find the absolute maximum value on (0, ∞) for f(x)=x^4 - 4x^3.\nselect the correct choice below and, if…

Explanation:

Step1: Find the derivative of the function

Using the power rule ( (x^n)^\prime=nx^{n - 1}), for (f(x)=x^4-4x^2), the derivative (f^\prime(x)=4x^3-8x). Factor out (4x): (f^\prime(x)=4x(x^2 - 2)=4x(x-\sqrt{2})(x + \sqrt{2})).

Step2: Find the critical points

Set (f^\prime(x)=0). Since (x\in(0,\infty)), we consider (4x(x-\sqrt{2})(x + \sqrt{2})=0). The critical point in the interval ((0,\infty)) is (x = \sqrt{2}) (because (x>0), we ignore (x=0) and (x=-\sqrt{2})).

Step3: Use the second - derivative test

Find the second - derivative. (f^{\prime\prime}(x)=12x^2-8). Substitute (x = \sqrt{2}) into (f^{\prime\prime}(x)): (f^{\prime\prime}(\sqrt{2})=12\times(\sqrt{2})^2-8=12\times2 - 8=16>0). So (x=\sqrt{2}) is a local minimum.

Step4: Analyze the behavior of the function as (x\to\infty)

As (x\to\infty), (y = f(x)=x^4-4x^2=x^2(x^2 - 4)\to\infty).

Answer:

B. There is no absolute maximum.