Semester Exam Self-Assessment Study Notes: Unit 5 Rational Exponents & Radical Functions

This document serves as a study guide and self-assessment for the Semester Exam, completed by student Carson Jackson.
Task: Solve identifying problems without assistance and mark confidence levels for each problem type.
Due Date: Monday.
Subject Area: Unit 5 - Rational Exponents & Radical Functions.

Problem Statement: Rewrite $16^{\frac{3}{4}}$ using radicals, then evaluate.
Concept: The rational exponent rule states $a^{\frac{m}{n}} = (\sqrt[n]{a})^m$ or $\sqrt[n]{a^m}$ .
Evaluation Process: * Step 1: Convert to radical form: $(\sqrt[4]{16})^3$ * Step 2: Evaluate the fourth root: Since $2 \times 2 \times 2 \times 2 = 16$ , $\sqrt[4]{16} = 2$ * Step 3: Cube the result: $2^3 = 8$
Student Status: Confident.

Problem Statement: Solve the equation $x^{\frac{3}{2}} = 9$
Theoretical Approach: To solve for $x$ , raise both sides of the equation to the reciprocal of the power, which is $\frac{2}{3}$ . * Equation: $(x^{\frac{3}{2}})^{\frac{2}{3}} = 9^{\frac{2}{3}}$ * Simplified: $x = (\sqrt[3]{9})^2$
Student Scratch Notes: * "X-9" * "19-3" * " $\sqrt{\sqrt{4}}=3$ " (Note: This appears to be a transcription error in the student's scratch work) * "33 27"
Student Status: Unsure.

Problem Statement: Simplify completely $\sqrt{72x^4}$
Simplification Process: * Factor the number into a perfect square and a remainder: $\sqrt{72} = \sqrt{36 \times 2}$ * Simplify the square root of the perfect square: $6\sqrt{2}$ * Simplify the variable component: $\sqrt{x^4} = x^2$ * Combine results: $6x^2\sqrt{2}$
Student Work Details: * Step-by-step noted: " $\sqrt{36 \cdot 2} = 6\sqrt{2}$ " * Variable result: " $x^2$ " * Final output listed: " $6x^2\sqrt{2}$ "
Student Status: Confident.

Problem Statement: Simplify and write with positive rational exponents: $(8x^3)^{\frac{1}{3}}$
Rule: The Power of a Product Rule applies $(ab)^n = a^n b^n$ .
Simplification Steps: * Apply exponent to the coefficient: $8^{\frac{1}{3}} = \sqrt[3]{8} = 2$ * Apply exponent to the variable: $(x^3)^{\frac{1}{3}} = x^{3 \times \frac{1}{3}} = x^1$ * Result: $2x$
Student Commentary: "I have no clue"
Student Status: Lost.

Problem Statement: Describe the domain, range, and starting point of $f(x) = \sqrt{x-4}$
Critical Values: * Domain: Setting the radicand to greater than or equal to zero ( $x - 4 \ge 0$ ) yields $x \ge 4$ . * Starting Point: Determined by the horizontal shift of 4 and no vertical shift, identified as $(4, 0)$ . * Range: Since the function is a positive square root with no vertical shift, $y \ge 0$ .
Student Annotations: * Domain identified as " $D= x \ge 4$ " * Starting point identified as " $SP (4, 0)$ "
Student Status: Confident.

Problem Statement: Describe the transformations from $y = \sqrt{x}$ to $y = -2\sqrt{x+1} + 3$
Identified Transformations: 1. Reflection: The negative sign in front of the 2 indicates a reflection across the x-axis. 2. Vertical Stretch: The coefficient 2 indicates a Vertical Stretch (VS) by a factor of 2. 3. Horizontal Shift: The " $x+1$ " inside the radical indicates a shift Left by 1 (L1). 4. Vertical Shift: The " $+3$ " indicates a shift Up by 3.
Student Abbreviations Used: * "VS x2" * "HS L1" * "UP 3"
Student Status: Confident.

Problem Statement: Describe the domain and end behavior of $f(x) = \sqrt[3]{x-5}$
Function Properties: * Domain ( $D$ ): For cube root functions, the domain is the set of all real numbers ( $R$ ). * Range ( $R$ ): The range is also all real numbers ( $R$ ). * End Behavior: * As $x \rightarrow \infty$ , $f(x) \rightarrow \infty$ * As $x \rightarrow -\infty$ , $f(x) \rightarrow -\infty$
Student Annotations: * Domain: " $D: R$ " * End behavior noted as: " $x \rightarrow \infty, f(x) \rightarrow \infty$ " and " $x \rightarrow -\infty, f(x) \rightarrow -\infty$ "
Student Status: Confident.

Problem Statement: Describe the transformations from $y = \sqrt[3]{x}$ to $y = \sqrt[3]{x+2}-1$
Identified Transformations: 1. Horizontal Shift: The " $x+2$ " indicates a shift to the left by 2. 2. Vertical Shift: The " $-1$ " indicates a shift down by 1.
Student Abbreviations Used: * "vs left x2" (Note: This uses "vs" likely standing for "vertical shift" incorrectly or simply a general notation for shift) * "down x1"
Student Status: Confident.

Problem Statement: Solve and check: $\sqrt{2x-1} = x-1$
Solving Process: * Step 1: Square both sides of the equation to eliminate the radical. * $(\sqrt{2x-1})^2 = (x-1)^2$ * $2x - 1 = x^2 - 2x + 1$ * Step 2: Move all terms to one side to set the quadratic equation to zero. * $x^2 - 4x + 2 = 0$ * Step 3: Solve using the Quadratic Formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ * $x = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(2)}}{2(1)}$ * $x = \frac{4 \pm \sqrt{16-8}}{2} = \frac{4 \pm \sqrt{8}}{2} = \frac{4 \pm 2\sqrt{2}}{2}$ * $x = 2 \pm \sqrt{2}$
Student Scratch Work: * " $x-1=(x-1)^2$ " * " $2x-1=x^2-2x+1$ " * " $x^2-4x+2=0$ " * Final solution written: " $x = 2 \pm \sqrt{2}$ "
Student Status: Confident.