The global maximum value of f(x) = log_{10}(4 | vedclass

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(B) Let $g(x) = 4x^3 - 12x^2 + 11x - 3$.Then $g'(x) = 12x^2 - 24x + 11$.We can write $g'(x) = 12(x^2 - 2x) + 11 = 12(x^2 - 2x + 1 - 1) + 11 = 12(x - 1

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$1 + \log_{10}3$

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(B) Let $g(x) = 4x^3 - 12x^2 + 11x - 3$.Then $g'(x) = 12x^2 - 24x + 11$.We can write $g'(x) = 12(x^2 - 2x) + 11 = 12(x^2 - 2x + 1 - 1) + 11 = 12(x - 1)^2 - 1$.For $x \in [2, 3]$,$(x - 1) \in [1, 2]$,so $(x - 1)^2 \in [1, 4]$.Thus,$g'(x) \geq 12(1) - 1 = 11 > 0$ for all $x \in [2, 3]$.Since $g'(x) > 0$,$g(x)$ is a strictly increasing function on $[2, 3]$.Consequently,$f(x) = \log_{10}(g(x))$ is also strictly increasing on $[2, 3]$.The global maximum value occurs at the right endpoint $x = 3$.$f(3) = \log_{10}(4(3)^3 - 12(3)^2 + 11(3) - 3) = \log_{10}(4(27) - 12(9) + 33 - 3) = \log_{10}(108 - 108 + 30) = \log_{10}(30)$.$f(3) = \log_{10}(10 	imes 3) = \log_{10}10 + \log_{10}3 = 1 + \log_{10}3$.

The global maximum value of $f(x) = \log_{10}(4x^3 - 12x^2 + 11x - 3)$,$x \in [2, 3]$,is

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