Solution: Let $ y = rac{3t}{4 - t^2} $. For $ t > 2 $, denominator $ 4 - t^2 < 0 $, so $ y < 0 $. Rewrite $ y = rac{3t}{-(t^2 - 4)} = -rac{3t}{(t - 2)(t + 2)} $. Let $ t = 2 + \epsilon $, $ \epsilon > 0 $, but instead, analyze $ y $: as $ t o 2^+ $, $ y o -\infty $; as $ t o \infty $, $ y o 0^- $. The minimum value of

["Understanding the Behavior of $ y = \dfrac{3t}{4 - t^2} $ for $ t > 2 $ – Analyzing the Minimum Value", "When analyzing the function $ y = \dfrac{3t}{4 - t^2} $, particularly for $ t > 2 $, we observe a critical shift in behavior at $ t = 2 $, where the denominator changes sign and fundamentally alters the value and trend of $ y $. This analysis reveals key insights useful in optimization and real-world modeling scenarios.", "---", "### Domain and Sign Analysis", "For $ t > 2 $, the denominator $ 4 - t^2 $ becomes negative because $ t^2 > 4 $. Specifically:", "$$\ny = \dfrac{3t}{4 - t^2} < 0 \quad \ ext{(since numerator > 0 and denominator < 0 for all } t > 2)\n$$", "As $ t $ approaches 2 from the right, $ 4 - t^2 \ o 0^- $, so:", "$$\n\lim_{t \ o 2^+} y = \lim_{t \ o 2^+} \dfrac{3t}{4 - t^2} = -\infty\n$$", "As $ t \ o \infty $, the degree of the denominator is higher than the numerator, so $ y \ o 0^- $, approaching zero from the negative side:", "$$\n\lim_{t \ o \infty} y = 0^-\n$$", "---", "### Rewriting the Expression", "To better understand critical values, rewrite:", "$$\ny = \dfrac{3t}{4 - t^2} = - \dfrac{3t}{t^2 - 4} = - \dfrac{3t}{(t - 2)(t + 2)}\n$$", "Let $ t = 2 + \epsilon $, with $ \epsilon > 0 $. While this substitution helps study behavior near $ t = 2 $, a more effective approach is to analyze $ y $ directly as a function of $ t $, using calculus to find any local extrema.", "---", "### Finding the Minimum Value Using Calculus", "Define $ y(t) = \dfrac{3t}{4 - t^2} $ for $ t > 2 $. Compute the derivative $ y'(t) $:", "$$\ny'(t) = \dfrac{(3)(4 - t^2) - 3t(-2t)}{(4 - t^2)^2} = \dfrac{12 - 3t^2 + 6t^2}{(4 - t^2)^2} = \dfrac{12 + 3t^2}{(4 - t^2)^2}\n$$", "Note: The numerator $ 12 + 3t^2 > 0 $ for all real $ t $, and the denominator $ (4 - t^2)^2 > 0 $ for $ t <br/>\ne 2 $. Since $ t > 2 $, $ 4 - t^2 < 0 $, but its square is positive.", "Therefore,", "$$\ny'(t) > 0 \quad \ ext{for all } t > 2\n$$", "The positive derivative implies that $ y(t) $ is strictly increasing on $ (2, \infty) $. This means:", "- $ y(t) $ starts from $ -\infty $ as $ t \ o 2^+ $\n- Increases toward $ 0^- $ as $ t \ o \infty $", "A strictly increasing function $ t > 2 $ has no local maximum, but it approaches a minimum value asymptotically at the left endpoint—though strictly speaking, $ y(t) $ does not attain a minimum, it approaches $ -\infty $. However, since $ y(t) $ increases continuously, the infimum of $ y $ on $ (2, \infty) $ is $ -\infty $, but the minimum finite value does not exist.", "But wait — reconsider: since $ y(t) $ is increasing and unbounded below, the smallest attained value does not exist in a finite range, yet we can interpret the least upper bound of lower values — which is still $ -\infty $. However, if the question refers to the minimum of $ -y $ or a peak behavior, note: $ y $ grows negatively toward zero. So $ y $ has no absolute minimum, but its behavior clarifies modeling constraints.", "But suppose the intent is to find the minimum of $ y $ in behavior near $ t = 2 $ and monotonic rise: since it increases from $ -\infty $ to $ 0^- $, the least value is approached but never reached — the function is unbounded below with strictly increasing trend.", "Wait — correction: since $ y(t) $ increases from $ -\infty $ to $ 0^- $, every real number less than 0 is taken exactly once for $ t > 2 $. Thus, $ y(t) $ has no minimum value, but its infimum is $ -\infty $.", "However, in practical modeling (e.g., physical constraints), if $ t > 2 $ represents a controlled regime, the "minimum attainable safe value" under increasing $ y $ would still be tending to $ -\infty $, meaning no strict lower bound.", "But if the expression was meant to model a maximum or peak, reconsider the sign: since $ y < 0 $, the function approaches zero from below — no maximum in $ (2, \infty) $.", "Recheck: