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📅 May 3, 2026 👤 scraface

May 03, 2026

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📌 Next, we set the derivative equal to zero to find the critical points:

📌 15x^2 - 30x + 10 = 0

📌 Divide the entire equation by 5:

📌 Using the quadratic formula \( x = rac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 3 \), \( b = -6 \), and \( c = 2 \):

📌 x = rac{-(-6) \pm \sqrt{(-6)^2 - 4 \cdot 3 \cdot 2}}{2 \cdot 3} = rac{6 \pm \sqrt{36 - 24}}{6} = rac{6 \pm \sqrt{12}}{6}

📌 Simplify \( \sqrt{12} = 2\sqrt{3} \):

📌 x = rac{6 \pm 2\sqrt{3}}{6} = 1 \pm rac{\sqrt{3}}{3}

📌 Thus, the critical points are:

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