[(x-1)e^{2y} + (x+1) = 0\]

\[e^{2y} = -\frac{x+1}{x-1} = \frac{x+1}{1-x}\]

Since \(|x| < 1\), we have \(1+x > 0\) and \(1-x > 0\), so \(\frac{1+x}{1-x} > 0\). Taking the natural logarithm of both sides:

\[2y = \ln\left(\frac{1+x}{1-x}\right)\]

\[y = \tanh^{-1} x = \frac{1}{2} \ln\left(\frac{1+x}{1-x}\right)\]

∎

Example 3: Let us prove \(\text{csch}^{-1} x = \ln\left(\frac{1}{x} + \frac{\sqrt{1+x^2}}{|x|}\right)\) for \(x \neq 0\).

Let \(y = \text{csch}^{-1} x\). Then, by definition:

\[x = \text{csch} \, y = \frac{1}{\sinh y} = \frac{2}{e^y - e^{-y}}\]

Thus:

\[x(e^y - e^{-y}) = 2\]

\[xe^y - xe^{-y} = 2\]

Multiplying both sides by \(e^y\):

\[xe^{2y} - x = 2e^y\]

\[xe^{2y} - 2e^y - x = 0\]

Since \(x \neq 0\), we can divide by \(x\):

\[e^{2y} - \frac{2}{x}e^y - 1 = 0\]

Let \(z = e^y\). Then:

\[z^2 - \frac{2}{x}z - 1 = 0\]

Using the quadratic formula:

\[z = \frac{\frac{2}{x} \pm \sqrt{\frac{4}{x^2} + 4}}{2} = \frac{1}{x} \pm \sqrt{\frac{1}{x^2} + 1} = \frac{1}{x} \pm \sqrt{\frac{1+x^2}{x^2}} = \frac{1}{x} \pm \frac{\sqrt{1+x^2}}{|x|}\]

Since \(z = e^y > 0\), we take the root that gives a positive value. We need:

\[e^y = \frac{1}{x} + \frac{\sqrt{1+x^2}}{|x|}\]

(The choice of sign depends on whether \(x > 0\) or \(x < 0\), but the formula with the plus sign works for both cases when we use \(|x|\) in the denominator.)

Taking the natural logarithm of both sides:

\[y = \text{csch}^{-1} x = \ln\left(\frac{1}{x} + \frac{\sqrt{1+x^2}}{|x|}\right)\]

∎

1.17.4 Relationships Between Hyperbolic and Inverse Hyperbolic Functions

The following identities relate hyperbolic functions of inverse hyperbolic functions:

(1) \(\sinh(\text{arccosh} \, x) = \sqrt{x^2-1}\) for \(|x| > 1\)

(2) \(\sinh(\text{arctanh} \, x) = \frac{x}{\sqrt{1-x^2}}\) for \(-1 < x < 1\)

(3) \(\cosh(\text{arcsinh} \, x) = \sqrt{1+x^2}\) for all \(x\)

(4) \(\cosh(\text{arctanh} \, x) = \frac{1}{\sqrt{1-x^2}}\) for \(-1 < x < 1\)

(5) \(\tanh(\text{arcsinh} \, x) = \frac{x}{\sqrt{1+x^2}}\) for all \(x\)

(6) \(\tanh(\text{arccosh} \, x) = \frac{\sqrt{x^2-1}}{x}\) for \(|x| > 1\)

Example Proof: Let us prove \(\sinh(\text{arccosh} \, x) = \sqrt{x^2-1}\) for \(|x| > 1\).

Let \(y = \text{arccosh} \, x\). Then, by definition:

\[x = \cosh y\]

Using the fundamental hyperbolic identity \(\cosh^2 y - \sinh^2 y = 1\):

\[\sinh^2 y = \cosh^2 y - 1 = x^2 - 1\]

Since \(|x| > 1\), we have \(x^2 - 1 > 0\). Also, for \(y = \text{arccosh} \, x\), we have \(y \geq 0\), so \(\sinh y \geq 0\). Therefore:

\[\sinh y = \sqrt{x^2-1}\]

\[\sinh(\text{arccosh} \, x) = \sqrt{x^2-1}\]

