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Home » You Won’t Believe What ‘Haveard’ Can Do—Watch Now! - Baxtercollege

You Won’t Believe What ‘Haveard’ Can Do—Watch Now! - Baxtercollege

📅 March 11, 2026 👤 scraface

Mar 11, 2026

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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?
Let \(x \to 0^+\), \(\sec x \to 1\), \(\csc x \to \infty\), so sum \(\...

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