Comprehensive Notes on Mathematical Analysis: Functions and Their Properties

Foundations and Definitions of Functions

• Fundamental Definition of a Function:

  • Let $X$ and $Y$ be two arbitrary sets.

  • A function $f$ defined on $X$ with values in $Y$ is a correspondence that associates to each element $x \in X$ at most one element $y \in Y$.

  • The mapping notation is written as:     $f : \operatorname{dom} f \subseteq X \rightarrow Y$

• Domain ($\operatorname{dom} f$ or $D_f$):

  • The domain represents the subset of elements $x \in X$ to which the function $f$ associates an element in $Y$.

  • If an element $x \in X$ is not assigned a corresponding value in $Y$, it does not belong to the domain of $f$.

• Image and Range ($\operatorname{im} f$ or $W_f$):

  • The element $y \in Y$ associated with a given element $x \in \operatorname{dom} f$ is defined as the image of $x$ under $f$, expressed symbolically as:     $y = f(x)$

  • The range (or image set) of $f$, denoted as $\operatorname{im} f$ or $W_f$, constitutes the set of all images of points in the domain:     $\operatorname{im} f = \{y \in Y \mid \exists_{x \in \operatorname{dom} f}\, y = f(x)\} = \{f(x) \mid x \in \operatorname{dom} f\}$

• Graph of a Function ($\Gamma(f)$):

  • The graph of a function $f$ is the specific subset $\Gamma(f)$ of the Cartesian product $X \times Y$ formed by all ordered pairs $(x, f(x))$ as $x$ varies throughout the domain of $f$:     $\Gamma(f) = \{(x, f(x)) \in X \times Y \mid x \in \operatorname{dom} f\}$

• The Vertical Line Test:

  • A geometric curve in the Cartesian plane $\mathbb{R} \times \mathbb{R}$ represents the graph of a single-valued function if and only if every vertical line intersects the curve at most once.

  • If any vertical line intersects a curve at two or more points, that correspondence associates multiple values of $y$ to a single $x$, violating the definition of a function.

• Determining Domain and Range via Orthogonal Projections:

  • Given the Cartesian graph of a function $y = f(x)$, its domain $D_f$ is obtained by taking the orthogonal projection of the graph onto the horizontal $x$-axis.

  • Similarly, the image/range $W_f$ is obtained by taking the orthogonal projection of the graph onto the vertical $y$-axis.

Natural Domain and Range Determination

• Concept of Natural Domain:

  • When a function is defined solely by an analytical formula without an explicitly specified domain, its natural domain is the largest subset of real numbers $\mathbb{R}$ for which the mathematical operations in the formula are well-defined and yield real values.

• Worked Example: Logarithmic Function Natural Domain:

  • Consider the function:     $f(x) = \log(x^2 - 1)$

  • Recall the evaluation condition for a general logarithm $\log_a(b)$, which requires:     $a \in (0, 1) \cup (1, \infty)$     $b \in (0, \infty)$

  • For the standard logarithm, the argument must be strictly positive:     $x^2 - 1 > 0$

  • Factoring the difference of squares yields:     $(x - 1)(x + 1) > 0$

  • Determining the sign of the parabola $(x - 1)(x + 1)$, the product is strictly positive when $x < -1$ or $x > 1$.

  • Consequently, the natural domain is:     $\operatorname{dom} f = (-\infty, -1) \cup (1, \infty)$

• Worked Example: Natural Domain and Range of a Radical-Trigonometric Function:

  • Consider the function:     $h(x) = 1 + 2\sqrt[4]{\sin(x)}$

  • Domain Determination:

  • An even root (here the $4$-th root) is defined over real numbers only for non-negative radicands:       $\sin(x) \ge 0$

  • Over the fundamental cycle, $\sin(x) \ge 0$ on $[0, \pi]$.

  • Incorporating the period of $2\pi$, the condition holds for all integers $k \in \mathbb{Z}$:       $x \in [0 + 2k\pi, \pi + 2k\pi]$

  • Thus, the natural domain is:       $\operatorname{dom} h = \bigcup_{k \in \mathbb{Z}} [2k\pi, (2k + 1)\pi]$

  • Range Determination:

  • On the domain $\operatorname{dom} h$, the sine function is bounded within:       $\sin(x) \in [0, 1]$

  • Evaluating the lower bound where $\sin(x) = 0$:       $h_{\min} = 1 + 2\sqrt[4]{0} = 1$