∎

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2. Definitions

• Function: A rule that assigns a single value \(f(x)\) in set \(Y\) to each \(x\) in domain \(D\).
• Domain: The set of all possible input values that a function can accept.
• Range: The set of all output values that a function produces as the input varies throughout the domain.
• Graph of a Function: The set of all points \((x, f(x))\) in the Cartesian plane where \(x\) is in the domain.
• Floor Function \(\lfloor x \rfloor\): The greatest integer less than or equal to \(x\).
• Ceiling Function \(\lceil x \rceil\): The smallest integer greater than or equal to \(x\).
• Increasing Function: A function \(f\) on interval \(I\) where \(x_1 < x_2\) implies \(f(x_1) \leq f(x_2)\).
• Strictly Increasing Function: A function \(f\) on interval \(I\) where \(x_1 < x_2\) implies \(f(x_1) < f(x_2)\).
• Decreasing Function: A function \(f\) on interval \(I\) where \(x_1 < x_2\) implies \(f(x_1) \geq f(x_2)\).
• Strictly Decreasing Function: A function \(f\) on interval \(I\) where \(x_1 < x_2\) implies \(f(x_1) > f(x_2)\).
• Monotone Function: A function that is either increasing or decreasing on its domain.
• Composition of Functions \((f \circ g)(x)\): The function defined by \(f(g(x))\), where we first apply \(g\) then apply \(f\) to the result.
• One-to-One Function (Injective): A function where \(f(x_1) \neq f(x_2)\) whenever \(x_1 \neq x_2\).
• Inverse Function \(f^{-1}\): The function that reverses \(f\), defined by \(f^{-1}(b) = a\) if and only if \(f(a) = b\).
• Even Function: A function where \(f(-x) = f(x)\) for all \(x\) in its domain (symmetric about the \(y\)-axis).
• Odd Function: A function where \(f(-x) = -f(x)\) for all \(x\) in its domain (symmetric about the origin).
• Polynomial: A function of the form \(p(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0\) where \(n\) is a non-negative integer.
• Degree of a Polynomial: The highest power \(n\) with non-zero coefficient \(a_n\).
• Linear Function: A polynomial of degree 1: \(f(x) = a_1 x + a_0\).
• Quadratic Function: A polynomial of degree 2: \(p(x) = ax^2 + bx + c\).
• Cubic Function: A polynomial of degree 3: \(p(x) = ax^3 + bx^2 + cx + d\).
• Rational Function: A function of the form \(f(x) = \frac{p(x)}{q(x)}\) where \(p\) and \(q\) are polynomials.
• Exponential Function: A function of the form \(f(x) = a^x\) where \(a > 0\) is the base.
• Natural Exponential Function: The exponential function with base \(e \approx 2.718\), written as \(f(x) = e^x\).
• Logarithm Function \(\log_a x\): The inverse of the exponential function \(a^x\), defined by \(y = \log_a x\) if and only if \(x = a^y\).
• Natural Logarithm \(\ln x\): The logarithm with base \(e\), written as \(\ln x = \log_e x\).
• Common Logarithm \(\log x\): The logarithm with base 10, written as \(\log x = \log_{10} x\).
• Periodic Function: A function where \(f(x + p) = f(x)\) for all \(x\) and some positive constant \(p\) called the period.
• Sine Function \(\sin \theta\): The trigonometric function defined as \(y/r\) where \((x, y)\) is on a circle of radius \(r\).
• Cosine Function \(\cos \theta\): The trigonometric function defined as \(x/r\) where \((x, y)\) is on a circle of radius \(r\).
• Tangent Function \(\tan \theta\): The trigonometric function defined as \(y/x\) or \(\sin \theta / \cos \theta\).
• Cosecant Function \(\csc \theta\): The reciprocal of sine, \(\csc \theta = 1/\sin \theta\).
• Secant Function \(\sec \theta\): The reciprocal of cosine, \(\sec \theta = 1/\cos \theta\).
• Cotangent Function \(\cot \theta\): The reciprocal of tangent, \(\cot \theta = 1/\tan \theta\).
• Inverse Sine \(\arcsin x\): The inverse of sine restricted to \([-\pi/2, \pi/2]\), giving values in \([-\pi/2, \pi/2]\).
• Inverse Cosine \(\arccos x\): The inverse of cosine restricted to \([0, \pi]\), giving values in \([0, \pi]\).
• Inverse Tangent \(\arctan x\): The inverse of tangent restricted to \((-\pi/2, \pi/2)\), giving values in \((-\pi/2, \pi/2)\).
• Hyperbolic Sine \(\sinh x\): Defined as \(\frac{e^x - e^{-x}}{2}\).
• Hyperbolic Cosine \(\cosh x\): Defined as \(\frac{e^x + e^{-x}}{2}\).
• Hyperbolic Tangent \(\tanh x\): Defined as \(\frac{\sinh x}{\cosh x} = \frac{e^x - e^{-x}}{e^x + e^{-x}}\).
• Inverse Hyperbolic Sine \(\sinh^{-1} x\): The inverse of \(\sinh x\), equal to \(\ln(x + \sqrt{x^2 + 1})\).
• Inverse Hyperbolic Cosine \(\cosh^{-1} x\): The inverse of \(\cosh x\) for \(x \geq 0\), equal to \(\ln(x + \sqrt{x^2 - 1})\) for \(x \geq 1\).
• Inverse Hyperbolic Tangent \(\tanh^{-1} x\): The inverse of \(\tanh x\), equal to \(\frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)\) for \(|x| < 1\).

