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Find the global maximum value of the function f(x) = log10(4x³ - 12x² + 11x - 3) over the interval x in [2, 3].

1. -(3/2)*log10(3)
2. 1 + log10(3)
3. log10(3)
4. (3/2)*log10(3)

Correct answer: 1 + log10(3)

Solution

g(x) = 4x³ - 12x² + 11x - 3. g'(x) = 12x² - 24x + 11. Setting g'(x) = 0: x = (24 +/- sqrt(576-528))/24 = (24 +/- sqrt(48))/24 = 1 +/- sqrt(48)/24 = 1 +/- (2*sqrt(3))/12... = 1 +/- sqrt(3)/6. sqrt(3)/6 ≈ 0.289. So x ≈ 1.289 or x ≈ 0.711. Both critical points are outside [2,3]. So g is monotone on [2,3]. g'(2) = 12*4 - 24*2 + 11 = 48 - 48 + 11 = 11 > 0. So g is increasing on [2,3]. Maximum at x=3: g(3) = 4*27 - 12*9 + 11*3 - 3 = 108 - 108 + 33 - 3 = 30. Maximum of f = log10(30) = log10(10*3) = 1 + log10(3).

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