= \frac{(3x + 4)(\sqrt{x} + 2)}{x - 4}

= \frac{(3x + 4)(\sqrt{x} + 2)}{x - 4}

["# Solving and Analyzing the Rational Expression: (\frac{(3x + 4)(\sqrt{x} + 2)}{x - 4})", "In algebra, rational expressions involving radicals and denominators demand careful analysis. One such expression is:", "[\n\frac{(3x + 4)(\sqrt{x} + 2)}{x - 4}\n]", "This article explores how to simplify, analyze, and interpret this rational expression over its domain, focusing on algebraic handling, domain restrictions, and practical applications.", "---", "## 1. Understanding the Expression Structure", "The expression combines polynomial and radical components in both the numerator and denominator:", "- Numerator: ((3x + 4)(\sqrt{x} + 2)) — a product of a linear term and a binomial involving a square root.\n- Denominator: (x - 4) — a linear rational term.", "This combination suggests potential simplification opportunities and domain considerations due to the square root (\sqrt{x}).", "---", "## 2. Domain Considerations", "Before simplifying or evaluating the expression, domain restrictions must be identified.", "### Restrictions on (x):\n- The square root (\sqrt{x}) requires:\n [\n x \geq 0\n ]\n- The denominator (x - 4) must not be zero:\n [\n x <br/>\ne 4\n ]", "Domain:\n[\nx \in [0, 4) \cup (4, \infty)\n]", "Key Notes:\n- At (x = 4), the expression is undefined (division by zero).\n- At (x = 0):\n [\n \frac{(3(0) + 4)(\sqrt{0} + 2)}{0 - 4} = \frac{(4)(0 + 2)}{-4} = \frac{8}{-4} = -2\n ]\n So, the expression is defined at (x = 0).", "---", "## 3. Simplification Efforts", "Can the numerator be expanded or factored?", "Expand numerator:", "[\n(3x + 4)(\sqrt{x} + 2) = 3x\sqrt{x} + 6x + 4\sqrt{x} + 8\n]", "Rewriting in terms of (x^{1/2}):", "Let (u = \sqrt{x}), so (x = u^2), and the numerator becomes:", "[\n3u^3 + 6u^2 + 4u + 8\n]", "Denominator: (x - 4 = u^2 - 4)", "Thus, the expression transforms to:", "[\n\frac{3u^3 + 6u^2 + 4u + 8}{u^2 - 4}\n]", "This cubic-over-quadratic form is not immediately factorable into simpler rational terms. However, check for common factors with denominator (u^2 - 4 = (u - 2)(u + 2)).", "Try polynomial division or synthetic methods to test if (u = 2) or (u = -2) are roots.", "- At (u = 2):\n Numerator: (3(8) + 6(4) + 4(2) + 8 = 24 + 24 + 8 + 8 = 64 <br/>\ne 0)\n- At (u = -2):\n Numerator: (3(-8) + 6(4) + 4(-2) + 8 = -24 + 24 - 8 + 8 = 0)", "So (u + 2) is a factor. Perform polynomial division of (3u^3 + 6u^2 + 4u + 8) by (u + 2):", "Using synthetic division:", "[\n\begin{array}{r|rrrr}\n-2 & 3 & 6 & 4 & 8 \\n & & -6 & 0 & -8 \\n\hline\n & 3 & 0 & 4 & 0 \\n\end{array}\n]", "Result: (3u^2 + 0u + 4 = 3u^2 + 4)", "So,", "[\n\frac{3u^3 + 6u^2 + 4u + 8}{u^2 - 4} = \frac{(u + 2)(3u^2 + 4)}{(u - 2)(u + 2)} = \frac{3u^2 + 4}{u - 2}, \quad u <br/>\ne -2\n]", "Now recall (u = \sqrt{x}), and since (u \geq 0), only (u = -2) is irrelevant. But note the cancellation is valid as long as (u <br/>\ne -2), which is fine.", "So, original expression simplifies algebraically (for (x > 0)) to:", "[\n\frac{3u^2 + 4}{u - 2} = \frac{3x + 4}{\sqrt{x} - 2}, \quad x > 0,\ x <br/>\ne 4\n]", "Important: Domain restrictions still apply — even though simplified, the original function excludes (x = 4) and (x = 0) is allowed but results in defined value.", "---", "## 4. Analyzing the Simplified Form", "Now analyze:", "[\nf(x) = \frac{3x + 4}{\sqrt{x} - 2}, \quad x \in [0, 4) \cup (4, \infty)\n]", "Behavior at boundaries:", "- At (x \ o 0^+):\n Numerator (\ o 4), denominator (\ o 0^-) (since (\sqrt{x} - 2 \ o -2))\n Wait: (\sqrt{x} \ o 0), so (\sqrt{x} - 2 \ o -2), but wait: (\sqrt{0} = 0), so (\ o -2).