Functions, Equations, and Symmetry in College Algebra

Solving Radical and Absolute Value Equations

Solving Equations Involving Square Roots

Equations with square roots often require isolating the radical and then squaring both sides to eliminate the root. Always check for extraneous solutions by substituting back into the original equation.

• Key Point: To solve $\sqrt{2x+15} - 6 = x + 6$, isolate the radical and square both sides:

$\sqrt{2x+15} - 6 = x + 6$ $\sqrt{2x+15} = x + 12$ $2x + 15 = (x + 12)^2$ $2x + 15 = x^2 + 24x + 144$ $0 = x^2 + 22x + 129$ $0 = (x + 7)(x + 3)$ $x = -7,\ -3$

• Verification: Substitute each solution into the original equation to check validity. Only $x = -3$ is valid.

• Extraneous Solutions: Squaring both sides can introduce solutions that do not satisfy the original equation.

Solving Absolute Value Equations

Absolute value equations often have two cases, corresponding to the positive and negative values inside the absolute value.

• Key Point: To solve $4|1 - \frac{3}{x}| = 7 + 10$:

$4|1 - \frac{3}{x}| = 17$ $|1 - \frac{3}{x}| = \frac{17}{4}$ Two cases: $1 - \frac{3}{x} = \frac{17}{4}$ or $1 - \frac{3}{x} = -\frac{17}{4}$ Solve each for $x$:

• $1 - \frac{3}{x} = \frac{17}{4} \implies x = -\frac{12}{13}$

• $1 - \frac{3}{x} = -\frac{17}{4} \implies x = \frac{12}{21} = \frac{4}{7}$

Functions and Their Properties

Determining if an Equation Defines a Function

A relation is a function if every input (x-value) corresponds to exactly one output (y-value).

• Example 1: $y^3 = 27 - x$ Solve for $y$: $y = \sqrt[3]{27 - x}$ Conclusion: Yes, $y$ is a function of $x$.

• Example 2: $x - 16 = 4y^2$ $y^2 = \frac{x - 16}{4}$ $y = \pm \sqrt{\frac{x - 16}{4}}$ Conclusion: No, $y$ is not a function of $x$ because each $x$ (for $x \geq 16$) gives two $y$ values.

Difference Quotient

Definition and Calculation

The difference quotient is a formula used to compute the average rate of change of a function over an interval. It is foundational for calculus.

• Formula: $\frac{f(x+h) - f(x)}{h}$

• Example: For $f(x) = -9x + 4$:

$f(x+h) = -9(x+h) + 4 = -9x - 9h + 4$ $f(x+h) - f(x) = (-9x - 9h + 4) - (-9x + 4) = -9h$ Difference quotient: $\frac{-9h}{h} = -9$

• Interpretation: For linear functions, the difference quotient equals the slope.

Symmetry: Even and Odd Functions

Definitions

• Even Function: $f(-x) = f(x)$ for all $x$ in the domain. The graph is symmetric about the y-axis.

• Odd Function: $f(-x) = -f(x)$ for all $x$ in the domain. The graph is symmetric about the origin.

Examples and Verification

• Example 1: $f(x) = x^3 - 5$ $f(-x) = (-x)^3 - 5 = -x^3 - 5$ $-f(x) = -(x^3 - 5) = -x^3 + 5$ Since $f(-x) \neq f(x)$ and $f(-x) \neq -f(x)$, the function is neither even nor odd.

• Example 2: $g(x) = x^2 + 7x$ $g(-x) = (-x)^2 + 7(-x) = x^2 - 7x$ Since $g(-x) \neq g(x)$ and $g(-x) \neq -g(x)$, the function is neither even nor odd.

• Example 3: $h(x) = x^3$ $h(-x) = (-x)^3 = -x^3 = -h(x)$ The function is odd.

Tables and Graphs of Functions

Function Table Example

Tables are useful for evaluating functions at specific values and observing patterns.

x	f(x) = x^2
-3	9
-2	4
-1	1
0	0
1	1
2	4
3	9
4	16

Piecewise Functions and Graphs

Piecewise functions are defined by different expressions over different intervals of the domain. The graph may "switch" at certain points.

• Example: For $f(x) = x^2$ for $x \geq -2$, and another rule for $x < -2$, the function switches at $x = -2$.

• Graph: The graph shows a parabola for $x \geq -2$ and another segment for $x < -2$.

Summary Table: Even vs. Odd Functions

Property	Even Function	Odd Function
Algebraic Test	$f(-x) = f(x)$	$f(-x) = -f(x)$
Graph Symmetry	y-axis	Origin
Example	$x^2$	$x^3$

Additional info:

• Some context and explanations have been expanded for clarity and completeness.

• Piecewise function details and graph interpretations are inferred from the notes and standard College Algebra topics.