g\left(\frac{2}{\sqrt{3}}\right) = \left(\frac{8}{3\sqrt{3}}\right) - 4\left(\frac{2}{\sqrt{3}}\right) + 2 = \frac{8 - 24 + 6\sqrt{3}}{3\sqrt{3}} = \frac{-16 + 6\sqrt{3}}{3\sqrt{3}}

["# Understanding the Evaluation of ( g\left(\frac{2}{\sqrt{3}}\right) ): A Step-by-Step Breakdown", "Mathematics often presents intricate expressions that require careful manipulation to simplify and interpret. One such expression is the evaluation of the function\n[ g\left(\frac{2}{\sqrt{3}}\right) = \left(\frac{8}{3\sqrt{3}}\right) - 4\left(\frac{2}{\sqrt{3}}\right) + 2. ]\nAt first glance, this may appear complex, especially with irrational terms like ( \sqrt{3} ). However, with systematic simplification and rationalization, we can express ( g\left(\frac{2}{\sqrt{3}}\right) ) in a clean and insightful form. This article provides a detailed explanation of how this expression simplifies, highlighting key algebraic techniques and the importance of rationalization.", "## The Original Expression", "We begin with:\n[\ng\left(\frac{2}{\sqrt{3}}\right) = \frac{8}{3\sqrt{3}} - 4\left(\frac{2}{\sqrt{3}}\right) + 2\n]", "This expression contains a term with a radical in the denominator: ( \frac{8}{3\sqrt{3}} ). To simplify, we rationalize the denominator, a standard procedure in algebra to eliminate square roots from denominators.", "## Step 1: Rationalizing ( \frac{8}{3\sqrt{3}} )", "[\n\frac{8}{3\sqrt{3}} = \frac{8}{3\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{8\sqrt{3}}{3 \cdot 3} = \frac{8\sqrt{3}}{9}\n]", "This step removes the radical from the denominator, yielding a cleaner term:\n[\n\frac{8}{3\sqrt{3}} = \frac{8\sqrt{3}}{9}\n]", "## Step 2: Simplifying the Second Term", "The second term is straightforward:\n[\n4\left(\frac{2}{\sqrt{3}}\right) = \frac{8}{\sqrt{3}}\n]", "Again, rationalizing:\n[\n\frac{8}{\sqrt{3}} = \frac{8\sqrt{3}}{3}\n]", "## Step 3: Rewriting the Full Expression", "Substituting the simplified terms back:\n[\ng\left(\frac{2}{\sqrt{3}}\right) = \frac{8\sqrt{3}}{9} - \frac{8}{\sqrt{3}} + 2 = \frac{8\sqrt{3}}{9} - \frac{8\sqrt{3}}{3} + 2\n]", "Now combine the terms with ( \sqrt{3} ).", "## Step 4: Combining like terms", "The coefficients of ( \sqrt{3} ) are ( \frac{8}{9} ) and ( -\frac{8}{3} ). Write them with a common denominator:\n[\n\frac{8}{9} - \frac{8}{3} = \frac{8}{9} - \frac{24}{9} = \frac{-16}{9}\n]", "So the expression becomes:\n[\n\frac{-16\sqrt{3}}{9} + 2\n]", "## Step 5: Expressing 2 with denominator 9", "[\n2 = \frac{18}{9}\n]", "Now combine:\n[\n\frac{-16\sqrt{3}}{9} + \frac{18}{9} = \frac{18 - 16\sqrt{3}}{9}\n]", "## Step 6: Final Rationalization Form (Optional Alternate Expression)", "We aim to write the entire expression in the form:\n[\ng\left(\frac{2}{\sqrt{3}}\right) = \frac{-16 + 6\sqrt{3}}{3\sqrt{3}}\n]", "Let’s verify this by manipulating ( \frac{18 - 16\sqrt{3}}{9} ) to match the target form.", "Multiply numerator and denominator by ( \sqrt{3} ) to rationalize the denominator:\n[\n\frac{18 - 16\sqrt{3}}{9} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{(18 - 16\sqrt{3})\sqrt{3}}{9\sqrt{3}} = \frac{18\sqrt{3} - 16 \cdot 3}{9\sqrt{3}} = \frac{18\sqrt{3} - 48}{9\sqrt{3}}\n]", "Simplify numerator:\n[\n\frac{18\sqrt{3} - 48}{9\sqrt{3}} = \frac{6(3\sqrt{3} - 8)}{9\sqrt{3}} = \frac{2(3\sqrt{3} - 8)}{3\sqrt{3}} = \frac{6\sqrt{3} - 16}{3\sqrt{3}} = \frac{-16 + 6\sqrt{3}}{3\sqrt{3}}\n]", "This matches the originally stated simplified form.", "## Why Rationalization Matters", "Rationalizing denominators simplifies expressions for comparison, integration, expansion, and interpretation—especially in calculus and advanced algebra. It ensures consistency and clarity across mathematical communication.", "## Summary", "Evaluating ( g\left(\frac{2}{\sqrt{3}}\right) ) involves:\n- Rationalizing radical denominators,\n- Combining like terms with care,\n- Expressing results in simplified fractional forms.", "The full simplification yields:\n[\ng\left(\frac{2}{\sqrt{3}}\right) = \frac{-16 + 6\sqrt{3}}{3\sqrt{3}}\n]\na compact representation perfect for analytical use.", "Understanding such transformations builds a strong foundation in algebraic manipulation and prepares learners for more complex functions and limits in upper-level mathematics.", "---", "Key takeaway: Complex-looking expressions often reduce elegantly with standard algebraic techniques—rationalization, combining like terms, and careful simplification."]