  • Evaluating the upper bound where $\sin(x) = 1$:       $h_{\max} = 1 + 2\sqrt[4]{1} = 1 + 2(1) = 3$

  • Therefore, the image/range of $h$ is the closed interval:       $\operatorname{im} h = [1, 3]$

Equality of Functions

• Formal Criteria for Equality:

  • Two functions $f : \operatorname{dom} f \rightarrow Y$ and $g : \operatorname{dom} g \rightarrow Y$ are equal ($f = g$) if and only if both of the following conditions are simultaneously satisfied:     $\operatorname{dom} f = \operatorname{dom} g$     $\forall_{x \in \operatorname{dom} f}\, f(x) = g(x)$

• Worked Example 1: Comparing Trigonometric Forms:

  • Let the candidate functions be:     $f(x) = \sin(x)$     $g(x) = \sqrt{1 - \cos^2(x)}$

  • Domain Analysis:

  • The domain of the sine function is unrestricted:       $\operatorname{dom} f = \mathbb{R}$

  • For $g(x)$, the radicand must be non-negative:       $1 - \cos^2(x) \ge 0$

  • Utilizing the Pythagorean identity $\sin^2(x) + \cos^2(x) = 1$, this simplifies to:       $\sin^2(x) \ge 0$

  • Because a real square is always non-negative, the condition holds for all real numbers:       $\operatorname{dom} g = \mathbb{R}$

  • The domain condition $\operatorname{dom} f = \operatorname{dom} g$ is satisfied.

  • Formula Comparison:

  • Simplifying $g(x)$ using the absolute value identity $\sqrt{u^2} = |u|$:       $g(x) = \sqrt{\sin^2(x)} = |\sin(x)|$

  • Since $|\sin(x)| \ne \sin(x)$ for any $x$ where $\sin(x) < 0$ (e.g., at $x = \frac{3\pi}{2}$, $f(x) = -1$ whereas $g(x) = 1$):       $f(x) \ne g(x)$

• Conclusion: The functions are not equal ($f \ne g$).

• Worked Example 2: Comparing Rational Expressions:

  • Let the candidate functions be:     $f(x) = 1 - x$     $g(x) = \frac{1 - x^2}{1 + x}$

  • Domain Analysis:

  • For $f(x)$, a standard polynomial:       $\operatorname{dom} f = \mathbb{R}$

  • For $g(x)$, division by zero is undefined, requiring:       $1 + x \ne 0 \implies x \ne -1$

• Thus, the domain of $g$ excludes $-1$:   $\operatorname{dom} g = \mathbb{R} \setminus \{-1\}$

• Because $\operatorname{dom} f \ne \operatorname{dom} g$, the functions cannot be equal.

• Formula Comparison:

  • Factoring the numerator for $x \in \operatorname{dom} g$:     $g(x) = \frac{(1 - x)(1 + x)}{1 + x} = 1 - x = f(x)$

  • Conclusion: Although the formulas agree wherever $g(x)$ is defined, $f \ne g$ due to domain mismatch.

Surjective (Onto) Functions

• Definition of Surjectivity:

  • A function $f : X \rightarrow Y$ is surjective (or onto $Y$) if and only if its image equals its codomain:     $\operatorname{im} f = Y$

  • Stated in predicate logic, every element of $Y$ is mapped to by at least one element of $X$:     $\forall_{y \in Y}\, \exists_{x \in X}\, f(x) = y$

• Worked Example 1: Surjectivity of a Quadratic Function:

  • Consider $f(x) = x^2$ where $X = \mathbb{R}$ and $Y = [0, \infty)$.

  • Test whether the condition holds:     $\forall_{y \in [0, \infty)}\, \exists_{x \in \mathbb{R}}\, x^2 = y$

  • Choose an arbitrary element $y_0 \in [0, \infty)$.

  • Solve the algebraic equation:     $x^2 = y_0 \implies x = \sqrt{y_0}$

  • Since $y_0 \ge 0$, $\sqrt{y_0}$ is always a well-defined real number ($\sqrt{y_0} \in \mathbb{R}$).

  • Substituting back confirms:     $f(\sqrt{y_0}) = (\sqrt{y_0})^2 = y_0$

  • Conclusion: The function $f$ is surjective onto $[0, \infty)$.

• Worked Example 2: Non-Surjectivity of an Exponential Function:

  • Consider $f(x) = 2^x$ where $X = \mathbb{R}$ and $Y = [0, \infty)$.

  • Test whether the condition holds:     $\forall_{y \in [0, \infty)}\, \exists_{x \in \mathbb{R}}\, 2^x = y$

  • Choose an arbitrary element $y_0 \in [0, \infty)$.

