Question

find the absolute maximum value on (0, ∞) for f(x) = 2x⁴ / eˣ. select the correct choice below and, if necessary, fill in the answer boxes to complete your choice. a. the absolute maximum is at x = (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 \(u = 2x^{4}\), \(u^\prime=8x^{3}\), \(v = e^{x}\), \(v^\prime=e^{x}\).

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Step2: Find the critical points

Set \(f^\prime(x)=0\). Since \(e^{x}>0\) for all \(x\), 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
• 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 \(x = 4.00\)