Question

find the absolute maximum value on (0, ∞) for f(x)=6x - 3x ln x.
select the correct choice below and, if necessary, fill in the answer boxes to complete your choice.
a. the absolute maximum is □ at x=□.
(round to two decimal places as needed.)
b. there is no absolute maximum.

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 \(x = 2.72\).