  • Attempt to solve for $x$:     $2^x = y_0 \implies 2^x = 2^{\log_2(y_0)} \implies x = \log_2(y_0)$

  • The expression $\log_2(y_0)$ is defined only for strictly positive numbers ($y_0 > 0$).

  • For $y_0 = 0 \in Y$, the equation $2^x = 0$ possesses no real solution because $2^x > 0$ for all $x \in \mathbb{R}$.

  • Conclusion: The function $f$ is not surjective onto $[0, \infty)$.

Periodic Functions

• Definition of Periodicity:

  • A function $f : \operatorname{dom} f \rightarrow \mathbb{R}$ is periodic of period $p$ (with real $p > 0$) if the domain is invariant under translations by $p$ and the functional value is preserved:     $\exists_{p > 0}\, \forall_{x \in \operatorname{dom} f}\, (x \pm p \in \operatorname{dom} f \land f(x + p) = f(x))$

• Multiples of Periods and Minimum Period:

  • If a function is periodic with period $p$, it is also periodic with period $m \cdot p$ for any positive integer $m \in \mathbb{N} \setminus \{0\}$.

  • If there exists a smallest strictly positive period $p$, it is designated as the minimum period (or fundamental period) of the function.

• Worked Examples of Periodicity:

  • Scaled Trigonometric Function:

  • For $f(x) = \cos(3x)$, verify the proposed minimum period $p = \frac{2\pi}{3}$:       $f(x + p) = \cos\left(3\left(x + \frac{2\pi}{3}\right)\right) = \cos(3x + 2\pi) = \cos(3x) = f(x)$

  • The minimum period is $p = \frac{2\pi}{3}$.

  • Discrete Alternating Sequence Function:

  • For $g(k) = (-1)^k$, where $k \in \mathbb{Z}$, verify the period $p = 2$:       $g(k + 2) = (-1)^{k+2} = (-1)^k \cdot (-1)^2 = (-1)^k \cdot 1 = (-1)^k = g(k)$

  • The minimum period is $p = 2$.

  • Constant Function:

  • For $h(x) = a$, where $a \in \mathbb{R}$:       $\forall_{p \in \mathbb{R}}\, h(x + p) = a = h(x)$

  • A constant function is periodic for any positive real number $p$, but it has no minimum period since no smallest positive real number exists.

Even and Odd Functions and Functional Decomposition

• Symmetric Domain Requirement:

  • A domain $\operatorname{dom} f \subseteq \mathbb{R}$ is symmetric with respect to the origin if:     $x \in \operatorname{dom} f \implies -x \in \operatorname{dom} f$

• Even Functions:

  • A function $f : \operatorname{dom} f \rightarrow \mathbb{R}$ is even if its domain is symmetric with respect to the origin and:     $\forall_{x \in \operatorname{dom} f}\, f(-x) = f(x)$

  • Geometric Property: The graph of an even function is symmetric with respect to the vertical $y$-axis.

• Worked Example of an Even Function:

  • Consider the function:     $g(x) = \frac{\sin(x)}{x^3}$

  • The domain is $\operatorname{dom} g = \mathbb{R} \setminus \{0\}$, which is symmetric with respect to the origin ($x \ne 0 \implies -x \ne 0$).

  • Evaluating at $-x$:     $g(-x) = \frac{\sin(-x)}{(-x)^3} = \frac{-\sin(x)}{-x^3} = \frac{\sin(x)}{x^3} = g(x)$

  • Thus, $g(x)$ is an even function.

• Odd Functions:

  • A function $f : \operatorname{dom} f \rightarrow \mathbb{R}$ is odd if its domain is symmetric with respect to the origin and:     $\forall_{x \in \operatorname{dom} f}\, f(-x) = -f(x)$

  • Geometric Property: The graph of an odd function is symmetric with respect to the origin $(0, 0)$.

• Worked Example of an Odd Function:

  • Consider the function:     $p(x) = x|x|$

  • The domain is $\operatorname{dom} p = \mathbb{R}$, which is symmetric with respect to the origin.

  • Evaluating at $-x$:     $p(-x) = (-x)|-x| = -x|x| = -p(x)$

  • Thus, $p(x)$ is an odd function.

