Expressing Functions Algebraically and Solving Cubic Equations for y page 11.5

Algebraic Expressibility of Functions

• Definition of a Function: A binary relation between two sets that associates each element of the first set (domain) to exactly one element of the second set (codomain). In two-variable equations involving $x$ and $y$, expressing $y$ as a function of $x$ ($y = f(x)$) requires isolating $y$ such that every valid input $x$ produces a single, unique output $y$.

• Odd vs. Even Powers in Implicit Equations:

  • Even Powers (e.g., $y^2$, $y^4$): Taking an even root (such as a square root) yields two solutions ($!\pm$), representing two distinct branches. Unless a principal branch is specified or the domain is restricted, an equation where $y$ is raised to an even power generally does not yield a single-valued function $y = f(x)$.

  • Odd Powers (e.g., $y^3$, $y^5$): Every real number has a unique real odd root (cube root, fifth root, etc.). Consequently, isolating $y$ from an odd-power term like $y^3$ yields a single, well-defined real-valued output for every real input, successfully defining a function $y = f(x)$.

Worked Example 1: Solving $x - 8y^3 = 0$ for $y$

• Problem Statement: Given the equation $x - 8y^3 = 0$, express $y$ as a function of $x$.

• Step-by-Step Algebraic Solution:

  • Step 1 (Isolate the cubic term): Add $8y^3$ to both sides of the equation to separate the variable terms:     $x = 8y^3$     or equivalently:     $8y^3 = x$

  • Step 2 (Isolate $y^3$): Divide both sides of the equation by the coefficient $8$:     $y^3 = \frac{x}{8}$

  • Step 3 (Apply the principal cube root): Take the real cube root of both sides to solve for $y$:     $y = \sqrt[3]{\frac{x}{8}}$

  • Step 4 (Simplify radicals): Use the quotient property of radicals $\sqrt[n]{\frac{a}{b}} = \frac{\sqrt[n]{a}}{\sqrt[n]{b}}$ to simplify the denominator:     $y = \frac{\sqrt[3]{x}}{\sqrt[3]{8}}$     Since $\sqrt[3]{8} = 2$, the simplified form is:     $y = \frac{\sqrt[3]{x}}{2}$

• Function Representation: Expressed in standard function notation $y = f(x)$:   $f(x) = \frac{\sqrt[3]{x}}{2}$   or equivalently:   $f(x) = \sqrt[3]{\frac{x}{8}}$

Worked Example 2: Solving $8x + 9y^3 = 3$ for $y$

• Problem Statement: Choose the function that expresses the relationship $8x + 9y^3 = 3$ as a function, $y = f(x)$, if possible.

• Option Analysis:

  • Option 1: "This equation cannot be expressed as a function" — Incorrect, because $y$ is raised to an odd power ($3$), ensuring a unique real cube root exists for every value of $x$.

  • Option 2: $f(x) = \frac{3 - 8x}{9}$ — Incorrect, this represents $y^3$ prior to taking the cube root of both sides.

  • Option 3: $f(x) = \sqrt{\frac{3 - 8x}{9}}, -\sqrt{\frac{3 - 8x}{9}}$ — Incorrect, this uses square roots and represents a two-valued relation rather than a single cube root function.

  • Option 4: $f(x) = \sqrt[3]{\frac{3 - 8x}{9}}$ — Correct solution.

• Step-by-Step Algebraic Solution:

  • Step 1 (Isolate the term containing $y$): Subtract $8x$ from both sides of the equation $8x + 9y^3 = 3$:     $9y^3 = 3 - 8x$

  • Step 2 (Isolate $y^3$): Divide both sides by $9$:     $y^3 = \frac{3 - 8x}{9}$

  • Step 3 (Extract the cube root): Take the cube root of both sides to isolate $y$:     $y = \sqrt[3]{\frac{3 - 8x}{9}}$

  • Step 4 (Write in function notation): Substitute $y = f(x)$:     $f(x) = \sqrt[3]{\frac{3 - 8x}{9}}$