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Home » \( v = t^2 + 10 \), so \( v' = 2t \). - SUpost

\( v = t^2 + 10 \), so \( v' = 2t \). - SUpost

📅 July 6, 2026 👤 scraface

Jul 06, 2026

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📌 We now find the maximum of $f(u)$ on this interval. Note that as $u \to 0^+$, $f(u) \to \infty$, but we must check whether a maximum occurs within the interval or at an endpoint. However, since all terms grow unbounded as $u \to 0$, but we must evaluate whether a minimum exists and whether the expression becomes large — yet we seek the **maximum**, which may occur near $u \to 0$, but let’s analyze behavior.

📌 But wait: as $u \to 0^+$, $\frac{1}{u^2} \to \infty$, so $f(u) \to \infty$. However, is this possible?

📌 Let’s test at $x = \frac{\pi}{4}$: $\sin x = \cos x = \frac{\sqrt{2}}{2}$, so $u = \frac{1}{2} \cdot \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2} = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{2}$

📌 So $u = \frac{1}{2}$ at $x = \frac{\pi}{4}$, and $f\left(\frac{1}{2}\right) = \frac{1}{(1/2)^2} + \frac{2}{1/2} + 1 + 2\cdot\frac{1}{2} = 4 + 4 + 1 + 1 = 10$

📌 Now check the behavior as $u \to 0^+$:

📌 \frac{1}{u^2} \to \infty, \quad \frac{2}{u} \to \infty, \quad 2u \to 0

📌 So $f(u) \to \infty$

📌 But this contradicts intuition — is the expression unbounded?

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