Question

find the absolute maximum value on (0, ∞) for f(x)=x^4 - 4x^3.
select the correct choice below and, if necessary, fill in the answer boxes to complete your choice.
a. the absolute maximum is □ at x = □.
b. there is no absolute maximum.

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.