---

3. Formulas

• Function Operations:
  • Sum: \((f + g)(x) = f(x) + g(x)\)
  • Difference: \((f - g)(x) = f(x) - g(x)\)
  • Product: \((fg)(x) = f(x)g(x)\)
  • Quotient: \((f/g)(x) = \frac{f(x)}{g(x)}\) where \(g(x) \neq 0\)
  • Composition: \((f \circ g)(x) = f(g(x))\)
  • Scalar Multiplication: \((cf)(x) = cf(x)\)
• Inverse Functions:
  • \((f^{-1} \circ f)(x) = x\) for all \(x\) in domain of \(f\)
  • \((f \circ f^{-1})(y) = y\) for all \(y\) in domain of \(f^{-1}\)
• Even and Odd Function Decomposition:
  • \(f_{\text{even}}(x) = \frac{f(x) + f(-x)}{2}\)
  • \(f_{\text{odd}}(x) = \frac{f(x) - f(-x)}{2}\)
• Power Functions:
  • \(x^m = \underbrace{x \cdot x \cdots x}_{m \text{ times}}\) for \(m \in \mathbb{N}\)
  • \(x^0 = 1\) for \(x \neq 0\)
  • \(x^{-m} = \frac{1}{x^m}\) for \(x \neq 0\), \(m \in \mathbb{N}\)
  • \(x^{1/m} = \sqrt[m]{x}\) (the \(m\)th root of \(x\))
  • \(x^{n/m} = (x^{1/m})^n\)
• Logarithm Properties:
  • Product Rule: \(\ln(x_1 x_2) = \ln(x_1) + \ln(x_2)\)
  • Quotient Rule: \(\ln\left(\frac{x_1}{x_2}\right) = \ln(x_1) - \ln(x_2)\)
  • Reciprocal Rule: \(\ln\left(\frac{1}{x}\right) = -\ln(x)\)
  • Power Rule: \(\ln(x^\alpha) = \alpha \ln(x)\)
  • Inverse Property: \(e^{\ln x} = x\) for \(x > 0\)
  • Inverse Property: \(\ln(e^x) = x\) for all \(x \in \mathbb{R}\)
  • Change-of-Base Formula: \(\log_a x = \frac{\ln(x)}{\ln(a)}\)
  • Exponential as Power of \(e\): \(a^x = e^{x \ln(a)}\)
• Trigonometric Identities:
  • Pythagorean Identities:
    • \(\sin^2 x + \cos^2 x = 1\)
    • \(1 + \tan^2 x = \sec^2 x\)
    • \(1 + \cot^2 x = \csc^2 x\)
  • Addition Formulas:
    • \(\sin(x \pm y) = \sin x \cos y \pm \cos x \sin y\)
    • \(\cos(x \pm y) = \cos x \cos y \mp \sin x \sin y\)
    • \(\tan(x \pm y) = \frac{\tan x \pm \tan y}{1 \mp \tan x \tan y}\)
  • Double-Angle Formulas:
    • \(\sin(2x) = 2 \sin x \cos x\)
    • \(\cos(2x) = \cos^2 x - \sin^2 x\)
    • \(\tan(2x) = \frac{2 \tan x}{1 - \tan^2 x}\)
  • Half-Angle Formulas:
    • \(\cos^2 x = \frac{1 + \cos(2x)}{2}\)
    • \(\sin^2 x = \frac{1 - \cos(2x)}{2}\)
  • Product-to-Sum Identities:
    • \(\cos x \cos y = \frac{1}{2}[\cos(x-y) + \cos(x+y)]\)
    • \(\sin x \sin y = \frac{1}{2}[\cos(x-y) - \cos(x+y)]\)
    • \(\sin x \cos y = \frac{1}{2}[\sin(x+y) + \sin(x-y)]\)
    • \(\cos x \sin y = \frac{1}{2}[\sin(x+y) - \sin(x-y)]\)
• Inverse Trigonometric Function Identities:
  • Complementary Angles:
    • \(\arcsin x + \arccos x = \frac{\pi}{2}\)
    • \(\arctan x + \text{arccot} \, x = \frac{\pi}{2}\)
    • \(\text{arcsec} \, x + \text{arccsc} \, x = \frac{\pi}{2}\)
  • Negative Arguments:
    • \(\arcsin(-x) = -\arcsin x\)
    • \(\arccos(-x) = \pi - \arccos x\)
    • \(\arctan(-x) = -\arctan x\)
• Hyperbolic Function Definitions:
  • \(\sinh x = \frac{e^x - e^{-x}}{2}\)
  • \(\cosh x = \frac{e^x + e^{-x}}{2}\)
  • \(\tanh x = \frac{e^x - e^{-x}}{e^x + e^{-x}}\)
  • \(\coth x = \frac{e^x + e^{-x}}{e^x - e^{-x}}\)
  • \(\text{sech} \, x = \frac{2}{e^x + e^{-x}}\)
  • \(\text{csch} \, x = \frac{2}{e^x - e^{-x}}\)
• Hyperbolic Function Identities:
  • Fundamental Identity: \(\cosh^2 x - \sinh^2 x = 1\)
  • Double-Angle Formulas:
    • \(\sinh(2x) = 2 \sinh x \cosh x\)
    • \(\cosh(2x) = \cosh^2 x + \sinh^2 x\)
    • \(\tanh(2x) = \frac{2 \tanh x}{1 + \tanh^2 x}\)
  • Half-Angle-Type Formulas:
    • \(\cosh^2 x = \frac{\cosh(2x) + 1}{2}\)
    • \(\sinh^2 x = \frac{\cosh(2x) - 1}{2}\)
  • Pythagorean-Type Identities:
    • \(\tanh^2 x = 1 - \text{sech}^2 x\)
    • \(\coth^2 x = 1 + \text{csch}^2 x\)
  • Addition Formulas:
    • \(\sinh(x \pm y) = \sinh x \cosh y \pm \cosh x \sinh y\)
    • \(\cosh(x \pm y) = \cosh x \cosh y \pm \sinh x \sinh y\)
    • \(\tanh(x \pm y) = \frac{\tanh x \pm \tanh y}{1 \pm \tanh x \tanh y}\)
• Inverse Hyperbolic Function Formulas:
  • \(\sinh^{-1} x = \ln(x + \sqrt{x^2 + 1})\) for all \(x\)
  • \(\cosh^{-1} x = \ln(x + \sqrt{x^2 - 1})\) for \(x \geq 1\)
  • \(\tanh^{-1} x = \frac{1}{2} \ln\left(\frac{1+x}{1-x}\right)\) for \(|x| < 1\)
  • \(\text{sech}^{-1} x = \ln\left(\frac{1 + \sqrt{1-x^2}}{x}\right)\) for \(0 < x \leq 1\)
  • \(\text{csch}^{-1} x = \ln\left(\frac{1}{x} + \frac{\sqrt{1+x^2}}{|x|}\right)\) for \(x \neq 0\)
  • \(\coth^{-1} x = \frac{1}{2} \ln\left(\frac{x+1}{x-1}\right)\) for \(|x| > 1\)
• Relationships Between Hyperbolic and Inverse Hyperbolic Functions:
  • \(\sinh(\text{arccosh} \, x) = \sqrt{x^2-1}\) for \(|x| > 1\)
  • \(\sinh(\text{arctanh} \, x) = \frac{x}{\sqrt{1-x^2}}\) for \(-1 < x < 1\)
  • \(\cosh(\text{arcsinh} \, x) = \sqrt{1+x^2}\)
  • \(\cosh(\text{arctanh} \, x) = \frac{1}{\sqrt{1-x^2}}\) for \(-1 < x < 1\)
  • \(\tanh(\text{arcsinh} \, x) = \frac{x}{\sqrt{1+x^2}}\)
  • \(\tanh(\text{arccosh} \, x) = \frac{\sqrt{x^2-1}}{x}\) for \(|x| > 1\)

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4. Examples

4.1. Prove an Inequality by Induction (Lab 4, Task 1.1)

Prove the following statement by induction: \(\sin^{2n}(\alpha) + \cos^{2n}(\alpha) \le 1\) for all \(n \in \mathbb{N}\).

Click to see the solution

Let \(P(n)\) be the statement \(\sin^{2n}(\alpha) + \cos^{2n}(\alpha) \le 1\).

1. Base Step (n=1):
   • For \(n=1\), the statement is \(\sin^2(\alpha) + \cos^2(\alpha) \le 1\).
   • By the fundamental Pythagorean identity, \(\sin^2(\alpha) + \cos^2(\alpha) = 1\).
   • Since \(1 \le 1\) is true, the base case holds.
2. Inductive Step:
   • Assumption (Inductive Hypothesis): Assume \(P(k)\) is true for some positive integer \(k\). That is, \(\sin^{2k}(\alpha) + \cos^{2k}(\alpha) \le 1\).
   • Goal: Prove that \(P(k+1)\) is true: \(\sin^{2(k+1)}(\alpha) + \cos^{2(k+1)}(\alpha) \le 1\).
   • Proof:
     • Consider the expression for \(P(k+1)\): \(\sin^{2k+2}(\alpha) + \cos^{2k+2}(\alpha) = \sin^{2k}(\alpha)\sin^2(\alpha) + \cos^{2k}(\alpha)\cos^2(\alpha)\).
     • Since \(0 \le \sin^2(\alpha) \le 1\) and \(0 \le \cos^2(\alpha) \le 1\), we can say:
       • \(\sin^{2k}(\alpha)\sin^2(\alpha) \le \sin^{2k}(\alpha)\).
       • \(\cos^{2k}(\alpha)\cos^2(\alpha) \le \cos^{2k}(\alpha)\).
     • Adding these two inequalities gives: \(\sin^{2k+2}(\alpha) + \cos^{2k+2}(\alpha) \le \sin^{2k}(\alpha) + \cos^{2k}(\alpha)\).
     • By our inductive hypothesis, we know that \(\sin^{2k}(\alpha) + \cos^{2k}(\alpha) \le 1\).
     • Therefore, by transitivity, \(\sin^{2k+2}(\alpha) + \cos^{2k+2}(\alpha) \le 1\). This proves \(P(k+1)\) is true.
3. Conclusion: By the principle of mathematical induction, the statement holds for all \(n \in \mathbb{N}\).

