Mathematical Functions

FUNCTIONS - Definition: A function is a correspondence between two sets, the domain and the range, such that for each value in the domain, there corresponds exactly one value in the range. - Example representation: $f: x \rightarrow y$

Domain and Range of a Function - Domain: The set of inputs, e.g., $\text{inputs} = \text{\texttt{\texttt{\texttt{{\texttt{{p, q, r}}}}}}}$ - Range: The set of outputs, e.g., $\text{outputs} = \text{\texttt{\texttt{\texttt{{\texttt{{1, 2, 3}}}}}}}$

TYPES OF FUNCTIONS

One-to-One Functions - A function $f: x \rightarrow y$ is one-to-one (injective) if: - Each image $y \text{ in } Y$ has exactly one $x \text{ in } X$ assigned to it. - Mathematically: If $f(a) = f(b)$ then $a = b$.

Example

• Determine whether the following functions are 1:1: 1. Function: $f(x) = 5x + 2$ - Let $a, b \text{ in } X$, - Then: - $f(a) = 5a + 2$ - $f(b) = 5b + 2$ - If $f(a) = f(b)$, then $5a + 2 = 5b + 2$ - This leads to: - $5a = 5b$ - Thus: $a = b$ - Conclusion: $f(x) = 5x + 2$ is a 1:1 function.

1. Function: $f(x) = x^2 + 3$ - Let $a, b \text{ in } X$, - Then: - $f(a) = a^2 + 3$ - $f(b) = b^2 + 3$ - If $f(a) = f(b)$, then $(a^2 + 3) = (b^2 + 3)$ - This implies: - $a^2 = b^2$ - Therefore: $a = \text{±}b$ - Conclusion: Since $a = b$ or $a = -b$, - $f(x) = x^2 + 3$ is not a 1:1 function.

Onto Functions - Definition: A function $f: x \rightarrow y$ is onto (surjective) if every $y \text{ in } Y$ is the image of some $x \text{ in } X$. - Alternatively, $f: x \rightarrow y$ is onto iff the range of f is the entire co-domain of f.

Composite Functions - If $f$ and $g$ are functions, the composite function denoted by $f \circ g$ is given by: - $(f \circ g)(x) = f[g(x)]$ - Likewise: - $(g \circ f)(x) = g[f(x)]$

Example

• Let $f(x) = x^2 + 4$ and $g(x) = \sqrt{3x} + 1$, find: 1. (i) $(f \circ g)(x)$: - $= f[g(x)] = f[\sqrt{3x} + 1]$ - $= (\sqrt{3x} + 1)^2 + 4$ - $= 3x + 2\sqrt{3x} + 1 + 4$ - $= 3x + 2\sqrt{3x} + 5$ 1. (ii) $(g \circ f)(x)$: - $= g[f(x)] = g[x^2 + 4]$ - $= \sqrt{3(x^2 + 4)} + 1$ - $= \sqrt{3}x^2 + 4\sqrt{3} + 1$ 1. (iii) $(f \circ f)(x)$: - $= f[f(x)] = f[x^2 + 4]$ - $= (x^2 + 4)^2 + 4$ - $= x^4 + 8x^2 + 16 + 4$ - $= x^4 + 8x^2 + 20$

Inverse Functions - A function can be represented by a set of ordered pairs. For example: - $f(x) = x + 3$ from set $A = \text{\texttt{\texttt{{1, 2, 3, 4}}}}$ to set $B = \text{\texttt{\texttt{{4, 5, 6, 7}}}}$ can be written as: - f = \big{(}(1, 4), (2, 5), (3, 6), (4, 7)\big{)} - To form the inverse function $f^{-1}$, interchange the first and second coordinates of each ordered pair: - f^{-1} = \big{(}(4, 1), (5, 2), (6, 3), (7, 4)\big{)} - Properties: - The domain of $f$ equals the range of $f^{-1}$ and vice versa. - To find the inverse of a function, express it in terms of $f(x)$. - For instance: - $f(x) = 3x$ - then $x = \frac{f(x)}{3}$ - Replace $x$ by $f(x)$ and $f(x)$ by $x$.

• Steps to find the inverse of a function: 1. Let $y = f(x)$ 2. Solve for $x$ 3. Replace $y$ by $x$ 4. Replace $x$ by $f^{-1}(x)$

Definition: - A function $f: X \rightarrow Y$ has an inverse $f^{-1}: Y \rightarrow X$ if it is one-to-one.

