Absolute Value Functions, Piecewise Graphing, and Exponents

Absolute Value Functions and Inequalities

• Definition: Plugging any real number into an absolute value function yields its non-negative (positive) magnitude.

  • Example: Plugging in $-2$ yields positive $2$ ($|-2| = 2$).

• Geometric Definition: The absolute value of $x$, denoted $|x|$, represents the distance between $x$ and the origin ($0$) on the real number line.

• Graph of Parent Absolute Value Function $y = |x|$:

  • Graph shape: Forms a V-shape.
  • Positive domain branch ($x \ge 0$): Graphs as the line $y = x$.
  • Negative domain branch ($x < 0$): Graphs as the line $y = -x$.
  • Domain: $(-\infty, \infty)$ (all real numbers).
  • Range: $[0, \infty)$ (starts at $0$ with a square bracket because $0$ is included, extending to positive infinity with a parenthesis).

• Graphing Absolute Value Inequalities:

  • Example problem: Graph the inequality $|x| \le 8$.
  • Boundary evaluation: Begins at $-8$ and ends at $8$ because $|-8| = 8$ and $|8| = 8$.
  • Plotting: Closed dots are placed at $-8$ and $8$ because the inequality is non-strict ($\le$).
  • Region: Includes all real numbers lying strictly between $-8$ and $8$.
  • Domain for this bounded graph: $[-8, 8]$.

• Converting Intervals to Absolute Value Inequalities:

  • Example problem: Express the interval $(-5, 1)$ as an absolute value inequality.
  • Number Line Representation: Open circles are plotted at $-5$ and $1$ because parentheses indicate non-inclusive boundary points ($x \neq -5$ and $x \neq 1$).
  • Step 1 (Find Center Point $h$): Calculate the midpoint between $-5$ and $1$:     $h = \frac{-5 + 1}{2} = -2$
  • Step 2 (Find Distance $d$): Determine the distance from center $-2$ to either endpoint:     $d = 1 - (-2) = 3 \quad \text{or} \quad d = -2 - (-5) = 3$
  • Step 3 (Formulate Inequality): Apply the distance formula $|x - h| < d$:     $|x - (-2)| < 3 \implies |x + 2| < 3$
  • Testing Values in $|x + 2| < 3$:
  • Testing interior point $x = -4$: $|-4 + 2| = |-2| = 2 < 3$ (Valid solution).
  • Testing boundary point $x = 1$: $|1 + 2| = |3| = 3$ (Boundary condition, strictly less than $3$ is required).
  • Testing boundary point $x = -5$: $|-5 + 2| = |-3| = 3$ (Boundary condition, strictly less than $3$ is required).

Solving Absolute Value Equations and Graph Transformations

• Solving Absolute Value Equations:

  • Example problem: Solve $3|t + 9| = 24$
  • Step 1: Isolate the absolute value expression by dividing both sides by $3$:     $|t + 9| = 8$
  • Step 2: Set up two independent linear equations corresponding to the positive and negative piecewise branches:     $t + 9 = 8 \implies t = -1$$t + 9 = -8 \implies t = -17$
  • Final Solutions: $t = -1$ and $t = -17$. These outputs represent standalone scalar solutions, not an ordered pair or interval.
  • Advanced Application Context: Absolute value equations require splitting into positive and negative cases. This concept appears frequently in Calculus 1 and Calculus 2, where forgotten positive/negative splits cause widespread test errors.

• General Transformations Formula $y = a|x - h| + k$:

  • Parent function: $y = |x|$.
  • Horizontal Shifts ($h$):
  • Shifts inside the absolute value bars operate inversely to intuitive sign expectation.
  • $|x + 4| = |x - (-4)|$ translates the graph horizontally $4$ units to the left.
  • Vertical Reflections ($a < 0$):
  • Placing a negative sign outside the bars ($-|x|$) reflects the graph across the x-axis (flips it upside down).
  • Vertical Stretch/Compression ($a$) and Vertical Shift ($k$):
  • Example: $y = 2|x| - 5$
  • The vertex shifts down $5$ units to $(0, -5)$.
  • Multiplying by $2$ vertically stretches the graph, making it narrower/skinnier.

Graphing Piecewise Functions

• Piecewise Function Definition and Structure:   $f(x) = \begin{cases} 2x + 7 & \text{for } x < -2 \\ 5 & \text{for } -2 \le x \le 3 \\ 8 - x & \text{for } x > 3 \end{cases}$

• Boundary Construction:

  • Vertical boundary lines are established at $x = -2$ and $x = 3$

• Plotting Individual Branches:

  • Branch 1 ($2x + 7$ for $x < -2$):
  • Plug in boundary value: $2(-2) + 7 = 3 \implies$ Open dot at $(-2, 3)$.
  • Behavior: Increasing linear segment on interval $(-\infty, -2)$.
  • Branch 2 ($5$ for $-2 \le x \le 3$):
  • Behavior: Horizontal line $y = 5$
  • Endpoints: Closed dots at $(-2, 5)$ and $(3, 5)$.
  • Behavior: Constant on interval $[-2, 3]$.
  • Branch 3 ($8 - x$ for $x > 3$):
  • Plug in boundary value: $8 - 3 = 5 \implies$ Open dot at $(3, 5)$.
  • Note: The open dot at $(3, 5)$ connects directly into the existing closed dot at $(3, 5)$ from Branch 2.
  • Behavior: Decreasing linear line on interval $(3, \infty)$.

