find the absolute maximum value on (0, ∞) for f(x) = 2x⁴ / eˣ. select the correct choice below and, if…

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 = 2x^{4}), (u^\prime=8x^{3}), (v = e^{x}), (v^\prime=e^{x}). [ \begin{align*} f^\prime(x)&=\frac{8x^{3}e^{x}-2x^{4}e^{x}}{(e^{x})^{2}}\ &=\frac{2x^{3}e^{x}(4 - x)}{e^{2x}}\ &=\frac{2x^{3}(4 - x)}{e^{x}} \end{align*} ]

Step2: Find the critical points

Set (f^\prime(x)=0). Since (e^{x}>0) for all (x), we set (2x^{3}(4 - x)=0). Solving (2x^{3}(4 - x)=0) gives (x = 0) or (x = 4). But (x\in(0,\infty)), so the critical point is (x = 4).

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

• For (0<x<4), (f^\prime(x)>0), so (f(x)) is increasing on ((0,4)).
• For (x>4), (f^\prime(x)<0), so (f(x)) is decreasing on ((4,\infty)).

Step4: Find the absolute maximum value

Evaluate (f(x)) at (x = 4): [ f(4)=\frac{2\times4^{4}}{e^{4}}=\frac{2\times256}{e^{4}}=\frac{512}{e^{4}}\approx\frac{512}{54.5982}\approx9.38 ]

Answer:

A. The absolute maximum is (9.38) at (x = 4.00)