Unit 1 Mathematics Review: Functions, Transformations, Equations, and Operations

Transformations of Parent Functions

Understanding function transformations requires identifying the parent function $f(x)$ and analyzing how horizontal shifts, vertical shifts, stretches, compressions, and reflections alter its graph. The general transformation model is given by $y = a \cdot f(b(x - h)) + k$, where:

• $a$ controls vertical stretching ($|a| > 1$), vertical compression ($0 < |a| < 1$), and reflection across the x-axis ($a < 0$).
• $h$ controls horizontal translation (shifted right if $h > 0$, left if $h < 0$).
• $k$ controls vertical translation (shifted up if $k > 0$, down if $k < 0$).

Problem Analysis and Transformations

• Function 1: $m(x) = |x + 3| - 2$

  • Parent Function: Absolute value function $f(x) = |x|$
  • Transformation Description: Translated left by $3$ units and translated down by $2$ units.
  • Vertex Shift: The parent vertex at $(0, 0)$ moves to $(-3, -2)$.

• Function 2: $F(x) = 3(x - 4)^2$

  • Parent Function: Quadratic function $f(x) = x^2$
  • Transformation Description: Translated right by $4$ units and vertically stretched by a factor of $3$.
  • Vertex Shift: The parent vertex at $(0, 0)$ moves to $(4, 0)$.

• Function 3: $y = \sqrt{x} - 3$

  • Parent Function: Square root function $f(x) = \sqrt{x}$
  • Transformation Description: Translated down by $3$ units.
  • Starting Point Shift: The initial endpoint at $(0, 0)$ moves to $(0, -3)$.

• Function 4: $g(x) = (x + 2)^3 - 1$

  • Parent Function: Cubic function $f(x) = x^3$
  • Transformation Description: Translated left by $2$ units and translated down by $1$ unit.
  • Inflection Point Shift: The point of inflection at $(0, 0)$ moves to $(-2, -1)$.

Algebraic Inverses of Functions

An inverse function $f^{-1}(x)$ reverses the operation of $f(x)$. Graphically, $f^{-1}(x)$ is a reflection of $f(x)$ across the line $y = x$.

Algebraic Procedure for Finding Inverses

1. Replace $f(x)$ with $y$.
2. Interchanged variables $x$ and $y$.
3. Solve the resulting equation explicitly for $y$.
4. Replace $y$ with inverse notation $f^{-1}(x)$.

Problems and Detailed Algebraic Solutions

• Function 5: $h(x) = 5x + 2$

  • Set $y = 5x + 2$
  • Swap variables: $x = 5y + 2$
  • Isolate $y$:     $x - 2 = 5y$$y = \frac{x - 2}{5}$
  • Result: $h^{-1}(x) = \frac{x - 2}{5}$

• Function 6: $g(x) = \frac{6x + 3}{4x - 2}$

  • Set $y = \frac{6x + 3}{4x - 2}$
  • Swap variables: $x = \frac{6y + 3}{4y - 2}$
  • Solve for $y$:     $x(4y - 2) = 6y + 3$$4xy - 2x = 6y + 3$$4xy - 6y = 2x + 3$$y(4x - 6) = 2x + 3$$y = \frac{2x + 3}{4x - 6}$
  • Result: $g^{-1}(x) = \frac{2x + 3}{4x - 6}$

• Function 7: $b(x) = \frac{1}{x - 3}$

  • Set $y = \frac{1}{x - 3}$
  • Swap variables: $x = \frac{1}{y - 3}$
  • Solve for $y$:     $y - 3 = \frac{1}{x}$$y = \frac{1}{x} + 3 = \frac{3x + 1}{x}$
  • Result: $b^{-1}(x) = \frac{3x + 1}{x}$

• Function 8: $w(x) = (x - 2)^2 + 4$

  • Desmos Graphing Instructions: To graph the inverse, plot the restricted domain function $y = (x - 2)^2 + 4$ for $x \ge 2$. Swap variables to get the parametric/implicit form $x = (y - 2)^2 + 4$, which simplifies algebraically to $w^{-1}(x) = \sqrt{x - 4} + 2$ for $x \ge 4$. Plot the line $y = x$ to verify symmetry.

Absolute Value Equations and Inequalities

Absolute value equations and inequalities require setting up separate cases based on the definition of absolute value $|A| = A$ if $A \ge 0$ and $|A| = -A$ if $A < 0$.

Core Principles

• Equation $|A| = C$: Requires $A = C$ or $A = -C$ (valid only if $C \ge 0$).
• Inequality $|A| < C$: Equivalent to $-C < A < C$.
• Inequality $|A| > C$: Equivalent to $A > C$ or $A < -C$.

Problems and Solutions

• Problem 9: Solve $-1 - |4 - 8b| = 12$

  • Isolate the absolute value expression:     $-|4 - 8b| = 13$$|4 - 8b| = -13$
  • Analysis: Absolute values represent non-negative distances ($|A| \ge 0$). An absolute value cannot equal a negative number.
  • Conclusion: No Solution.