4.2. Prove Divisibility by Induction (Lab 4, Task 1.2)

Show that \(5(2^{3n-2}) + 3^{3n-1}\) is divisible by 19 for all \(n \in \mathbb{N}\).

Click to see the solution

Let \(P(n) = 5(2^{3n-2}) + 3^{3n-1}\).

1. Base Step (n=1):
   • For \(n=1\), \(P(1) = 5(2^{3(1)-2}) + 3^{3(1)-1} = 5(2^1) + 3^2 = 10 + 9 = 19\).
   • Since \(19\) is divisible by 19, the base case is true.
2. Inductive Step:
   • Assumption (Inductive Hypothesis): Assume \(P(k)\) is divisible by 19 for some positive integer \(k\). That is, \(5(2^{3k-2}) + 3^{3k-1} = 19m\) for some integer \(m\).
   • Goal: Prove that \(P(k+1) = 5(2^{3(k+1)-2}) + 3^{3(k+1)-1}\) is divisible by 19.
   • Proof:
     • \(P(k+1) = 5(2^{3k+1}) + 3^{3k+2} = 5(2^3 \cdot 2^{3k-2}) + 3^3 \cdot 3^{3k-1}\).
     • \(= 8 \cdot [5(2^{3k-2})] + 27 \cdot [3^{3k-1}]\).
     • From the hypothesis, we can write \(5(2^{3k-2}) = 19m - 3^{3k-1}\). Substitute this into the expression:
     • \(= 8(19m - 3^{3k-1}) + 27 \cdot 3^{3k-1}\)
     • \(= 8 \cdot 19m - 8 \cdot 3^{3k-1} + 27 \cdot 3^{3k-1}\)
     • \(= 8 \cdot 19m + 19 \cdot 3^{3k-1}\)
     • \(= 19(8m + 3^{3k-1})\).
   • Since \(m\) and \(k\) are integers, \((8m + 3^{3k-1})\) is an integer. Thus, \(P(k+1)\) is a multiple of 19.
3. Conclusion: By the principle of mathematical induction, the statement is true for all \(n \in \mathbb{N}\).

4.3. Find Inverse Functions (Lab 4, Task 3)

Find \(f^{-1}(x)\) and identify the domain and range of \(f^{-1}\) for the following functions:

1. \(f(x) = x^5\)
2. \(f(x) = (2x^3 + 1)^{1/5}\)
3. \(f(x) = \frac{\sqrt{x}}{\sqrt{x}-3}\)

Click to see the solution

The domain of \(f^{-1}\) is the range of \(f\), and the range of \(f^{-1}\) is the domain of \(f\).

(a) \(f(x) = x^5\)

• Inverse: Let \(y=x^5\). Solving for \(x\) gives \(x = y^{1/5}\). So, \(f^{-1}(x) = x^{1/5}\).
• Domain and Range of \(f\): \(D_f = (-\infty, \infty)\), \(R_f = (-\infty, \infty)\).
• Domain and Range of \(f^{-1}\): \(D_{f^{-1}} = (-\infty, \infty)\), \(R_{f^{-1}} = (-\infty, \infty)\).

(b) \(f(x) = (2x^3 + 1)^{1/5}\)

• Inverse: Let \(y=(2x^3 + 1)^{1/5}\). Then \(y^5 = 2x^3+1 \implies y^5-1 = 2x^3 \implies x^3 = \frac{y^5-1}{2} \implies x = \sqrt{\frac{y^5-1}{2}}\). So, \(f^{-1}(x) = \sqrt{\frac{x^5-1}{2}}\).
• Domain and Range of \(f\): \(D_f = (-\infty, \infty)\), \(R_f = (-\infty, \infty)\).
• Domain and Range of \(f^{-1}\): \(D_{f^{-1}} = (-\infty, \infty)\), \(R_{f^{-1}} = (-\infty, \infty)\).

(c) \(f(x) = \frac{\sqrt{x}}{\sqrt{x}-3}\)

• Inverse: Let \(y=\frac{\sqrt{x}}{\sqrt{x}-3}\). Then \(y(\sqrt{x}-3) = \sqrt{x} \implies y\sqrt{x}-3y=\sqrt{x} \implies y\sqrt{x}-\sqrt{x}=3y \implies \sqrt{x}(y-1)=3y \implies \sqrt{x} = \frac{3y}{y-1} \implies x = \left(\frac{3y}{y-1}\right)^2\). So, \(f^{-1}(x) = \left(\frac{3x}{x-1}\right)^2\).
• Domain and Range of \(f\): Domain requires \(x \ge 0\) and \(\sqrt{x} \neq 3 \implies x \neq 9\). So \(D_f = [0, 9) \cup (9, \infty)\). The range is \((-\infty, 0] \cup (1, \infty)\).
• Domain and Range of \(f^{-1}\): \(D_{f^{-1}} = (-\infty, 0] \cup (1, \infty)\), \(R_{f^{-1}} = [0, 9) \cup (9, \infty)\).

4.4. Properties of Functions (Lab 4, Task 4)

Given \(f(x) = x^2\), choose suitable domains and codomains to make the function:

1. Surjective
2. Injective
3. Bijective

Click to see the solution

Let the function be defined as \(f: A \to B\), where A is the domain and B is the codomain. The range of \(f(x)=x^2\) is \([0, \infty)\).

(a) Surjective: A function is surjective if its codomain equals its range.

• We can keep the standard domain \(A = \mathbb{R}\).
• We must set the codomain \(B\) to be the range, so \(B = [0, \infty)\).
• Example: \(f: \mathbb{R} \to [0, \infty)\) is surjective.

(b) Injective: A function is injective (one-to-one) if it passes the horizontal line test. The function \(f(x)=x^2\) fails this test on \(\mathbb{R}\).

• We must restrict the domain to a part where the function is strictly increasing or decreasing.
• We can choose the domain \(A = [0, \infty)\) (or \(A = (-\infty, 0]\)).
• The codomain B can be \(\mathbb{R}\) or any superset of the range.
• Example: \(f: [0, \infty) \to \mathbb{R}\) is injective.

(c) Bijective: A function is bijective if it is both surjective and injective.

• We must combine the restrictions from the previous parts.
• We must restrict the domain to make it injective, e.g., \(A = [0, \infty)\).
• We must set the codomain to equal the range to make it surjective, e.g., \(B = [0, \infty)\).
• Example: \(f: [0, \infty) \to [0, \infty)\) is bijective. Another example is \(f: (-\infty, 0] \to [0, \infty)\).

