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Home » So the critical points are: - DevRocket

So the critical points are: - DevRocket

📅 May 3, 2026 👤 scraface

May 03, 2026

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📌 Substitute $ x^2 - 1 $:

📌 g(x^2 - 1) = \dfrac{5}{3}(x^2 - 1)^3 - 6(x^2 - 1)^2 + \dfrac{43}{3}(x^2 - 1) - 15.

📌 \boxed{\dfrac{5}{3}x^6 - 5x^4 + \dfrac{50}{3}x^2 - 18}.

📌 $$Question: A particle physicist models the trajectory of a neutrino through a detector as the line $ y = \frac{3}{4}x + 2 $. Find the point on this trajectory that is closest to the detector's origin $ (0, 0) $.

📌 Solution: To find the point on the line $ y = \frac{3}{4}x + 2 $ closest to $ (0, 0) $, we minimize the square of the distance function:

📌 D(x) = x^2 + \left( \frac{3}{4}x + 2 \right)^2

📌 D(x) = x^2 + \frac{9}{16}x^2 + 3x + 4 = \frac{25}{16}x^2 + 3x + 4

📌 D'(x) = \frac{50}{16}x + 3 = \frac{25}{8}x + 3 = 0 \Rightarrow x = -\frac{24}{25}

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