So the only way to resolve this is to define theQuestion: A chemical engineer models the concentration of a reactant in a reactor over time with the function $ C(t) = rac{3t + 2}{t^2 + 4} $. For what values of $ t $ is the concentration increasing?

["Title: When is Reactant Concentration Increasing? Analyzing the Function $ C(t) = \frac{3t + 2}{t^2 + 4} $", "Understanding how the concentration of a reactant changes over time is crucial in chemical engineering for optimizing reactor performance and ensuring safety. Our goal is to determine for which values of time $ t $ the concentration $ C(t) $ is increasing. This analysis hinges on computing the first derivative of $ C(t) $ and identifying where it is positive.", "### Understanding the Model\nThe concentration is modeled by\n[\nC(t) = \frac{3t + 2}{t^2 + 4}\n]\nHere, the numerator $ 3t + 2 $ represents net reaction growth influenced by input rates and reaction stoichiometry, while the denominator $ t^2 + 4 $ accounts for inhibition due to dispersion, heat effects, or geometric constraints—common in real reactors.", "### Step 1: Differentiate $ C(t) $ Using the Quotient Rule\nThe derivative of a quotient $ \frac{u}{v} $ is\n[\nC'(t) = \frac{u'v - uv'}{v^2}\n]\nLet:\n- $ u = 3t + 2 $ → $ u' = 3 $\n- $ v = t^2 + 4 $ → $ v' = 2t $", "Applying the quotient rule:\n[\nC'(t) = \frac{3(t^2 + 4) - (3t + 2)(2t)}{(t^2 + 4)^2}\n]", "### Step 2: Expand and Simplify the Numerator\nCompute the numerator:\n[\n3(t^2 + 4) = 3t^2 + 12\n]\n[\n(3t + 2)(2t) = 6t^2 + 4t\n]\nSo,\n[\n\ ext{Numerator} = (3t^2 + 12) - (6t^2 + 4t) = -3t^2 - 4t + 12\n]\nThus,\n[\nC'(t) = \frac{-3t^2 - 4t + 12}{(t^2 + 4)^2}\n]", "### Step 3: Analyze the Sign of $ C'(t) $\nSince the denominator $ (t^2 + 4)^2 $ is always positive for all real $ t $, the sign of $ C'(t) $ depends solely on the numerator:\n[\n-3t^2 - 4t + 12 > 0\n]\nMultiply both sides by $-1$ (reversing the inequality):\n[\n3t^2 + 4t - 12 < 0\n]", "### Step 4: Solve the Quadratic Inequality\nFind the roots of $ 3t^2 + 4t - 12 = 0 $ using the quadratic formula:\n[\nt = \frac{-4 \pm \sqrt{4^2 - 4(3)(-12)}}{2(3)} = \frac{-4 \pm \sqrt{16 + 144}}{6} = \frac{-4 \pm \sqrt{160}}{6}\n]\nSimplify $ \sqrt{160} = \sqrt{16 \ imes 10} = 4\sqrt{10} $, so:\n[\nt = \frac{-4 \pm 4\sqrt{10}}{6} = \frac{-2 \pm 2\sqrt{10}}{3}\n]\nLet:\n- $ t_1 = \frac{-2 - 2\sqrt{10}}{3} \approx \frac{-2 - 6.32}{3} \approx -2.77 $\n- $ t_2 = \frac{-2 + 2\sqrt{10}}{3} \approx \frac{-2 + 6.32}{3} \approx 1.44 $", "Since the parabola $ 3t^2 + 4t - 12 $ opens upward, the expression is negative between the roots:\n[\nC'(t) > 0 \quad \ ext{when} \quad t \in \left( \frac{-2 - 2\sqrt{10}}{3},\ \frac{-2 + 2\sqrt{10}}{3} \right)\n]\nApproximate interval: $ t \in (-2.77,\ 1.44) $", "### Step 5: Interpret the Result in Context\nIn chemical engineering applications, time $ t $ typically starts at injection or reactor startup (often $ t \geq 0 $). Thus, within physical time domains ($ t \geq 0 $), the concentration increases for:\n[\nt \in \left[0,\ \frac{-2 + 2\sqrt{10}}{3}\right) \approx [0,\ 1.44)\n]", "### Conclusion\nThe reactant concentration $ C(t) $ increases as time progresses within the interval determined by when the derivative changes sign. Engineering models like this help predict optimal operation windows and detect inefficiencies or hazardous buildup periods.", "Keywords: chemical reactor analysis, concentration dynamics, $ C(t) = \frac{3t + 2}{t^2 + 4} $, derivative, increasing concentration, chemical engineering, differential modeling, process optimization.\nMeta Description: Find when reactant concentration rises in a chemical reactor modeled by $ C(t) = \frac{3t + 2}{t^2 + 4} $. Learn when $ t \in \left( \frac{-2 - 2\sqrt{10}}{3}, \frac{-2 + 2\sqrt{10}}{3} \right) $, and how this informs real-world reactor design."]