4.5. Prove Hyperbolic and Inverse Function Identities (Chapter 2, Example 1)

Prove the following identities involving hyperbolic and inverse hyperbolic/trigonometric functions.

1. \(\sinh(\text{arccosh } x) = \sqrt{x^2 - 1}; \quad |x| \ge 1\)
2. \(\sinh(\text{arctanh } x) = \frac{x}{\sqrt{1-x^2}}; \quad -1 < x < 1\)
3. \(\cosh(\text{arcsinh } x) = \sqrt{1+x^2}\)
4. \(\cosh(\text{arctanh } x) = \frac{1}{\sqrt{1-x^2}}; \quad -1 < x < 1\)
5. \(\tanh(\text{arcsinh } x) = \frac{x}{\sqrt{1+x^2}}\)
6. \(\tanh(\text{arccosh } x) = \frac{\sqrt{x^2-1}}{x}; \quad |x| \ge 1\)

Click to see the solution

The proofs for (1), (3), and (6) are shown in other tasks. The others follow a similar pattern. Let’s prove (2) as an example.

1. Let \(y = \text{arctanh } x\). By definition, this means \(x = \tanh y\).
2. Use hyperbolic identities. We know the identity \(\text{sech}^2 y = 1 - \tanh^2 y\). Since \(\cosh y = \frac{1}{\text{sech } y}\), we have \(\cosh^2 y = \frac{1}{1 - \tanh^2 y} = \frac{1}{1-x^2}\).
3. Find \(\cosh y\). Since the range of arctanh is all real numbers, but \(\cosh y\) is always positive, we take the positive root: \(\cosh y = \frac{1}{\sqrt{1-x^2}}\).
4. Find \(\sinh y\). We know \(\tanh y = \frac{\sinh y}{\cosh y}\), so \(\sinh y = \tanh y \cosh y\).
5. Substitute: \(\sinh y = x \cdot \frac{1}{\sqrt{1-x^2}} = \frac{x}{\sqrt{1-x^2}}\).
6. Conclusion: Since \(y = \text{arctanh } x\), we have shown \(\sinh(\text{arctanh } x) = \frac{x}{\sqrt{1-x^2}}\).

4.6. Floor and Ceiling Functions (Chapter 2, Example 2)

Explain the floor function (\(\lfloor x \rfloor\)) and the ceiling function (\(\lceil x \rceil\)) and provide examples for the inputs 2.4, 1.9, 0, -1.2, 2, 0.2, and -0.3.

Click to see the solution

1. Floor Function \(\lfloor x \rfloor\):
   • Definition: The floor function, also called the greatest integer function, gives the greatest integer that is less than or equal to \(x\).
   • Examples:
     • \(\lfloor 2.4 \rfloor = 2\)
     • \(\lfloor 1.9 \rfloor = 1\)
     • \(\lfloor 0 \rfloor = 0\)
     • \(\lfloor -1.2 \rfloor = -2\)
     • \(\lfloor 2 \rfloor = 2\)
     • \(\lfloor 0.2 \rfloor = 0\)
     • \(\lfloor -0.3 \rfloor = -1\)
2. Ceiling Function \(\lceil x \rceil\):
   • Definition: The ceiling function, also called the least integer function, gives the smallest integer that is greater than or equal to \(x\).
   • Examples:
     • \(\lceil 2.4 \rceil = 3\)
     • \(\lceil 1.9 \rceil = 2\)
     • \(\lceil 0 \rceil = 0\)
     • \(\lceil -1.2 \rceil = -1\)
     • \(\lceil 2 \rceil = 2\)
     • \(\lceil 0.2 \rceil = 1\)
     • \(\lceil -0.3 \rceil = 0\)

4.7. Composition of Functions (Chapter 2, Example 3)

Let \(f(x) = \sqrt{x}\) and \(g(x) = x+1\). Find the compositions \((f \circ g)(x)\), \((g \circ f)(x)\), \((f \circ f)(x)\), and \((g \circ g)(x)\) and their respective domains.

Click to see the solution

1. \((f \circ g)(x)\):
   • \((f \circ g)(x) = f(g(x)) = f(x+1) = \sqrt{x+1}\).
   • Domain: For the output to be real, we need \(x+1 \ge 0 \implies x \ge -1\). Domain is \([-1, \infty)\).
2. \((g \circ f)(x)\):
   • \((g \circ f)(x) = g(f(x)) = g(\sqrt{x}) = \sqrt{x}+1\).
   • Domain: For the input of \(f(x)\) to be valid, we need \(x \ge 0\). Domain is \([0, \infty)\).
3. \((f \circ f)(x)\):
   • \((f \circ f)(x) = f(f(x)) = f(\sqrt{x}) = \sqrt{\sqrt{x}} = x^{1/4}\).
   • Domain: For the input of the outer \(f\) to be valid, we need \(f(x) \ge 0\), which is always true for \(\sqrt{x}\). For the input of the inner \(f\), we need \(x \ge 0\). Domain is \([0, \infty)\).
4. \((g \circ g)(x)\):
   • \((g \circ g)(x) = g(g(x)) = g(x+1) = (x+1)+1 = x+2\).
   • Domain: The domain of linear functions is all real numbers. Domain is \((-\infty, \infty)\).

4.8. Domain and Range of Common Functions (Chapter 2, Example 4)

State the domain and range for the following common functions: \(y=x^2\), \(y=1/x\), \(y=\sqrt{x}\), \(y=\sqrt{4-x}\), \(y=\sqrt{4-x^2}\).

Click to see the solution

1. \(y = x^2\) (Parabola):
   • Domain: The function is defined for all real numbers. Domain is \((-\infty, \infty)\).
   • Range: The square of any real number is non-negative. Range is \([0, \infty)\).
2. \(y = 1/x\) (Hyperbola):
   • Domain: The function is defined for all real numbers except where the denominator is zero. Domain is \((-\infty, 0) \cup (0, \infty)\).
   • Range: The function can take any real value except zero. Range is \((-\infty, 0) \cup (0, \infty)\).
3. \(y = \sqrt{x}\) (Square Root):
   • Domain: The expression under the square root must be non-negative. Domain is \([0, \infty)\).
   • Range: The principal square root is always non-negative. Range is \([0, \infty)\).
4. \(y = \sqrt{4-x}\):
   • Domain: We require \(4-x \ge 0 \implies x \le 4\). Domain is \((-\infty, 4]\).
   • Range: The output is always non-negative. Range is \([0, \infty)\).
5. \(y = \sqrt{4-x^2}\) (Semicircle):
   • Domain: We require \(4-x^2 \ge 0 \implies x^2 \le 4 \implies -2 \le x \le 2\). Domain is \([-2, 2]\).
   • Range: The minimum value is 0 (at \(x=\pm 2\)) and the maximum value is \(\sqrt{4-0}=2\) (at \(x=0\)). Range is $$.