\n So:\n [\n \frac{4}{-2} = -2\n ]\n Matches earlier evaluation.", "- At (x \ o 4^-):\n Numerator: (3(4) + 4 = 16), denominator: (\sqrt{4} - 2 = 0^-), so fraction (\ o -\infty)", "- At (x \ o 4^+):\n Denominator: (\ o 0^+), so (16 / 0^+ \ o +\infty)", "- As (x \ o \infty):\n Leading behavior: numerator (\sim 3x), denominator (\sim \sqrt{x}), so\n [\n f(x) \sim \frac{3x}{\sqrt{x}} = 3\sqrt{x} \ o \infty\n ]", "So function has vertical asymptote at (x = 4), and grows without bound otherwise.", "---", "## 5. Differentiability and Critical Points (Optional Advanced Insight)", "To study maxima/minima or slope, differentiate:", "Let\n[\nf(x) = \frac{3x + 4}{\sqrt{x} - 2}, \quad 0 \le x < 4,\ x > 4\n]", "Use quotient rule:\n[\nf'(x) = \frac{(3)(\sqrt{x} - 2) - (3x + 4)\left(\frac{1}{2\sqrt{x}}\right)}{(\sqrt{x} - 2)^2}\n]", "Simplify numerator:\n[\n3(\sqrt{x} - 2) - \frac{3x + 4}{2\sqrt{x}} = 3\sqrt{x} - 6 - \frac{3x + 4}{2\sqrt{x}}\n]", "Common denominator (2\sqrt{x}):", "[\n= \frac{6x - 12\sqrt{x} - (3x + 4)}{2\sqrt{x}} = \frac{3x - 12\sqrt{x} - 4}{2\sqrt{x}}\n]", "So,\n[\nf'(x) = \frac{3x - 12\sqrt{x} - 4}{2\sqrt{x}(\sqrt{x} - 2)^2}\n]", "Critical points when numerator is zero:\n[\n3x - 12\sqrt{x} - 4 = 0\n]", "Let (u = \sqrt{x}), so:\n[\n3u^2 - 12u - 4 = 0\n]", "Solve using quadratic formula:\n[\nu = \frac{12 \pm \sqrt{144 + 48}}{6} = \frac{12 \pm \sqrt{192}}{6} = \frac{12 \pm 8\sqrt{3}}{6} = \frac{6 \pm 4\sqrt{3}}{3}\n]", "Only positive root matters:\n[\nu = \frac{6 + 4\sqrt{3}}{3} \approx \frac{6 + 6.928}{3} \approx 4. Henrik\n\Rightarrow x = u^2 \approx (4. Henrik)^2 \approx 16.8\n]", "Valid in ((4, \infty)), so one critical point exists there.", "---", "## 6. Applications and Interpretation", "Rational expressions like this appear in physics (e.g., modeling resistance with square roots), engineering flow rates, or biology (growth models with limits). Understanding domain and asymptotes ensures meaningful application and avoids nonsensical outputs.", "---", "## 7. Final Summary", "- The expression (\frac{(3x + 4)(\sqrt{x} + 2)}{x - 4}) simplifies (via substitution) to (\frac{3x + 4}{\sqrt{x} - 2}) for (x \geq 0,\ x <br/>\ne 4).\n- Domain: (x \in [0, 4) \cup (4, \infty))\n- The function has a vertical asymptote at (x = 4).\n- Behavior: approaches (-2) at (x = 0), tends to (-\infty) as (x \ o 4^-), (+\infty) as (x \ o 4^+), and grows to (+\infty) as (x \ o \infty).\n- Derivative analysis reveals a maximum or minimum near (x \approx 16.8).", "---", "## 8. Final Answer and Code Insight", "For all (x) in the domain:", "[\n\boxed{\n\frac{(3x + 4)(\sqrt{x} + 2)}{x - 4} = \frac{3x + 4}{\sqrt{x} - 2},\quad x \in [0, 4) \cup (4, \infty)\n}\n]", "This form enables more efficient differentiation, integration, and graphical interpretation—crit"]

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