• Intervals of Behavior:

  • Increasing: $(-\infty, -2)$
  • Constant: $[-2, 3]$
  • Decreasing: $(3, \infty)$

• Intercept Analysis:

  • $y$--intercept: Evaluate $f(0)$. Since $x = 0$ lies within $-2 \le x \le 3$, $f(0) = 5$. The $y$-intercept is $(0, 5)$.
  • $x$--intercepts: Set individual function pieces equal to $0$:
  1. Top linear equation: $2x + 7 = 0 \implies 2x = -7 \implies x = -\frac{7}{2} \implies \left(-\frac{7}{2}, 0\right)$
  2. Bottom linear equation: $8 - x = 0 \implies x = 8 \implies (8, 0)$

Properties of Exponents and Common Algebraic Pitfalls

• Fundamental Rules of Exponents:

  • Product Rule: Same bases multiplied requiring exponent addition:     $x^a \cdot x^b = x^{a+b} \quad \text{Example: } x^2 \cdot x^3 = x^5$
  • Quotient Rule: Same bases divided requiring exponent subtraction:     $\frac{x^a}{x^b} = x^{a-b} \quad \text{Example: } \frac{x^5}{x^2} = x^3$
  • Power Rule: Outer exponent multiplying inner exponent:     $(x^a)^b = x^{a \cdot b} \quad \text{Example: } (x^3)^2 = x^6$
  • Power of a Product Rule: Exponent distributes to coefficient and variable:     $(ab)^n = a^n b^n \quad \text{Example: } (3x)^2 = 3^2 x^2 = 9x^2 \quad (\text{Incorrect form: } 3x^2)$
  • Negative Exponents: Inverse placement rule:     $x^{-n} = \frac{1}{x^n} \quad \text{Example: } 2^{3-4} = 2^{-1} = \frac{1}{2}$

• "Freshman's Dream" Algebraic Error:

  • Incorrect Statement: $(x + 2)^3 = x^3 + 2^3$
  • Correct Expansion: Summed quantities raised to a power must be expanded fully:     $(x + 2)^3 = (x + 2)(x + 2)(x + 2)$
  • Contextual Note: Mark Civic, a former math instructor, coined the term "Freshman's Dream" for this error. Mark Civic subsequently left teaching to manage a Raising Cane's chicken restaurant.

Simplification of Complex Algebraic Expressions

• Problem 1: Simplify $\frac{40 v w^2}{48 v^4 w^{-1}}$

  • Step 1: Simplify scalar fraction: $\frac{40}{48} = \frac{5}{6}$
  • Step 2: Apply quotient rule to variable $v$: $1 - 4 = -3 \implies v^{-3}$
  • Step 3: Apply quotient rule to variable $w$: $2 - (-1) = 2 + 1 = 3 \implies w^3$
  • Step 4: Write intermediate expression: $\frac{5 v^{-3} w^3}{6}$
  • Step 5: Move negative exponent $v^{-3}$ to denominator: $\frac{5 w^3}{6 v^3}$

• Problem 2: Simplify $\left( \frac{4 x^{-1} y^3}{3 x y^{-5}} \right)^{-2}$

  • Step 1: Simplify terms inside parentheses:
  • $x$ exponents: $-1 - 1 = -2 \implies x^{-2}$
  • $y$ exponents: $3 - (-5) = 3 + 5 = 8 \implies y^8$
  • Inner expression: $\frac{4 x^{-2} y^8}{3}$
  • Step 2: Invert inner expression to convert outer exponent from $-2$ to $+2$:     $\left( \frac{3}{4 x^{-2} y^8} \right)^2 = \left( \frac{3 x^2}{4 y^8} \right)^2$
  • Step 3: Distribute power of $2$ to every factor in numerator and denominator:     $\frac{3^2 (x^2)^2}{4^2 (y^8)^2} = \frac{9 x^4}{16 y^{16}}$

• Monomial Power Distribution vs Polynomial FOIL Expansion:

  • Single term monomial expressions raised to powers require direct power distribution: $(3x^2)^2 = 9x^4$
  • Multi-term polynomial expressions require binomial expansion/FOIL: $(3x^2 + 1)^2 = (3x^2 + 1)(3x^2 + 1)$

Questions and Course Logistics

• Questions & Discussion:

  • Question: Why does the interval for range have a bracket at $0$ but a parenthesis at infinity?
  • Response: Range starts at $0$ with a bracket $[0, \infty)$ because $0$ is an attainable output value of the absolute value function.
  • Question: Why start at $-8$ for graphing $|x| \le 8$?
  • Response: Plugging $-8$ into absolute value yields positive $8$, satisfying the boundary condition $|-8| = 8$
  • Question: Why are open circles used on interval $(-5, 1)$?
  • Response: Parentheses signify strict inequalities ($<$ or $>$), excluding endpoints $-5$ and $1$ from the solution set.
  • Question: Where does $3$ come from in $|x + 2| < 3$?
  • Response: $3$ represents the distance from center point $-2$ to either endpoint ($-5$ or $1$).
  • Question: Can $v$ and $w$ flip positions in $\frac{5 w^3}{6 v^3}$?
  • Response: No, $v^3$ must remain in denominator because its exponent was $-3$, whereas $w^3$ remains in numerator because its exponent was positive.
  • Question: Is FOIL required for simplifying $(3x^2)^2$?
  • Response: No, FOIL applies only when terms are separated by addition or subtraction within the quantity, such as $$(3x^2 + 1)^2$.

• Course Logistics:

  • Homework assignment covering sections 1.4 and 1.5 is due on Friday.
  • Future course sequence option: Math 168 is offered in the spring term.
  • Calculus course assignments: Calculus courses are typically instructed by Justin.