earlier derivative showed $ y'(t) > 0 $, so strictly increasing — no local minimum or maximum.", "Therefore, there is no minimum value of $ y $ for $ t > 2 $; $ y(t) \ o -\infty $ as $ t \ o 2^+ $, and $ y(t) \ o 0^- $ as $ t \ o \infty $.", "But this contradicts typical regression or optimization questions. Let’s reassess the intent.", "---", "### Correct Interpretation: Find the Minimum of $ y $ in Context", "Even though $ y(t) $ is increasing, suppose the model assumes $ t $ is bounded above due to physical limits. However, mathematically, on $ (2, \infty) $, $ y $ has no minimum — it can be arbitrarily negative near $ t = 2 $.", "But let's compute $ y(t) $ at a sample point, say $ t = 3 $:", "$$\ny(3) = \dfrac{3 \cdot 3}{4 - 9} = \dfrac{9}{-5} = -1.8\n$$", "At $ t = 4 $: $ y = \dfrac{12}{4 - 16} = \dfrac{12}{-12} = -1 $\nAt $ t = 10 $: $ y = \dfrac{30}{4 - 100} = \dfrac{30}{-96} \approx -0.3125 $\nAs $ t \ o \infty $, $ y \ o 0^- $", "So $ y $ increases: $ -\infty < -1.8> -1 > -0.3125> \dots $", "Therefore, the infimum is $ -\infty $, but no minimum exists.", "However, if the function were $ y = \dfrac{3t}{t^2 - 4} $, then $ y > 0 $ for $ t > 2 $, and $ y \ o 0^+ $, $ y(3) = \dfrac{9}{9 - 4} = \dfrac{9}{5} $, increasing to 0 — still no minimum.", "But in our case, since $ y < 0 $ and increasing, minimum value does not exist in $ \mathbb{R} $.", "But wait — perhaps the expression $ y = \dfrac{3t}{4 - t^2} $ is meant to represent a quasi-sphere radius or deviation, and the minimum of $ |y| $ or maximum magnitude is sought.", "Alternatively, suppose we are to find the global minimum of $ y $ on $ t > 2 $, but again, $ y \ o -\infty $, so no finite minimum.", "Conclusion: The function $ y = \dfrac{3t}{4 - t^2} $ for $ t > 2 $ has no minimum value — it decreases without bound near $ t = 2 $ and increases toward 0. The infimum is $ -\infty $. However, if the questionAsked for the minimum of $ -y $, that would be a maximum of $ y $, which does not exist (since $ y $ increases to 0).", "But likely, the intent was to find the minimum of the expression in simplified form or asymptotic behavior — yet mathematically, the function increases from $ -\infty $ to 0.", "Revised interpretation: Perhaps “minimum value” refers to the least upper bound of the limit inferior, but strictly, the function has no minimum.", "However, in applied settings, if modeled as $ y < 0 $, a "minimum" may be misinterpreted — but strictly, no value is the smallest.", "But suppose the question meant: What is the infimum of $ y $ for $ t > 2 $?", "Answer: $ -\infty $", "But that’s not informative.", "Alternatively, suppose the expression was $ y = \dfrac{3t}{t^2 - 4} $, then $ y > 0 $, increases to 0? No — $ t=3 $: $ 9/5=1.8 $, $ t=10 $: $ 10/96\approx 0.104 $, so decreasing to 0 — maximum at $ t=2^+ $, minimum $ 0 $, not attained.", "But our function drops below zero.", "After careful reconsideration, the key insight is:", "> Since $ y(t) $ is strictly increasing on $ (2, \infty) $ with $ \lim_{t \ o 2^+} y = -\infty $ and $ \lim_{t \ o \infty} y = 0^- $, the function has no minimum value — the set $ { y(t) \mid t > 2 } = (-\infty, 0) $, which contains no least element.", "But if the problem seeks the minimum of $ y $ in a compact interval, say $ [2+\delta, M] $, it would be near $ t=2 $. But unbounded below.", "Final Answer:", "[\n\boxed{-\infty}\n]", "While $ y $ does not attain a minimum, it decreases without bound as $ t \ o 2^+ $. In applied optimization, this indicates an unstable or exploding low boundary—critical in modeling constraints where $ t > 2 $ must be bounded away from 2 to avoid divergence.", "---", "### Summary", "- $ y = \dfrac{3t}{4 - t^2} $ is defined for $ t > 2 $\n- $ y < 0 $ due to negative denominator\n- $ y(t) $ is strictly increasing on $ (2, \infty) $\n- $ \lim_{t \ o 2^+} y = -\infty $, $ \lim_{t \ o \infty} y = 0^- $\n- The function has no minimum value; its infimum is $ -\infty $", "This analysis is vital in fields like control theory, thermodynamics, or econometrics where positive deviations dominate, but here the negative output signals a model in quadrature requiring deviation from equilibrium."]