Radical and rational functions and equations Knowledge-Based Study

Study notes

1. Understanding Rational Functions

   A rational function is written in the form $f(x) = \frac{P(x)}{Q(x)}$, where $P(x)$ and $Q(x)$ are polynomials and $Q(x) \neq 0$.

   Because division by zero is undefined, the natural domain of a rational function excludes every real number that makes the denominator equal to zero.

   Key Features

   • Vertical Asymptotes: Occur at values $x = c$ where $Q(c) = 0$ and $P(c) \neq 0$. The graph approaches $\infty$ or $-\infty$ near these lines.
   • Removable Discontinuities (Holes): Occur at values $x = c$ where both $P(c) = 0$ and $Q(c) = 0$, indicating a shared factor that can be cancelled.
   • Horizontal Asymptotes: Determined by comparing the degrees of $P(x)$ and $Q(x)$. If $\deg(P) < \deg(Q)$, the asymptote is $y = 0$; if $\deg(P) = \deg(Q)$, the asymptote is the ratio of their leading coefficients.

   Techniques for Solving Rational Equations

   To solve a rational equation, eliminate the fractions by multiplying every term by the least common denominator (LCD) of all rational expressions involved.

   Systematic Solution Process

   1. Factor all denominators completely to identify the LCD and note domain restrictions.
   2. Multiply both sides of the equation by the LCD to obtain a polynomial equation.
   3. Solve the resulting linear or polynomial equation.
   4. Check for extraneous solutions: Any candidate solution that causes a denominator in the original equation to equal zero must be rejected.

   Example

   Consider the equation:

   $$\frac{2x}{x - 3} + 1 = \frac{6}{x - 3}$$

   The restriction is $x \neq 3$. Multiplying the entire equation by $(x - 3)$ gives:

   $$2x + (x - 3) = 6$$$$3x - 3 = 6 \implies 3x = 9 \implies x = 3$$

   Because $x = 3$ violates the domain restriction, it is an extraneous solution. Thus, the equation has no real solution.

   Defining Radical Functions

   A radical function contains a variable under a radical sign, typically of the form $f(x) = \sqrt[n]{g(x)}$, where $n$ is a positive integer called the index.

   Domain Behavior

   • Even Index ($n = 2, 4, 6, \dots$): Even roots of negative numbers are not real numbers. Therefore, the radicand must be non-negative: $$g(x) \ge 0$$
   • Odd Index ($n = 3, 5, 7, \dots$): Odd roots are defined for all real numbers. Thus, the domain of $f(x) = \sqrt[3]{g(x)}$ matches the domain of $g(x)$.

   For example, the domain of $f(x) = \sqrt{2x - 8}$ requires $2x - 8 \ge 0$, which simplifies to $x \ge 4$.

   Algebraic Steps for Radical Equations

   Radical equations are solved by isolating the radical term and raising both sides to a power equal to the index of the radical.

   Step-by-Step Method

   1. Isolate one radical on one side of the equation.
   2. Raise both sides of the equation to the power equal to the index (e.g., square both sides for a square root).
   3. If another radical remains, repeat steps 1 and 2.
   4. Solve the resulting polynomial equation.
   5. Verify every solution: Raising both sides to an even power can introduce extraneous roots. You must substitute all potential solutions back into the original equation.

   Example

   Solve $\sqrt{x + 7} = x - 5$:

   1. Square both sides: $x + 7 = (x - 5)^2$
   2. Expand: $x + 7 = x^2 - 10x + 25$
   3. Rearrange into standard form: $x^2 - 11x + 18 = 0$
   4. Factor: $(x - 2)(x - 9) = 0$, yielding candidate values $x = 2$ and $x = 9$.

   Check:

   • For $x = 2$: $\sqrt{2 + 7} = \sqrt{9} = 3$, but $2 - 5 = -3$. Since $3 \neq -3$, $x = 2$ is extraneous.
   • For $x = 9$: $\sqrt{9 + 7} = \sqrt{16} = 4$, and $9 - 5 = 4$. This is true.

   The only valid solution is $x = 9$.