4.9. Derive the Logarithmic Form of \(\sinh^{-1}(x)\) (Chapter 2, Example 5)

Prove the first formula: \(\sinh^{-1} x = \ln (x + \sqrt{x^2 + 1})\).

Click to see the solution

1. Let \(y = \sinh^{-1}(x)\). By definition, \(x = \sinh(y)\).
2. Use the exponential definition of \(\sinh(y)\): \(x = \frac{e^y - e^{-y}}{2}\).
3. Multiply by 2 and then by \(e^y\) to clear the denominator and negative exponent: \(2x = e^y - e^{-y} \implies 2xe^y = e^{2y} - 1\).
4. Rearrange into a quadratic equation in terms of \(e^y\): \((e^y)^2 - 2x(e^y) - 1 = 0\).
5. Let \(z = e^y\). The equation is \(z^2 - 2xz - 1 = 0\). Solve for \(z\) using the quadratic formula:
   • \(z = \frac{-(-2x) \pm \sqrt{(-2x)^2 - 4(1)(-1)}}{2} = \frac{2x \pm \sqrt{4x^2+4}}{2} = x \pm \sqrt{x^2+1}\).
6. Since \(z=e^y\), it must be positive. The term \(\sqrt{x^2+1}\) is always greater than \(|x|\). Therefore, the root \(x - \sqrt{x^2+1}\) is always negative. We must take the positive root.
   • \(e^y = x + \sqrt{x^2+1}\).
7. Take the natural logarithm of both sides to solve for \(y\): \(y = \ln(x + \sqrt{x^2+1})\).

4.10. Prove a Hyperbolic-Inverse Trig Identity (Chapter 2, Example 6)

Prove the identity \(\sinh(\text{arccosh}(x)) = \sqrt{x^2-1}\) for \(|x| \ge 1\).

Click to see the solution

1. Set up the variables: Let \(y = \text{arccosh}(x)\). By the definition of the inverse hyperbolic cosine, this means \(x = \cosh(y)\).
2. Use the fundamental hyperbolic identity: \(\cosh^2(y) - \sinh^2(y) = 1\).
3. Solve for \(\sinh(y)\):
   • \(\sinh^2(y) = \cosh^2(y) - 1\).
   • Substitute \(x = \cosh(y)\): \(\sinh^2(y) = x^2 - 1\).
4. Take the square root:
   • \(\sinh(y) = \pm\sqrt{x^2-1}\).
5. Determine the sign: The range of the arccosh function is \(y \ge 0\). For non-negative values of \(y\), the function \(\sinh(y)\) is also non-negative. Therefore, we must take the positive root.
   • \(\sinh(y) = \sqrt{x^2-1}\).
6. Substitute back: Since \(y = \text{arccosh}(x)\), we have \(\sinh(\text{arccosh}(x)) = \sqrt{x^2-1}\).

4.11. Prove Inverse Trigonometric Identities (Chapter 2, Example 7)

Prove the following identities for inverse trigonometric functions:

1. \(\arcsin(x) + \arccos(x) = \frac{\pi}{2}\) for \(x \in [-1, 1]\).
2. \(\arcsin(-x) = -\arcsin(x)\) for \(x \in [-1, 1]\).

Click to see the solution

(a) \(\arcsin(x) + \arccos(x) = \frac{\pi}{2}\)

1. Let \(\theta = \arcsin(x)\). By definition, this means \(\sin(\theta) = x\) and \(\theta \in [-\pi/2, \pi/2]\).
2. Using the co-function identity, we know that \(\cos(\frac{\pi}{2} - \theta) = \sin(\theta)\).
3. Therefore, \(\cos(\frac{\pi}{2} - \theta) = x\).
4. Since \(\theta \in [-\pi/2, \pi/2]\), the angle \((\frac{\pi}{2} - \theta)\) is in the interval \([0, \pi]\), which is the range of the arccos function.
5. From the definition of arccos, it follows that \(\arccos(x) = \frac{\pi}{2} - \theta\).
6. Substituting back \(\theta = \arcsin(x)\), we get \(\arccos(x) = \frac{\pi}{2} - \arcsin(x)\), which rearranges to the desired identity.

(b) \(\arcsin(-x) = -\arcsin(x)\)

1. Let \(\theta = \arcsin(x)\). This means \(\sin(\theta) = x\) and \(\theta \in [-\pi/2, \pi/2]\).
2. The sine function is an odd function, so \(\sin(-\theta) = -\sin(\theta) = -x\).
3. Since \(\theta \in [-\pi/2, \pi/2]\), the angle \(-\theta\) is also in the interval \([-\pi/2, \pi/2]\).
4. From the definition of arcsin, it follows that \(\arcsin(-x) = -\theta\).
5. Substituting back \(\theta = \arcsin(x)\) gives \(\arcsin(-x) = -\arcsin(x)\).

4.12. Derive the Logarithmic Form of \(\tanh^{-1}(x)\) (Chapter 2, Example 8)

Prove the third formula: \(\tanh^{-1} x = \frac{1}{2}\ln \left(\frac{1+x}{1-x}\right)\).

Click to see the solution

1. Let \(y = \tanh^{-1}(x)\). By definition, \(x = \tanh(y)\).
2. Use the exponential definition: \(x = \frac{e^y - e^{-y}}{e^y + e^{-y}}\).
3. Multiply numerator and denominator by \(e^y\): \(x = \frac{e^{2y}-1}{e^{2y}+1}\).
4. Solve for \(e^{2y}\):
   • \(x(e^{2y}+1) = e^{2y}-1\).
   • \(xe^{2y}+x = e^{2y}-1\).
   • \(x+1 = e^{2y}-xe^{2y} = e^{2y}(1-x)\).
   • \(e^{2y} = \frac{1+x}{1-x}\).
5. Take the natural logarithm of both sides to solve for \(y\):
   • \(2y = \ln\left(\frac{1+x}{1-x}\right)\).
   • \(y = \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)\).

4.13. Prove an Inverse Trigonometric Identity (Chapter 2, Example 9)

Prove the following identity: \(\arcsin(\frac{1}{x}) = \text{arccsc}(x)\) for \(x \in (-\infty, -1] \cup [1, \infty)\).

