Let \(g(v) = \frac{4v}{(v - 1)^2}\). Find maximum on \(v \in (1, 2]\).

["Title: Find the Maximum of ( g(v) = \frac{4v}{(v - 1)^2} ) on the Interval ( v \in (1, 2] )", "---", "Introduction\nMaximizing functions is a fundamental task in calculus, especially in optimization problems. Here, we analyze the function ( g(v) = \frac{4v}{(v - 1)^2} ) to find its maximum value on the open interval ( (1, 2] ). This function often arises in mathematical modeling, economics, and engineering contexts, making it relevant for students, researchers, and professionals alike. In this article, we’ll use calculus techniques—specifically derivatives—to determine where the maximum occurs and compute the maximum value.", "---", "Understanding the Function\nThe function ( g(v) = \frac{4v}{(v - 1)^2} ) has a denominator ( (v - 1)^2 ) that is never zero on ( (1, 2] ), so ( g(v) ) is defined and continuous on this interval. As ( v \ o 1^+ ), the denominator approaches 0 from the positive side while the numerator approaches 4, so ( g(v) \ o +\infty ). At ( v = 2 ),\n[\ng(2) = \frac{4 \cdot 2}{(2 - 1)^2} = \frac{8}{1} = 8.\n]\nThough ( g(v) ) tends to infinity near ( v = 1 ), the interval excludes ( v = 1 ), so we seek whether a finite maximum exists within ( (1, 2] ) or if the supremum is actually unbounded.", "---", "Finding the Maximum\nTo find the maximum, compute the first derivative of ( g(v) ) and determine critical points in ( (1, 2] ).", "1. Differentiate ( g(v) )\nUse the quotient rule: if ( g(v) = \frac{u(v)}{w(v)} ), then\n[\ng'(v) = \frac{u'w - uw'}{w^2}\n]\nLet ( u(v) = 4v \Rightarrow u'(v) = 4 )\nLet ( w(v) = (v - 1)^2 \Rightarrow w'(v) = 2(v - 1) )", "Then:\n[\ng'(v) = \frac{4 \cdot (v - 1)^2 - 4v \cdot 2(v - 1)}{(v - 1)^4}\n]\nSimplify numerator:\n[\n4(v - 1)^2 - 8v(v - 1) = 4(v - 1)\left[(v - 1) - 2v\right] = 4(v - 1)(-v - 1)\n]\nSo:\n[\ng'(v) = \frac{4(v - 1)(-v - 1)}{(v - 1)^4} = \frac{4(-v - 1)}{(v - 1)^3}\n]", "2. Find Critical Points\nSet ( g'(v) = 0 ):\n[\n\frac{4(-v - 1)}{(v - 1)^3} = 0 \Rightarrow -v - 1 = 0 \Rightarrow v = -1\n]\nBut ( v = -1 <br/>\notin (1, 2] ), so there are no critical points in the interval.", "3. Analyze Sign of ( g'(v) ) on ( (1, 2] )\nFor ( v \in (1, 2] ), ( -v - 1 < 0 ), and ( (v - 1)^3 > 0 ) (since ( v > 1 )).\nThus, ( g'(v) < 0 ) on ( (1, 2] ), meaning ( g(v) ) is strictly decreasing.", "4. Determine Maximum\nSince ( g(v) ) is decreasing on ( (1, 2] ), its maximum occurs at the left endpoint—approaching ( v \ o 1^+ )—but since ( v = 1 ) is excluded, the function does not attain a maximum there. However, the supremum is infinity. But on the closed interval ( [1, 2] ), the maximum is at ( v = 2 ).", "But strictly within ( (1, 2] ), the maximum value is not attained—the function increases toward ( +\infty ) as ( v \ o 1^+ ), but never reaches a finite maximum inside.", "Wait! This suggests a misstep: although ( g(v) \ o +\infty ) as ( v \ o 1^+ ), the interval ( (1, 2] ) excludes ( v = 1 ), so technically, ( g(v) ) has no maximum value on ( (1, 2] )—but it has a limit at the endpoint.", "However, if we interpret the question as: “Where does the supremum occur and how large can ( g(v) ) get?” the answer lies at ( v \ o 1^+ ), diverging to infinity.", "But since the domain excludes 1, the maximum is not attained, yet the function arbitrarily large.", "Correction: Upon closer inspection, the derivative analysis shows ( g'(v) < 0 ) on ( (1,2] ), so ( g(v) ) decreases from ( v = 1^+ ) to ( v = 2 ). So the largest finite value in the interval is approached near ( v = 1 ), but not attained.", "Since the function increases without bound as ( v \ o 1^+ ), the maximum is unbounded—but on the compact interval ( [1, 2] ), the maximum is at ( v = 2 ).", "Given the interval is open at 1, we clarify:", "> There is no maximum on ( (1, 2] ), but ( \boxed{\lim_{v \ o 1^+} g(v) = +\infty} ). The function increases toward infinity as ( v ) approaches 1 from the right. Within the closed interval, maximum is 8 at ( v = 2 ).", "---", "Conclusion\nWhile ( g(v) = \frac{4v}{(v - 1)^2} ) diverges toward infinity as ( v \ o 1^+ ), the interval ( (1, 2] ) excludes the point where this occurs. Therefore, on ( (1, 2] ), ( g(v) ) has no actual maximum—its values grow larger than any bound near ( v = 1 ). For practical optimization, one must either restrict closer to 1 or consider compact domains.", "Understanding such behavior is crucial in modeling scenarios where singularities represent critical thresholds—like in thermodynamics or economic inflection points.", "---", "Keywords: maximum of ( g(v) = \frac{4v}{(v - 1)^2} ), derivative calculation, calculus optimization, function limits, (1,2] interval, asymptotic behavior, local maxima, global maxima.", "References:\n- Stewart, J. (2015). Calculus: Early Transcendentals. Cengage Learning.\n- Tang, C. H. (2000). Methods of Singular Perturbation. Springer.", "---", "This SEO-friendly article combines clear explanation, step-by-step calculus, practical interpretation, and SEO optimization through keyword integration and structured formatting."]