Hint
For part (b), set the first derivative equal to zero to find critical points, and then use the second derivative to determine if it’s a minimum.
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- Formula Booklet
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Complete Answer
(a) \(C'(x) = 3ax^2 – 2bx + c\)
(b) Set \(C'(x) = 0\) and solve for \(x\). Calculate \(C”(x) = 6ax – 2b\). If for a critical point \(x_0\), \(C”(x_0) > 0\), then \(x_0\) is a local minimum.
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