Find the absolute maximum value on \( (0, \infty) \) for \( f(x)=6 x-2 x \ln x \). Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice. A. The absolute maximum is \( \square \) at \( x=\square \). (Round to two decimal places as needed.) B. There is no absolute maximum.

To find the absolute maximum of the function \( f(x) = 6x - 2x \ln x \) on the interval \( (0, \infty) \), we first need to find its critical points. We do this by taking the derivative of \( f(x) \) and setting it to zero. The derivative is: \[ f'(x) = 6 - 2 \ln x - 2 \] Simplifying this gives us: \[ f'(x) = 4 - 2 \ln x \] Setting the derivative equal to zero: \[ 4 - 2 \ln x = 0 \implies 2 \ln x = 4 \implies \ln x = 2 \implies x = e^2 \] Next, we need to evaluate \( f(x) \) at this critical point as well as the behavior as \( x \) approaches 0 and infinity. 1. Evaluate at \( x = e^2 \): \[ f(e^2) = 6(e^2) - 2(e^2) \ln(e^2) = 6e^2 - 2e^2 \cdot 2 = 6e^2 - 4e^2 = 2e^2 \] 2. As \( x \to 0^+ \): \[ f(x) \to 0 \quad \text{(as } 6x \to 0 \text{ and } -2x \ln x \to 0 \text{)} \] 3. As \( x \to \infty \): \[ f(x) \to \infty \quad \text{(since } 6x \text{ dominates )} \] Thus, the function grows without bound as \( x \to \infty \). Since \( f(e^2) = 2e^2 \) is finite and \( f(x) \) increases to infinity as \( x \to \infty \), the function does not attain an absolute maximum on this interval. Therefore, the correct option is: B. There is no absolute maximum.