Question

find the absolute maximum value on (0, ∞) for f(x) = 5x^6 / e^x. 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 \((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\). Here \(u = 5x^{6}\), \(u^\prime=30x^{5}\), \(v = e^{x}\), \(v^\prime=e^{x}\). Then \(f^\prime(x)=\frac{30x^{5}e^{x}-5x^{6}e^{x}}{(e^{x})^{2}}=\frac{5x^{5}e^{x}(6 - x)}{e^{2x}}=\frac{5x^{5}(6 - x)}{e^{x}}\)

Step2: Find the critical points

Set \(f^\prime(x)=0\). Since \(e^{x}>0\) for all \(x\), we solve \(5x^{5}(6 - x)=0\). The solutions are \(x = 0\) and \(x = 6\). But \(x\in(0,\infty)\), so we consider \(x = 6\)

Step3: Use the second - derivative test or analyze the sign of \(f^\prime(x)\)

For \(x\in(0,6)\), let's take a test point \(x = 1\). Then \(f^\prime(1)=\frac{5\times1^{5}(6 - 1)}{e^{1}}=\frac{25}{e}>0\). For \(x\in(6,\infty)\), let's take a test point \(x = 7\). Then \(f^\prime(7)=\frac{5\times7^{5}(6 - 7)}{e^{7}}=\frac{-5\times7^{5}}{e^{7}}<0\)

Step4: Find the function value at \(x = 6\)

\(f(6)=\frac{5\times6^{6}}{e^{6}}\approx\frac{5\times46656}{403.4288}\approx578.71\)

Answer:

A. The absolute maximum is \(578.71\) at \(x = 6\)