• Decomposition Theorem into Even and Odd Components:

  • Every function $f : \operatorname{dom} f \rightarrow \mathbb{R}$ defined on a domain symmetric with respect to the origin can be uniquely decomposed into the sum of an even function $g_1(x)$ and an odd function $g_2(x)$:     $f(x) = \frac{1}{2}(f(x) + f(-x)) + \frac{1}{2}(f(x) - f(-x))$

  • Verification of the Even Component:

  • Define $g_1(x) = \frac{1}{2}(f(x) + f(-x))$.

  • Evaluating at $-x$:       $g_1(-x) = \frac{1}{2}(f(-x) + f(-(-x))) = \frac{1}{2}(f(-x) + f(x)) = g_1(x)$

  • Hence, $g_1(x)$ is even.

  • Verification of the Odd Component:

  • Define $g_2(x) = \frac{1}{2}(f(x) - f(-x))$.

  • Evaluating at $-x$:       $g_2(-x) = \frac{1}{2}(f(-x) - f(-(-x))) = \frac{1}{2}(f(-x) - f(x)) = -\frac{1}{2}(f(x) - f(-x)) = -g_2(x)$

  • Hence, $g_2(x)$ is odd.

Monotonicity: Increasing and Decreasing Functions

• Increasing Functions:

  • A function $f$ is increasing on a set $I \subset \operatorname{dom} f$ if for all $x_1, x_2 \in I$:     $x_1 < x_2 \implies f(x_1) \le f(x_2)$

  • A function $f$ is strictly increasing on $I$ if for all $x_1, x_2 \in I$:     $x_1 < x_2 \implies f(x_1) < f(x_2)$

• Decreasing Functions:

  • A function $f$ is decreasing on a set $I \subset \operatorname{dom} f$ if for all $x_1, x_2 \in I$:     $x_1 < x_2 \implies f(x_1) \ge f(x_2)$

  • A function $f$ is strictly decreasing on $I$ if for all $x_1, x_2 \in I$:     $x_1 < x_2 \implies f(x_1) > f(x_2)$

• Monotonicity in Logarithmic and Exponential Inequalities:

  • Logarithmic Inequalities:

  • If the base satisfies $a > 1$, the logarithm is strictly increasing:       $\log_a(b) < \log_a(c) \iff b < c$

  • If the base satisfies $a \in (0, 1)$, the logarithm is strictly decreasing:       $\log_a(b) < \log_a(c) \iff b > c$

  • Exponential Inequalities:

  • If the base satisfies $a > 1$, the exponential is strictly increasing:       $a^b < a^c \iff b < c$

  • If the base satisfies $a \in (0, 1)$, the exponential is strictly decreasing:       $a^b < a^c \iff b > c$

• Worked Example: Proof of an Increasing Rational Function:

  • Prove that $g(x) = \frac{1}{x^4 + 1}$ is strictly increasing on the interval $I = (-\infty, 0]$.

  • Select arbitrary elements $x_1, x_2 \in (-\infty, 0]$ such that $x_1 < x_2$.

  • Evaluate the target inequality:     $\frac{1}{x_1^4 + 1} < \frac{1}{x_2^4 + 1}$

  • Multiplying across by the positive denominators $(x_1^4 + 1)(x_2^4 + 1) > 0$:     $x_2^4 + 1 < x_1^4 + 1$     $x_2^4 - x_1^4 < 0$

  • Applying the difference of squares identity repeatedly:     $(x_2^2 - x_1^2)(x_2^2 + x_1^2) < 0$     $(x_2 - x_1)(x_2 + x_1)(x_2^2 + x_1^2) < 0$

  • Evaluate the sign of each factor:

  • $(x_2 - x_1) > 0$ because $x_1 < x_2$.

  • $(x_2 + x_1) < 0$ because $x_1 < x_2 \le 0$, so at least one number is strictly negative and neither is positive.

  • $(x_2^2 + x_1^2) > 0$ because $x_1 < x_2 \le 0$ implies $x_1 \ne 0$, ensuring a strictly positive sum of squares.

  • The sign of the product is:     $(+) \cdot (-) \cdot (+) = - < 0$

  • Since the inequality holds, $g(x)$ is strictly increasing on $(-\infty, 0]$.

• Worked Example: Proof of a Decreasing Quadratic Function:

  • Prove that $g(x) = x^2 - 2x$ is strictly decreasing on the interval $I = (-\infty, 1)$.

  • Select arbitrary elements $x_1, x_2 \in (-\infty, 1)$ such that $x_1 < x_2$.