Examples - Find the inverse for: 1. $f(x) = x + 2$: - $y = x + 2$ leads to: $x = y - 2$ - Therefore: $f^{-1}(x) = x - 2$ 2. $f(x) = 2x^3 + 1$: - $y = 2x^3 + 1$ leads to: $2x^3 = y - 1$ - Thus: $x^3 = \frac{y-1}{2}$ - This implies: $x = \sqrt[3]{\frac{y-1}{2}}$; - Therefore: $f^{-1}(x) = \sqrt[3]{\frac{x-1}{2}}$ 3. $f(x) = 2x^3$: - $y = 2x^3$ - Then: $x = \frac{y}{2}$ - Thus: $f^{-1}(x) = \sqrt[3]{\frac{x}{2}}$

Odd and Even Functions ### Definitions: - A function $f(x)$ is said to be even if: - $f(a) = f(-a)$ for every $a$; thus $f(-a) - f(a) = 0$. - Example: $f(x) = x^2$ is an even function. - A function is said to be odd if: - $f(-a) = -f(a)$ or $[f(-a) + f(a) = 0]$ - Example: $f(x) = x^3$ is an odd function.

Examples: - Assess if the following functions are odd or even: 1. Function: $f(x) = x^5$ - Since $f(a) = a^5$ and $f(-a) = (-a)^5 = -a^5$ - Therefore, $f(a) = -f(-a)$, thus, $f(x)$ is odd. 1. Function: $f(x) = x^2$ - $f(a) = a^2$ and $f(-a) = (-a)^2 = a^2$ - $f(-a) = f(a)$, thus, $f(x)$ is even.

Linear and Quadratic Functions ### Linear Inequalities: - An inequality states a relationship between two mathematical expressions, written as follows: - Less than: < - Less than or equal to: $\text{≤}$ Greater than: > - Greater than or equal to: $\text{≥}$

Example: - Solve: - 3x + 7 > x - 8 - or $5x - 7y \text{ ≤ } 4$ - etc.

Important Note: - Inequalities can be simplified similarly to equations, but remember that if multiplying both sides by -1, the inequality sign must be reversed.

Absolute Value - The absolute value of a number $X$, denoted by $|x|$, is defined as follows: - $|x| = \begin{cases} x, &amp; \text{if } x \geq 0 \ -x, &amp; \text{if } x &lt; 0 \end{cases}$

Example: - $|3| = 3, \text{ and } |-5| = 5$

Properties: 1. $|-a| = |a|$ 2. $|ab| = |a||b|$ and $|a + b| \leq |a| + |b|$ (Triangle Inequality)

Equations and Inequalities with Absolute Values: - If $D$ is positive: - $|x| = D \text{ iff } x = D \text{ or } x = -D$ - |x| < D \text{ iff } -D < x < D

Example Problem: - Solve: $|2x + 5| = 3$ - Split into two cases: 1. $2x + 5 = 3 \rightarrow 2x = -2 \rightarrow x = -1$ 2. $2x + 5 = -3 \rightarrow 2x = -8 \rightarrow x = -4$ The complete set of solutions is: $\text{Solution Set} = \text{\texttt{\texttt{\texttt{{-1, -4}}}}}$

Quadratic Equations - A quadratic equation is defined as any equation that can be expressed in the form: - $ax^2 + bx + c = 0$ - where $a, b, c$ are real numbers, and $a \neq 0$.

Methods of Solving Quadratics: 1. Factorization Method - via Zero Product Property: - Example: To solve $x^2 - x = 0$, factor to get: $x(x - 1) = 0$ - Solutions are: $x = 0 \text{ or } x = 1$

Example Problem: - Solve for $x$: $x^2 - x = 12$, 1. Rearranging gives: $x^2 - x - 12 = 0$ 2. Factorization gives: $(x - 4)(x + 3) = 0$ 3. Apply Zero Product Property to find: $x = 4 \text{ or } x = -3$

1. Completing Square Method: - Example: To solve $x^2 = 16$, 1. Rearranging gives: $x^2 - 16 = 0$ 2. Factoring leads to: $(x - 4)(x + 4) = 0$ 3. Find solutions: $x = 4 \text{ or } x = -4$

2. Quadratic Formula: - $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Example Problem Using Quadratic Formula: - Solve: $x^2 - 6x - 16 = 0$ - Identify: $a = 1, b = -6, c = -16$, - Substituting in formula: $x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-16)}}{2(1)}$ - Simplifying gives us: $x = 8 \text{ or } x = -2$

Discriminant Analysis: - Depending on the discriminant ($b^2 - 4ac$): 1. If > 0, two unequal real solutions. 2. If = 0, one unique solution. 3. If < 0, no real solutions.

Roots Outcomes: - Using roots $\alpha$ and $\beta$ from quadratic equation $ax^2 + bx + c = 0$: - Sum of Roots: $\alpha + \beta = -\frac{b}{a}$ - Product of Roots: $\alpha\beta = \frac{c}{a}$

Example - Roots Evaluation: - Given roots from $5x^2 - 2x + 7 = 0$, - Calculate sum and product using the formulas derived.