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1) Find the largest x such that \(3\sqrt{3x^2+2x+36}+\sqrt{3x^2+2x-36}=40.\)

2) Find the value of x that maximizes \(f(x) = \log_2 (-20x + 12\sqrt{x}).\)

 Jul 21, 2019
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1) x = 2

 

2) The maximum occurs at x = 1/9.

 Nov 27, 2019

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