• Problem 10: Solve $8|x + 7| - 3 = 5$

  • Isolate the absolute value expression:     $8|x + 7| = 8$$|x + 7| = 1$
  • Split into two cases:     $x + 7 = 1 \quad \text{or} \quad x + 7 = -1$
  • Solve each equation:     $x = -6 \quad \text{or} \quad x = -8$
  • Solution: $x \in \{-8, -6\}$

• Problem 11: Solve $10 + 4|1 - 8r| < 74$

  • Isolate the absolute value expression:     $4|1 - 8r| < 64$$|1 - 8r| < 16$
  • Set up three-part inequality:     $-16 < 1 - 8r < 16$
  • Subtract $1$ from all parts:     $-17 < -8r < 15$
  • Divide all parts by $-8$ (reversing inequality symbols):     $\frac{17}{8} > r > -\frac{15}{8}$$-\frac{15}{8} < r < \frac{17}{8}$
  • Solution Interval: \left(-\frac{15}{8}, \frac{17}{8}\right)

• Problem 12: Solve $-6 + 7|6v - 5| \ge 29$

  • Isolate the absolute value expression:     $7|6v - 5| \ge 35$$|6v - 5| \ge 5$
  • Split into disjunctive inequalities:     $6v - 5 \ge 5 \quad \text{or} \quad 6v - 5 \le -5$
  • Solve each inequality:     $6v \ge 10 \implies v \ge \frac{5}{3}$$6v \le 0 \implies v \le 0$
  • Solution Interval: $(-\infty, 0] \cup \left[\frac{5}{3}, \infty\right)$

• Problem 13: Model Airplane Engine Weight Inequality

  • Parameter Definition:
  • Ideal weight $w_0 = 33.86\,\text{ounces}$
  • Maximum allowable weight variation $\delta = 0.05\,\text{ounce}$
  • Actual weight $w$
  • Absolute Value Inequality:     $|w - 33.86| \le 0.05$
  • Equivalent Range:     $-0.05 \le w - 33.86 \le 0.05$$33.81\,\text{ounces} \le w \le 33.91\,\text{ounces}$

Systems of Equations and Inequalities

Systems of equations can be solved using substitution or elimination methods. Systems of inequalities define bounded or unbounded regions in the Cartesian plane.

Problems and Step-by-Step Solutions

• Problem 14: Solve the Linear System   $\begin{cases} -9 + 5y = -4x \\ -11x = -20 + 9y \end{cases}$

  • Standard form rearrangement:     $4x + 5y = 9$$-11x - 9y = -20$
  • Elimination method:     Multiply equation 1 by $9$: $36x + 45y = 81$     Multiply equation 2 by $5$: $-55x - 45y = -100$     Add equations: $-19x = -19 \implies x = 1$
  • Solve for $y$:     $4(1) + 5y = 9 \implies 5y = 5 \implies y = 1$
  • Solution Point: $(1, 1)$

• Problem 15: Solve the Linear System   $\begin{cases} 4x + 2y = 10 \\ x - y = 13 \end{cases}$

  • Express $x$ from equation 2: $x = y + 13$
  • Substitute into equation 1:     $4(y + 13) + 2y = 10$$4y + 52 + 2y = 10$$6y = -42 \implies y = -7$
  • Solve for $x$:     $x = -7 + 13 = 6$
  • Solution Point: $(6, -7)$

• Problem 16: Solve the Non-linear System   $\begin{cases} y = x^2 - 1 \\ x + y = 11 \end{cases}$

  • Substitute $y = x^2 - 1$ into equation 2:     $x + (x^2 - 1) = 11$$x^2 + x - 12 = 0$
  • Factor quadratic:     $(x + 4)(x - 3) = 0$$x = -4 \quad \text{or} \quad x = 3$
  • Compute corresponding $y$ values:     For $x = -4$: $y = (-4)^2 - 1 = 15$     For $x = 3$: $y = 3^2 - 1 = 8$
  • Solution Points: $(-4, 15)$ and $(3, 8)$

• Problem 17: Graphing the Inequality System   $\begin{cases} 5x + 2y > -6 \\ x + 2y \ge 2 \end{cases}$

  • Convert to slope-intercept form:     Line 1: $y > -\frac{5}{2}x - 3$ (Dashed boundary line, shade region above).     Line 2: $y \ge -\frac{1}{2}x + 1$ (Solid boundary line, shade region above).

• Problem 18: Solution Point Verification for $(2, 3)$   Given System:   $\begin{cases} x + 2y \ge -6 \\ 3x - 2y > -2 \end{cases}$

  • Substitute $x = 2$ and $y = 3$ into inequality 1:     $2 + 2(3) \ge -6 \implies 8 \ge -6$ (Satisfied)
  • Substitute $x = 2$ and $y = 3$ into inequality 2:     $3(2) - 2(3) > -2 \implies 0 > -2$ (Satisfied)
  • Conclusion: Point $(2, 3)$ is a valid solution to the system.

