find the absolute maximum value on (0, ∞) for f(x)=6x - 3x ln x.\nselect the correct choice below and, if…

Explanation:

Step1: Find the first derivative

Use the product rule ((uv)^\prime = u^\prime v+uv^\prime) where (u = - 3x), (v=\ln x), (u^\prime=-3), (v^\prime=\frac{1}{x}) and the derivative of (y = 6x) is (y^\prime=6). The derivative of (f(x)=6x - 3x\ln x) is (f^\prime(x)=6-(3\ln x + 3x\times\frac{1}{x})=6-(3\ln x + 3)=3 - 3\ln x).

Step2: Find the critical points

Set (f^\prime(x)=0), so (3-3\ln x = 0). Divide both sides by (3): (1-\ln x=0), then (\ln x = 1). Using the property (y = \ln x\Leftrightarrow x = e^{y}), we get (x = e\approx2.72).

Step3: Find the second derivative

Differentiate (f^\prime(x)=3 - 3\ln x). The derivative of (3) is (0) and the derivative of (-3\ln x) is (-\frac{3}{x}). So (f^{\prime\prime}(x)=-\frac{3}{x}). When (x = e), (f^{\prime\prime}(e)=-\frac{3}{e}<0), so (x = e) is a local maximum.

Step4: Find the function value at the critical point

Substitute (x = e) into (f(x)): (f(e)=6e-3e\ln e). Since (\ln e = 1), then (f(e)=6e-3e=3e\approx8.15).

Answer:

A. The absolute maximum is (8.15) at (x = 2.72).