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You Won’t Believe What You’ll Find on readcomiconline.to – Explore Now! - Appfinity Technologies

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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C'(t) = rac{(3)(t^2 + 4) - (3t + 2)(2t)}{(t^2 + 4)^2}
3(t^2 + 4) - 2t(3t + 2) = 3t^2 + 12 - (6t^2 + 4t) = -3t^2 - 4t + 12
C'(t) = rac{-3t^2 - 4t + 12}{(t^2 + 4)^2}
To find where $ C(t) $ is increasing, we solve:
-3t^2 - 4t + 12 > 0
This is a quadratic inequality. First, find the roots:
-3t^2 - 4t + 12 = 0 \Rightarrow 3t^2 + 4t - 12 = 0
t = rac{-4 \pm \sqrt{16 + 144}}{6} = rac{-4 \pm \sqrt{160}}{6} = ra...
So the critical points are:
t_1 = rac{-2 - 2\sqrt{10}}{3}, \quad t_2 = rac{-2 + 2\sqrt{10}}{3}

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