Exhaustive Study Notes on Domain, Function Combinations, Compositions, and Piecewise Functions

Domain Restrictions and Interval Notation

• The domain of a function represents all valid input values ($x$) that can be plugged into the function without causing undefined operations.

• Two primary conditions impose restrictions on the domain of a function:

  • Fractions: Division by zero is undefined in mathematics. Any value of $x$ that causes a denominator to equal zero must be excluded from the domain.
  • Even Roots: Taking an even root (such as a square root) of a negative number yields imaginary numbers, which are excluded when working with real-valued functions. The expression inside an even root (the radicand) must always be greater than or equal to zero.

• Example 1: Rational Function

  • Function: $f(x) = \frac{5x}{x^2 - 4}$
  • The numerator ($5x$) does not affect the domain.
  • To find excluded values, set the denominator equal to zero:     $x^2 - 4 = 0$$x^2 = 4$$x = \pm 2$
  • Excluded values: $x = -2$ and $x = 2$.
  • Interval notation representation: $(-\infty, -2) \cup (-2, 2) \cup (2, \infty)$.
  • Plain English description: All real numbers except $x = -2$ and $x = 2$.

• Example 2: Polynomial Function

  • Function with no fractions and no even roots.
  • Since there are no division operations or radical expressions, no real values cause restrictions.
  • Domain: $(-\infty, \infty)$.

• Example 3: Radical Function

  • Function: $f(x) = \sqrt{3x + 12}$
  • The radicand must be non-negative:     $3x + 12 \ge 0$$3x \ge -12$$x \ge -4$
  • Domain in interval notation: $[-4, \infty)$.
  • Inequality Rule: When dividing or multiplying an inequality by a negative number, the inequality sign must be flipped (e.g., $-2x > 4 \implies x < -2$).

• Example 4: Radical Function in a Denominator

  • Function: $f(x) = \frac{A}{\sqrt{x - 4}}$
  • A square root normally requires the interior expression to be $\ge 0$. However, because it is located in the denominator, it cannot equal zero.
  • Combined restriction:     $x - 4 > 0$$x > 4$
  • The value $x = 4$ must be excluded with a parenthesis because plugging in $4$ produces $\sqrt{0} = 0$ in the denominator.
  • Domain in interval notation: $(4, \infty)$.

Algebra of Functions

• Functions can be combined using basic arithmetic operations:

  • Sum: $(f + g)(x) = f(x) + g(x)$
  • Difference: $(f - g)(x) = f(x) - g(x)$
  • Product: $(f \cdot g)(x) = f(x) \cdot g(x)$
  • Quotient: $\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)}$, provided $g(x) \neq 0$

• Polynomial Combination Example

  • Given $f(x) = x - 5$ and $g(x) = x^2 - 1$:
  • Sum:     $(f + g)(x) = (x - 5) + (x^2 - 1) = x^2 + x - 6$
  • Domain: $(-\infty, \infty)$
  • Difference:     $(f - g)(x) = (x - 5) - (x^2 - 1) = x - 5 - x^2 + 1 = -x^2 + x - 4$
  • Note: Enclose the subtracted function in parentheses to ensure the negative sign is distributed to every term.
  • Domain: $(-\infty, \infty)$
  • Product:     $(f \cdot g)(x) = (x - 5)(x^2 - 1) = x^3 - x - 5x^2 + 5 = x^3 - 5x^2 - x + 5$
  • Domain: $(-\infty, \infty)$
  • Quotient:     $\left(\frac{f}{g}\right)(x) = \frac{x - 5}{x^2 - 1} = \frac{x - 5}{(x - 1)(x + 1)}$
  • Setting the denominator to zero: $x^2 - 1 = 0 \implies x = 1, -1$
  • Excluded values: $x = -1$ and $x = 1$
  • Domain: $(-\infty, -1) \cup (-1, 1) \cup (1, \infty)$

• Radical Combination Example

  • Given $f(x) = \sqrt{x + 3}$ and $g(x) = \sqrt{x - 2}$:
  • Sum:     $(f + g)(x) = \sqrt{x + 3} + \sqrt{x - 2}$
  • Restrictions: $x + 3 \ge 0 \implies x \ge -3$ AND $x - 2 \ge 0 \implies x \ge 2$
  • Overlapping domain for both expressions to be defined: $[2, \infty)$
  • Difference:     $(f - g)(x) = \sqrt{x + 3} - \sqrt{x - 2}$
  • Domain remains restricted by both individual radicals: $[2, \infty)$
  • Product:     $(f \cdot g)(x) = \sqrt{x + 3} \cdot \sqrt{x - 2} = \sqrt{(x + 3)(x - 2)}$
  • Rule for multiplying radicals: $\sqrt{a} \cdot \sqrt{b} = \sqrt{a \cdot b}$
  • Testing intervals for $(x + 3)(x - 2) \ge 0$:
    • Critical numbers: $x = -3$ and $x = 2$
    • Interval $(-\infty, -3]$: Test $x = -98 \implies (-)(-)=+$ (Valid)
    • Interval $[-3, 2]$: Test $x = 0 \implies (+)(-)=-$ (Invalid)
    • Interval $[2, \infty)$: Test $x = 98 \implies (+)(+)=+$ (Valid)
  • Domain: $(-\infty, -3] \cup [2, \infty)$
  • Quotient:     $\left(\frac{f}{g}\right)(x) = \frac{\sqrt{x + 3}}{\sqrt{x - 2}}$
  • Numerator restriction: $x \ge -3$
  • Denominator restriction: $x - 2 > 0 \implies x > 2$ (cannot equal zero)
  • Combined Domain: $(2, \infty)$

Composite Functions

• A composite function combines two functions by evaluating one function inside another: $(f \circ g)(x) = f(g(x))$.

