Question

find the absolute maximum value on $(0,\infty)$ for $f(x)=\frac{3x^{4}}{e^{x}}$.
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.

Explanation:

Step1: Find the derivative of \(f(x)\)

Use the quotient rule \((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\), where \(u = 3x^{4}\), \(u^\prime=12x^{3}\), \(v = e^{x}\), \(v^\prime=e^{x}\).

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Step2: Find the critical points

Set \(f^\prime(x)=0\). Since \(e^{x}>0\) for all \(x\in(0,\infty)\), we solve \(3x^{3}(4 - x)=0\).
The solutions of \(3x^{3}(4 - x)=0\) in the interval \((0,\infty)\) are \(x = 4\) (because \(x = 0\) is not in the open - interval \((0,\infty)\)).

Step3: Analyze the sign of \(f^\prime(x)\)

• When \(0
• When \(x>4\), \(f^\prime(x)<0\). So \(f(x)\) is decreasing on the interval \((4,\infty)\).

Step4: Evaluate \(f(x)\) at the critical point

Substitute \(x = 4\) into \(f(x)\):

$$ f(4)=\frac{3\times4^{4}}{e^{4}}=\frac{3\times256}{e^{4}}\approx\frac{768}{54.5982}\approx14.06 $$

Step5: Analyze the behavior as \(x\to0^{+}\) and \(x\to\infty\)

• As \(x\to0^{+}\), \(f(x)=\frac{3x^{4}}{e^{x}}\to0\) (since \(x^{4}\to0\) and \(e^{x}\to1\)).
• As \(x\to\infty\), use L'Hopital's rule (repeatedly). \(\lim_{x\to\infty}\frac{3x^{4}}{e^{x}}\), applying L'Hopital's rule 4 times:

\(\lim_{x\to\infty}\frac{3x^{4}}{e^{x}}=\lim_{x\to\infty}\frac{12x^{3}}{e^{x}}=\lim_{x\to\infty}\frac{36x^{2}}{e^{x}}=\lim_{x\to\infty}\frac{72x}{e^{x}}=\lim_{x\to\infty}\frac{72}{e^{x}} = 0\)

Answer:

A. The absolute maximum is \(14.06\) at \(x = 4.00\)