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

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}). [ \begin{align*} f^\prime(x)&=\frac{12x^{3}\cdot e^{x}-3x^{4}\cdot e^{x}}{(e^{x})^{2}}\ &=\frac{3x^{3}e^{x}(4 - x)}{e^{2x}}\ &=\frac{3x^{3}(4 - x)}{e^{x}} \end{align*} ]

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