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Home » Set numerator of \( I''(t) = 0 \): - NBX Soluciones

Set numerator of \( I''(t) = 0 \): - NBX Soluciones

📅 May 3, 2026 👤 scraface

May 03, 2026

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📌 Solution: Recognize it as a difference of cubes: $ (2a)^3 - (3b)^3 = 8a^3 - 27b^3 $. Alternatively, expand directly:

📌 2a \cdot 4a^2 + 2a \cdot 6ab + 2a \cdot 9b^2 - 3b \cdot 4a^2 - 3b \cdot 6ab - 3b \cdot 9b^2 = 8a^3 + 12a^2b + 18ab^2 - 12a^2b - 18ab^2 - 27b^3.

📌 Combine like terms: $ 8a^3 - 27b^3 $. The expanded form is $ oxed{8a^3 - 27b^3} $.

📌 Question: Two biotechnology strains grow according to $ y = 3x + 2 $ and $ y = -2x + 12 $. Find their intersection point.

📌 Solution: Set $ 3x + 2 = -2x + 12 $. Solve: $ 5x = 10 $, so $ x = 2 $. Substitute into $ y = 3(2) + 2 = 8 $. The intersection is $ oxed{(2, 8)} $.Question: Let $ g(x) $ be a polynomial such that $ g(2x + 3) = 4x^2 + 12x + 5 $. Find $ g(x^2 - 1) $.

📌 Solution: Let $ u = 2x + 3 $. Then $ x = rac{u - 3}{2} $. Substitute into $ g(u) = 4\left(rac{u - 3}{2}

📌 ight)^2 + 12\left(rac{u - 3}{2}

📌 ight) + 5 $. Simplify: $ g(u) = (u - 3)^2 + 6(u - 3) + 5 = u^2 - 6u + 9 + 6u - 18 + 5 = u^2 - 4 $. Thus, $ g(x^2 - 1) = (x^2 - 1)^2 - 4 = x^4 - 2x^2 + 1 - 4 = x^4 - 2x^2 - 3 $. Final answer: $oxed{x^4 - 2x^2 - 3}$

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