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Home » Discover the Ultimate Ground-In Pool That Transforms Your Backyard Forever! - American Beagle Club

Discover the Ultimate Ground-In Pool That Transforms Your Backyard Forever! - American Beagle Club

📅 February 27, 2026 👤 scraface

Feb 27, 2026

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g'(v) = \frac{4(v - 1)^2 - 4v \cdot 2(v - 1)}{(v - 1)^4} = \frac{4(v -...
For \(v > 1\), \(g'(v) < 0\), so \(g(v)\) is decreasing on \((1, 2]\)....
Wait — since derivative is negative, function decreases from \(v = 1^+...
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 infi...
Thus, \((\sec x + \csc x)^2\) is unbounded on \(0 < x < \frac{\pi}{2}\...
But wait — is that correct?

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