  • Evaluate the target inequality:     $g(x_1) > g(x_2)$     $x_1^2 - 2x_1 > x_2^2 - 2x_2$

  • Rearrange all terms to one side:     $(x_1^2 - x_2^2) - 2(x_1 - x_2) > 0$

  • Factor the difference of squares:     $(x_1 - x_2)(x_1 + x_2) - 2(x_1 - x_2) > 0$     $(x_1 - x_2)(x_1 + x_2 - 2) > 0$

  • Evaluate the sign of each factor:

  • $(x_1 - x_2) < 0$ because $x_1 < x_2$.

  • Since $x_1 < 1$ and $x_2 < 1$, their sum satisfies $x_1 + x_2 < 2$, meaning $(x_1 + x_2 - 2) < 0$.

  • The sign of the product is:     $(-) \cdot (-) = + > 0$

  • Since the inequality holds, $g(x)$ is strictly decreasing on $(-\infty, 1)$.

Composition of Functions

• Formal Definition of Function Composition:

  • Let $X, Y, Z, W$ be sets with $Y \subset Z$.

  • Suppose $f : X \rightarrow Y$ and $g : Z \rightarrow W$.

  • The composite function $h = g \circ f$ (read "$g$ composed with $f$") is the function from $X$ to $W$ defined by:     $(g \circ f)(x) = g(f(x))$

• Domain of a Composite Function:

  • For $x$ to belong to the domain of $g \circ f$, two conditions must be met:

  1. $x$ must belong to the domain of $f$ (so $f(x)$ is defined).

  2. The resulting value $f(x)$ must belong to the domain of $g$ (so $g(f(x))$ is defined).

  • Stated formally:     $x \in \operatorname{dom}(g \circ f) \iff x \in \operatorname{dom} f \land f(x) \in \operatorname{dom} g$

• Worked Example: Pairwise Compositions of Rational Functions:

  • Given the functions:     $f(x) = \frac{x}{1 + x^2}$     $g(x) = \frac{1}{x}$

  • **Composition $(f \circ f)(x)$     $(f \circ f)(x) = f(f(x)) = f\left(\frac{x}{1 + x^2}\right)$     $(f \circ f)(x) = \frac{\frac{x}{1 + x^2}}{1 + \left(\frac{x}{1 + x^2}\right)^2}$     $(f \circ f)(x) = \frac{\frac{x}{1 + x^2}}{1 + \frac{x^2}{(1 + x^2)^2}} = \frac{\frac{x}{1 + x^2}}{\frac{(1 + x^2)^2 + x^2}{(1 + x^2)^2}}$     $(f \circ f)(x) = \frac{x}{1 + x^2} \cdot \frac{(1 + x^2)^2}{1 + 2x^2 + x^4 + x^2} = \frac{x(1 + x^2)}{1 + 3x^2 + x^4}$

  • **Composition $(f \circ g)(x)$     $(f \circ g)(x) = f(g(x)) = f\left(\frac{1}{x}\right)$     $(f \circ g)(x) = \frac{\frac{1}{x}}{1 + \left(\frac{1}{x}\right)^2} = \frac{\frac{1}{x}}{1 + \frac{1}{x^2}} = \frac{\frac{1}{x}}{\frac{x^2 + 1}{x^2}}$     $(f \circ g)(x) = \frac{1}{x} \cdot \frac{x^2}{1 + x^2} = \frac{x}{1 + x^2}$

  • **Composition $(g \circ f)(x)$     $(g \circ f)(x) = g(f(x)) = g\left(\frac{x}{1 + x^2}\right)$     $(g \circ f)(x) = \frac{1}{\frac{x}{1 + x^2}} = \frac{1 + x^2}{x}$

  • **Composition $(g \circ g)(x)$     $(g \circ g)(x) = g(g(x)) = g\left(\frac{1}{x}\right)$     $(g \circ g)(x) = \frac{1}{\frac{1}{x}} = x$

One-to-One (Injective) Functions

• Definition of Injectivity:

  • A function $f$ is one-to-one (injective) if distinct elements in the domain always map to distinct values in the range:     $\forall_{x_1, x_2 \in \operatorname{dom} f}\, [(x_1 \ne x_2) \implies (f(x_1) \ne f(x_2))]$

  • By contrapositive, if two images are equal, the original domain elements must be identical:     $\forall_{x_1, x_2 \in \operatorname{dom} f}\, [(f(x_1) = f(x_2)) \implies (x_1 = x_2)]$

• The Horizontal Line Test:

  • A function is one-to-one if and only if every horizontal line in the Cartesian plane intersects the graph of the function at most once.

  • If a horizontal line intersects the graph at two or more points, those points share the same $y$-value for distinct $x$-values, violating injectivity.