Piecewise Functions and Graph Evaluation

Piecewise functions apply different rule formulas depending on the input variable $x$ value relative to defined domain intervals.

Problems and Evaluations

• Problem 7: Piecewise Definition and Graph Features   $f(x) = \begin{cases} -\frac{1}{3}x & \text{if } x < -3 \\ x + 2 & \text{if } -3 \le x \le 0 \\ (x - 2)^2 & \text{if } x > 0 \end{cases}$

  • Sub-function 1 ($x < -3$): Linear branch with slope $-\frac{1}{3}$ extending to the left with an open circle endpoint at $(-3, 1)$.
  • Sub-function 2 ($-3 \le x \le 0$): Linear segment from $(-3, -1)$ to $(0, 2)$ with closed endpoints at both boundaries.
  • Sub-function 3 ($x > 0$): Parabolic branch opening upwards with vertex at $(2, 0)$ and open circle at $(0, 4)$.

• Problem 19: Piecewise Definition   $f(x) = \begin{cases} |x - 2| & \text{if } x \le 0 \\ -|x + 3| & \text{if } x < -1 \\ (x - 2)^2 + 1 & \text{if } x > 0 \end{cases}$

• Problem 20: Piecewise Definition   $w(x) = \begin{cases} 2 & \text{if } x < -1 \\ x - 2 & \text{if } -1 \le x \le 2 \\ 3 & \text{if } x \ge 2 \end{cases}$

• Problem 21: Numerical Evaluation of Piecewise Function $w(x)$   Given Function:   $w(x) = \begin{cases} 3x - 1 & \text{if } x < -1 \\ x & \text{if } -1 \le x \le 2 \\ 2 & \text{if } x \ge 2 \end{cases}$

  • Part a: Evaluate $f(2)$ (or $w(2)$):     $x = 2$ falls into the domain interval $-1 \le x \le 2$, using rule w(x) = x$.\n    w(2) = 2\n - Part b: Evaluate w(-2) + 2w(0):\n    For x = -2 < -1$, use rule$w(x) = 3x - 1:\n    w(-2) = 3(-2) - 1 = -7\n    For x = 0$in$-1 \le x \le 2$, use rule$w(x) = x:\n    w(0) = 0\n    Calculate expression:\n    w(-2) + 2w(0) = -7 + 2(0) = -7\n\n- Problem 22: Graphical Function Evaluation\n - Part a: Determine f(1)$by reading the y-coordinate corresponding to$x = 1 on the graph.\n - Part b: Determine f(3)$by reading the y-coordinate corresponding to$x = 3 on the graph.\n - Part c: Calculate f(-2) + 2f(1) using evaluated values from the graph.\n\n# Function Operations and Compositions\n\nGiven base functions:\nf(x) = 2x^2 + x - 1\ng(x) = x^2 + 6x - 5\nh(x) = 2x + 1\nj(x) = x - 3\n\n### Problem Computations\n\n- Problem 23: Perform Product (h \cdot g)(x)\n - Expansion:\n    (h \cdot g)(x) = (2x + 1)(x^2 + 6x - 5)\n    (h \cdot g)(x) = 2x(x^2 + 6x - 5) + 1(x^2 + 6x - 5)\n    (h \cdot g)(x) = 2x^3 + 12x^2 - 10x + x^2 + 6x - 5\n - Combine like terms:\n    (h \cdot g)(x) = 2x^3 + 13x^2 - 4x - 5\n\n- Problem 24: Perform Subtraction f(x) - g(x)\n - Expansion:\n    f(x) - g(x) = (2x^2 + x - 1) - (x^2 + 6x - 5)\n    f(x) - g(x) = 2x^2 + x - 1 - x^2 - 6x + 5\n - Combine like terms:\n    f(x) - g(x) = x^2 - 5x + 4\n\n- Problem 25: Perform Composition (h \circ j)(x)\n - Substitute j(x) = x - 3$into$h(x) = 2x + 1:\n    (h \circ j)(x) = h(j(x)) = 2(x - 3) + 1\n - Simplify expression:\n    (h \circ j)(x) = 2x - 6 + 1 = 2x - 5\n\n- Problem 26: Evaluate Composite Value h(g(2))\n - Inner evaluation g(2):\n    g(2) = (2)^2 + 6(2) - 5 = 4 + 12 - 5 = 11\n - Outer evaluation h(11):\n    h(11) = 2(11) + 1 = 22 + 1 = 23\n - Final Value: h(g(2)) = 23\n\n- Problem 27: Evaluate Fractional Input g\left(\frac{1}{2}\right)\n - Substitute x = \frac{1}{2}$into$g(x) = x^2 + 6x - 5:\n    g\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 + 6\left(\frac{1}{2}\right) - 5\n    g\left(\frac{1}{2}\right) = \frac{1}{4} + 3 - 5\n    g\left(\frac{1}{2}\right) = \frac{1}{4} - 2 = -\frac{7}{4}\n - Final Value: g\left(\frac{1}{2}\right) = -\frac{7}{4}$$