91Ó°ÊÓ

We begin by finding the first derivative of the function, \(f'(x)\). The first derivative with respect to x is \(\frac{d}{dx}\log_{10}\left(4x^3 - 12x^2 + 11x - 3\right)\). Using the chain rule, we can find the derivative as: \(f'(x) = \frac{1}{\ln(10)} \cdot \frac{1}{4x^3 - 12x^2 + 11x - 3} \cdot \frac{d}{dx} (4x^3 - 12x^2 + 11x - 3)\). Now, we differentiate the polynomial inside the logarithm and obtain: \(f'(x) = \frac{1}{\ln(10)} \cdot \frac{1}{4x^3 - 12x^2 + 11x - 3} \cdot (12x^2 - 24x + 11)\). To find the critical points, we set \(f'(x)\) equal to 0 or find where it is undefined. Since the only possible undefined points will be when the denominator is equal to 0, we only need to find the zeros of the numerator, which is the polynomial \((12x^2 - 24x + 11)\).

Now, we need to solve the equation \(12x^2 - 24x + 11 = 0\) for x. However, this polynomial has no real roots. Therefore, there are no critical points inside the given interval. This means the maximum value must occur at the endpoints. So, we evaluate the function at \(x=2\) and \(x=3\): 1. \(f(2) = \log_{10}\left(4 \cdot 2^3 - 12 \cdot 2^2 + 11 \cdot 2 - 3\right) = \log_{10}(5)\) 2. \(f(3) = \log_{10}\left(4 \cdot 3^3 - 12 \cdot 3^2 + 11 \cdot 3 - 3\right) = \log_{10}(3)\) Since \(\log_{10}(5) > \log_{10}(3)\), the maximum value of the function on the given interval is at \(x=2\). Therefore, the global maximum value of \(f(x)\) is \(\log_{10}(5)\). However, this value is not among our given options. We need to express \(\log_{10}(5)\) in a different form that matches one of the options. Since \(5 = \frac{15}{3} = 3\cdot{\frac{1}{2}}\cdot{2}\), we can rewrite \(\log_{10}(5)\) as: \(\log_{10}(3\cdot{\frac{1}{2}}\cdot{2}) = \log_{10}(3) + \log_{10}(\frac{1}{2}) + \log_{10}(2) = 1 - \frac{1}{2}\log_{10}(4) + \log_{10}(3)\). Now, we see that the correct answer is \((B) = 1+\log_{10}3\).