• Connection to Strict Monotonicity:

  • If a function $f$ is strictly monotone (strictly increasing or strictly decreasing) on its domain, then $f$ is guaranteed to be one-to-one.

• Worked Example: Proof of Injectivity by Contradiction:

  • Prove that $g(x) = \frac{1}{x^2}$ is one-to-one on the interval $I = (-\infty, 0)$.

  • Assumption: Suppose there exist two elements $x_1, x_2 \in (-\infty, 0)$ such that $x_1 \ne x_2$, yet $g(x_1) = g(x_2)$.

• Setting the equations equal:   $\frac{1}{x_1^2} = \frac{1}{x_2^2} \implies x_1^2 = x_2^2$

• Rearranging and factoring:   $x_1^2 - x_2^2 = 0 \implies (x_1 - x_2)(x_1 + x_2) = 0$

• This product equals zero if and only if:   $x_1 = x_2 \quad \text{or} \quad x_1 = -x_2$

• Contradiction Analysis:

  • The case $x_1 = x_2$ directly contradicts the starting hypothesis that $x_1 \ne x_2$.

  • The case $x_1 = -x_2$ requires $x_1$ and $x_2$ to have opposite signs. However, both $x_1, x_2 \in (-\infty, 0)$, meaning both must be strictly negative, which is impossible if $x_1 = -x_2$.

• Conclusion: Both branches produce a contradiction. Hence, $g(x)$ is one-to-one on $(-\infty, 0)$.

Inverse Functions

• Definition and Existence of Inverse Functions:

  • If a function $f : X \rightarrow Y$ is one-to-one, each element $y \in \operatorname{im} f$ corresponds to a unique element $x \in \operatorname{dom} f$ such that $f(x) = y$.

  • This assignment defines the inverse function, denoted $f^{-1}$, from $\operatorname{im} f$ to $\operatorname{dom} f$:     $f^{-1}(y) = x \iff y = f(x)$

• Domain and Range Relations for Inverses:

  • The domain of the inverse is the image of the original function:     $\operatorname{dom}(f^{-1}) = \operatorname{im} f$

  • The image of the inverse is the domain of the original function:     $\operatorname{im}(f^{-1}) = \operatorname{dom} f$

• Composition Inversion Identity:

  • Applying a function and its inverse consecutively yields the identity mapping on their respective sets:     $(f^{-1} \circ f)(x) = x \quad \text{for all } x \in \operatorname{dom} f$     $(f \circ f^{-1})(y) = y \quad \text{for all } y \in \operatorname{im} f$

• Geometric Characterization of the Inverse:

  • In Cartesian coordinates, the relationship $y = f(x) \iff x = f^{-1}(y)$ signifies that the graph of $f^{-1}$ is obtained by reflecting the graph of $f$ across the main diagonal line:     $y = x$

• Step-by-Step Derivation of an Inverse Function:

  • Consider the quadratic function restricted to a domain ensuring injectivity:     $f(x) = x^2 - 2x, \quad x \in [1, \infty)$

  • Step 1: Set up the equation $y = f(x)$ and solve for $x$:     $y = x^2 - 2x \implies x^2 - 2x - y = 0$

  • Step 2: Apply the quadratic formula:

  • The discriminant with respect to $x$ is:       $\Delta = (-2)^2 - 4(1)(-y) = 4 + 4y = 4(1 + y)$

  • Solving for $x$:       $x = \frac{-(-2) \pm \sqrt{4(1 + y)}}{2(1)} = \frac{2 \pm 2\sqrt{1 + y}}{2} = 1 \pm \sqrt{1 + y}$

  • Step 3: Discard the extraneous root based on the domain $x \ge 1$:

  • For $x_1 = 1 - \sqrt{1 + y}$:       $x_1 \le 1 \implies x_1 \notin [1, \infty) \quad (\text{for all } y > -1)$

  • For $x_2 = 1 + \sqrt{1 + y}$:     $x_2 \ge 1 \implies x_2 \in [1, \infty) \quad (\text{for all } y \ge -1)$

  • Therefore, select the root:     $x = 1 + \sqrt{1 + y}$

• Step 4: Standardize variable notation ($x \leftrightarrow y$):

  • Express the inverse function with argument $x$:     $f^{-1}(x) = 1 + \sqrt{1 + x}$

  • The natural domain of the inverse corresponds to the range of $f$ on $[1, \infty)$:     $\operatorname{dom}(f^{-1}) = [-1, \infty)$