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Let \(x\) be a real number such that \(x^2 + 7x + 12 \le 0.\) Find the largest possible value of \(x^2 + 5x + 6.\)

 May 20, 2019
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\(x^2+7x+12 \leq 0\\ (x+3)(x+4)\leq 0\\ x \in [-4,-3]\)

 

\(x^2 + 5x + 6 = \left(x - \dfrac 5 2\right)^2 -\dfrac 1 4\\~\\ \text{The interval of interest is entirely to the left of the vertex of this parabola}\\ \text{and thus the value is increasing as we head towards -4}\\ \text{So the function evaluated at the endpoint, -4, will provide the maximum value}\\ \left . x^2+5x+6 \right|_{x=-4} = 16-20+6 = 2\)

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 May 20, 2019

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