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Home » $24 = 2^3 \cdot 3$, - Veritas Home Health

$24 = 2^3 \cdot 3$, - Veritas Home Health

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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📌 For \(v > 1\), \(g'(v) < 0\), so \(g(v)\) is decreasing on \((1, 2]\). Hence, maximum at \(v \to 1^+\)? But let’s check value at endpoint.

📌 Wait — since derivative is negative, function decreases from \(v = 1^+\) to \(v = 2\), so maximum occurs as \(v \to 1^+\), but \(v > 1\), and we want finite maximum.

📌 But observe: as \(v \to 1^+\), g(v) → ∞? Let’s test:

📌 At \(v = 1.1\), \(g = 4(1.1)/(0.1)^2 = 4.4 / 0.01 = 440\)

📌 At \(v = 2\), \(g = 8 / (1)^2 = 8\)

📌 Wait — this suggests decreasing, but 440 > 8, so yes, decreases.

📌 But we seek the maximum value. Since \(g(v)\) decreases from near infinity (as \(v \to 1^+\)) to 8 at \(v = 2\), but \(v = 1\) is not included. However, as \(x \to 0^+\) or \(\frac{\pi}{2}^-\), one of \(\sin x\) or \(\cos x \to 0\), so \(\sec x\) or \(\csc x \to \infty\), so \(f(x) \to \infty\).

📌 Thus, \((\sec x + \csc x)^2\) is unbounded on \(0 < x < \frac{\pi}{2}\), and has no maximum.

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