Click to see the solution

1. Let \(\theta = \arcsin(\frac{1}{x})\). By the definition of the arcsin function, this means \(\sin(\theta) = \frac{1}{x}\).
2. Use the reciprocal identity. The definition of the cosecant function is \(\csc(\theta) = \frac{1}{\sin(\theta)}\).
3. Substitute. From step 1, we can substitute \(\frac{1}{x}\) for \(\sin(\theta)\), which gives \(\csc(\theta) = \frac{1}{1/x} = x\).
4. Apply the inverse cosecant function. The statement \(\csc(\theta) = x\) is, by definition, equivalent to \(\theta = \text{arccsc}(x)\).
5. Conclusion. We started with \(\theta = \arcsin(\frac{1}{x})\) and have shown that \(\theta = \text{arccsc}(x)\). Therefore, the two expressions must be equal.

4.14. List Logarithmic Forms of Inverse Hyperbolic Functions (Chapter 2, Example 10)

List the logarithmic form for each of the six inverse hyperbolic functions and their respective domains.

Click to see the solution

1. \(\sinh^{-1} x = \ln (x + \sqrt{x^2 + 1}), \quad -\infty < x < \infty\).
2. \(\cosh^{-1} x = \ln (x + \sqrt{x^2 - 1}), \quad x \ge 1\).
3. \(\tanh^{-1} x = \frac{1}{2}\ln \left(\frac{1+x}{1-x}\right), \quad |x| < 1\).
4. \(\text{sech}^{-1} x = \ln \left(\frac{1 + \sqrt{1-x^2}}{x}\right), \quad 0 < x \le 1\).
5. \(\text{csch}^{-1} x = \ln \left(\frac{1}{x} + \frac{\sqrt{1+x^2}}{|x|}\right), \quad x \neq 0\).
6. \(\coth^{-1} x = \frac{1}{2}\ln \left(\frac{x+1}{x-1}\right), \quad |x| > 1\).

4.15. Derive the Logarithmic Form of \(\text{csch}^{-1}(x)\) (Chapter 2, Example 11)

Prove the fifth formula: \(\text{csch}^{-1} x = \ln \left(\frac{1}{x} + \frac{\sqrt{1+x^2}}{|x|}\right)\).

Click to see the solution

1. Let \(y = \text{csch}^{-1}(x)\). By definition, \(x = \text{csch}(y)\).
2. Use the exponential definition: \(x = \frac{1}{\sinh y} = \frac{2}{e^y - e^{-y}}\).
3. Solve for \(e^y\):
   • \(e^y - e^{-y} = \frac{2}{x}\).
   • Multiply by \(e^y\): \(e^{2y} - 1 = \frac{2}{x}e^y\).
   • Rearrange into a quadratic equation: \((e^y)^2 - \frac{2}{x}(e^y) - 1 = 0\).
4. Let \(z=e^y\). Solve \(z^2 - \frac{2}{x}z - 1 = 0\) using the quadratic formula:
   • \(z = \frac{2/x \pm \sqrt{4/x^2+4}}{2} = \frac{1}{x} \pm \sqrt{\frac{1}{x^2}+1} = \frac{1}{x} \pm \frac{\sqrt{1+x^2}}{x^2} = \frac{1}{x} \pm \frac{\sqrt{1+x^2}}{|x|}\).
5. Since \(z=e^y\) must be positive, we must choose the correct root. The term \(\frac{\sqrt{1+x^2}}{|x|}\) is always positive. If we take the minus sign, the result could be negative (e.g., if x is positive and small). The plus sign guarantees a positive result.
   • \(e^y = \frac{1}{x} + \frac{\sqrt{1+x^2}}{|x|}\).
6. Take the natural logarithm: \(y = \ln(\frac{1}{x} + \frac{\sqrt{1+x^2}}{|x|})\).

4.16. Proof by Induction (Test 1, Task 1a)

Prove by induction the following statement: \(1^3 + 2^3 + 3^3 + \dots + n^3 = \frac{n^2(n+1)^2}{4}\) for all \(n \ge 1\).

Click to see the solution

Let \(P(n)\) be the statement \(1^3 + 2^3 + \dots + n^3 = \frac{n^2(n+1)^2}{4}\).

1. Base Step (n=1):
   • LHS = \(1^3 = 1\).
   • RHS = \(\frac{1^2(1+1)^2}{4} = \frac{1 \cdot 4}{4} = 1\).
   • Since LHS = RHS, the statement \(P(1)\) is true.
2. Inductive Step:
   • Assumption (Inductive Hypothesis): Assume \(P(k)\) is true for some positive integer \(k\). That is, \(1^3 + \dots + k^3 = \frac{k^2(k+1)^2}{4}\).
   • Goal: Prove that \(P(k+1)\) is true: \(1^3 + \dots + (k+1)^3 = \frac{(k+1)^2(k+2)^2}{4}\).
   • Proof:
     • Start with the left-hand side of the goal: LHS = \((1^3 + \dots + k^3) + (k+1)^3\).
     • Substitute the inductive hypothesis: LHS = \(\frac{k^2(k+1)^2}{4} + (k+1)^3\).
     • Find a common denominator and factor out \((k+1)^2\): \(= \frac{k^2(k+1)^2 + 4(k+1)^3}{4} = \frac{(k+1)^2 [k^2 + 4(k+1)]}{4}\).
     • Simplify the expression in the brackets: \(= \frac{(k+1)^2 [k^2 + 4k + 4]}{4}\).
     • Recognize the perfect square: \(= \frac{(k+1)^2 (k+2)^2}{4}\).
   • This is the right-hand side of the goal. Thus, \(P(k+1)\) is true.
3. Conclusion: By the principle of mathematical induction, the statement is true for all integers \(n \ge 1\).

4.17. Proof by Induction (Divisibility) (Test 1, Task 1b)

Prove by induction that \(4 | (9^n + 3)\) for all \(n \ge 0\) (i.e., 4 is a factor of \(9^n+3\)).

Click to see the solution

Let \(P(n)\) be the statement that \(9^n + 3\) is divisible by 4.

1. Base Step (n=0):
   • For \(n=0\), we have \(9^0 + 3 = 1 + 3 = 4\).
   • Since 4 is divisible by 4, the statement \(P(0)\) is true.
2. Inductive Step:
   • Assumption (Inductive Hypothesis): Assume \(P(k)\) is true for some integer \(k \ge 0\). That is, \(9^k + 3\) is divisible by 4, which means we can write \(9^k + 3 = 4m\) for some integer \(m\).
   • Goal: Prove that \(P(k+1)\) is true: \(9^{k+1} + 3\) is divisible by 4.
   • Proof:
     • Consider the expression \(9^{k+1} + 3 = 9 \cdot 9^k + 3\).
     • From the hypothesis, we can express \(9^k\) as \(9^k = 4m - 3\). Substitute this into the expression:
     • \(= 9(4m - 3) + 3\)
     • \(= 36m - 27 + 3\)
     • \(= 36m - 24\)
     • \(= 4(9m - 6)\).
   • Since \(m\) is an integer, \((9m - 6)\) is also an integer. Thus, \(9^{k+1} + 3\) is a multiple of 4. This shows \(P(k+1)\) is true.
3. Conclusion: By the principle of mathematical induction, the statement is true for all integers \(n \ge 0\).