• Step-by-Step Method for Determining Composite Domain:

  1. Find the domain of the inner function $g(x)$.
  2. Find the domain of the simplified composite function $f(g(x))$.
  3. Find the intersection (overlap) of both domains.

• Polynomial Composite Example

  • Given $f(x) = 5x - 3$ and $g(x) = 3x^2 - 2$:
  • Finding $(f \circ g)(x)$:     $(f \circ g)(x) = f(g(x)) = 5(3x^2 - 2) - 3 = 15x^2 - 10 - 3 = 15x^2 - 13$
  • Domain of inner function $g(x)$: $(-\infty, \infty)$
  • Domain of composite $15x^2 - 13$: $(-\infty, \infty)$
  • Intersection Domain: $(-\infty, \infty)$
  • Evaluation at $x = 1$: $(f \circ g)(1) = 15(1)^2 - 13 = 2$
  • Finding $(g \circ f)(x)$:     $(g \circ f)(x) = g(f(x)) = 3(5x - 3)^2 - 2$$= 3(25x^2 - 30x + 9) - 2 = 75x^2 - 90x + 27 - 2 = 75x^2 - 90x + 25$
  • Domain of inner function $f(x)$: $(-\infty, \infty)$
  • Domain of composite $75x^2 - 90x + 25$: $(-\infty, \infty)$
  • Intersection Domain: $(-\infty, \infty)$
  • Evaluation at $x = 1$: $(g \circ f)(1) = 75(1)^2 - 90(1) + 25 = 10$

• Rational Composite Example

  • Given $f(x) = \frac{4}{x + 2}$ and $g(x) = \frac{1}{x}$:
  • Finding $(f \circ g)(x)$:     $(f \circ g)(x) = f\left(\frac{1}{x}\right) = \frac{4}{\frac{1}{x} + 2}$
  • Simplify complex fraction by multiplying numerator and denominator by $x$:       $\frac{4 \cdot x}{\left(\frac{1}{x} + 2\right) \cdot x} = \frac{4x}{1 + 2x}$
  • Domain Determination:
  1. Inner function $g(x) = \frac{1}{x}$ requires $x \neq 0$. Domain: $(-\infty, 0) \cup (0, \infty)$.
  2. Composite function $\frac{4x}{1 + 2x}$ requires $1 + 2x \neq 0 \implies 2x \neq -1 \implies x \neq -\frac{1}{2}$.
  3. Intersection of restrictions ($x \neq 0$ and $x \neq -\frac{1}{2}$):
     • Domain: $(-\infty, -\frac{1}{2}) \cup (-\frac{1}{2}, 0) \cup (0, \infty)$

Linear Equations and Piecewise Functions

• Linear Equations Review

  • Given points: $(-2, 4)$ and $(1, -1)$.
  • Slope ($m$):     $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-1 - 4}{1 - (-2)} = \frac{-5}{3} = -\frac{5}{3}$
  • Point-Slope Form ($y - y_1 = m(x - x_1)$):     $y - 4 = -\frac{5}{3}(x - (-2)) \implies y - 4 = -\frac{5}{3}(x + 2)$
  • Slope-Intercept Form ($y = mx + b$):     $y - 4 = -\frac{5}{3}x - \frac{10}{3}$$y = -\frac{5}{3}x - \frac{10}{3} + \frac{12}{3}$$y = -\frac{5}{3}x + \frac{2}{3}$

• Piecewise Functions

  • Function defined in rules by domain intervals:     $f(x) = \begin{cases} -3 & \text{if } x < 0 \\ x & \text{if } x \ge 0 \end{cases}$
  • Evaluations:
  • $f(-100)$: Since $-100 < 0$, use top rule $f(x) = -3 \implies f(-100) = -3$
  • $f(0)$: Since $0 \ge 0$, use bottom rule $f(x) = x \implies f(0) = 0$
  • $f(125)$: Since $125 \ge 0$, use bottom rule $f(x) = x \implies f(125) = 125$
  • Graphing the Piecewise Function:
  • For $x \ge 0$: Graph line $y = x$ starting at $(0, 0)$ with a solid closed circle, extending right through $(1, 1)$, $(2, 2)$, etc.
  • For $x < 0$: Graph horizontal line $y = -3$ extending left to $\infty$. Place an open circle at $(0, -3)$ to indicate $x = 0$ is excluded from this branch.