4.18. Find Minimum, Maximum, Infimum, and Supremum of Sets (Test 1, Task 2)

Let \(A = [1, 5)\) and \(B = (1, 5) \cup \{6, 8\}\). Fill in the blanks (put X if there isn’t an answer):

1. max(A), min(A), sup(A), inf(A)
2. max(B), min(B), sup(B), inf(B)

Click to see the solution

(a) For \(A = [1, 5)\)

• max(A) = X: The maximum is the largest element in the set. Since 5 is not included, there is no largest element.
• min(A) = 1: The minimum is the smallest element in the set. Since 1 is included, it is the minimum.
• sup(A) = 5: The supremum is the least upper bound. 5 is an upper bound, and any number smaller than 5 is not.
• inf(A) = 1: The infimum is the greatest lower bound. It is equal to the minimum when a minimum exists.

(b) For \(B = (1, 5) \cup \{6, 8\}\)

• max(B) = 8: The largest element in the set is 8.
• min(B) = X: The minimum would be the smallest element. The set gets arbitrarily close to 1 but does not include 1, so there is no minimum.
• sup(B) = 8: The supremum is the least upper bound, which is the maximum when a maximum exists.
• inf(B) = 1: The infimum is the greatest lower bound. The set of lower bounds is \((-\infty, 1]\), and the greatest of these is 1.

4.19. Complex Number Arithmetic (Test 1, Task 3a)

Let \(z_1 = -1 + i\) and \(z_2 = -2 - 4i\). Find the following:

1. \(2z_1 - 3z_2\)
2. \(\frac{z_1+z_2}{z_1^2}\)

Click to see the solution

(a) \(2z_1 - 3z_2\)

1. Substitute the values: \(2(-1 + i) - 3(-2 - 4i)\).
2. Distribute the scalars: \((-2 + 2i) - (-6 - 12i)\).
3. Combine real and imaginary parts: \((-2 + 6) + (2i + 12i) = 4 + 14i\).

(b) \(\frac{z_1+z_2}{z_1^2}\)

1. Calculate the numerator: \(z_1 + z_2 = (-1 + i) + (-2 - 4i) = -3 - 3i\).
2. Calculate the denominator: \(z_1^2 = (-1 + i)^2 = (-1)^2 + 2(-1)(i) + i^2 = 1 - 2i - 1 = -2i\).
3. Perform the division: \(\frac{-3 - 3i}{-2i}\).
4. Multiply by the conjugate of the denominator: \(\frac{-3 - 3i}{-2i} \cdot \frac{2i}{2i} = \frac{-6i - 6i^2}{-4i^2}\).
5. Simplify: \(\frac{-6i - 6(-1)}{-4(-1)} = \frac{6 - 6i}{4} = \frac{6}{4} - \frac{6}{4}i = \frac{3}{2} - \frac{3}{2}i\).

Answer:

1. \(4 + 14i\)
2. \(\frac{3}{2} - \frac{3}{2}i\)

4.20. Trigonometric and Exponential Forms of a Complex Number (Test 1, Task 3b)

Find the trigonometric and exponential forms of \(z_1 = -1 + i\).

Click to see the solution

1. Find the Modulus (r): The modulus is the distance from the origin to the point \((-1, 1)\) in the complex plane.
   • \(r = |z_1| = \sqrt{(-1)^2 + 1^2} = \sqrt{1+1} = \sqrt{2}\).
2. Find the Argument (\(\theta\)): The argument is the angle the vector makes with the positive real axis.
   • The point \((-1, 1)\) is in the second quadrant.
   • The reference angle is \(\arctan\left(\frac{|1|}{|-1|}\right) = \arctan(1) = \pi/4\).
   • In the second quadrant, the angle is \(\theta = \pi - (\text{reference angle}) = \pi - \pi/4 = 3\pi/4\).
3. Write the Trigonometric Form: The form is \(z = r(\cos\theta + i\sin\theta)\).
   • \(z_1 = \sqrt{2}\left(\cos\left(\frac{3\pi}{4}\right) + i\sin\left(\frac{3\pi}{4}\right)\right)\).
4. Write the Exponential Form: The form is \(z = re^{i\theta}\).
   • \(z_1 = \sqrt{2}e^{i3\pi/4}\).

Answer:

• Trigonometric: \(\sqrt{2}(\cos(3\pi/4) + i\sin(3\pi/4))\)
• Exponential: \(\sqrt{2}e^{i3\pi/4}\)

4.21. Prove a Function is One-to-One and Find its Inverse (Test 1, Task 4)

Let \(f(x) = \frac{x-2}{2x-3}\). Does \(f\) represent a one-to-one function (prove)? If Yes, Find \(f^{-1}\).

Click to see the solution

Part 1: Prove the function is one-to-one

1. Definition of one-to-one: A function \(f\) is one-to-one if for any \(a, b\) in the domain of \(f\), the condition \(f(a)=f(b)\) implies that \(a=b\).
2. Set up the equation: Assume \(f(a) = f(b)\) for some \(a\) and \(b\) in the domain.
   • \(\frac{a-2}{2a-3} = \frac{b-2}{2b-3}\).
3. Solve for \(a\):
   • Cross-multiply: \((a-2)(2b-3) = (b-2)(2a-3)\).
   • Expand both sides: \(2ab - 3a - 4b + 6 = 2ab - 3b - 4a + 6\).
   • Subtract \(2ab\) and 6 from both sides: \(-3a - 4b = -3b - 4a\).
   • Add \(4a\) and \(4b\) to both sides: \(a = b\).
4. Conclusion: Since \(f(a)=f(b)\) implies \(a=b\), the function is one-to-one.

Part 2: Find the inverse function \(f^{-1}\)

1. Replace \(f(x)\) with y: \(y = \frac{x-2}{2x-3}\).
2. Solve for x in terms of y:
   • \(y(2x-3) = x-2\).
   • \(2xy - 3y = x-2\).
   • \(2xy - x = 3y - 2\).
   • Factor out x: \(x(2y-1) = 3y-2\).
   • \(x = \frac{3y-2}{2y-1}\).
3. Swap x and y to get the inverse function:
   • \(y = \frac{3x-2}{2x-1}\).

Answer: Yes, the function is one-to-one. The inverse is \(f^{-1}(x) = \frac{3x-2}{2x-1}\).