Question

find the absolute maximum and minimum values of the following function on the given interval. identify the points on the graph where the absolute extrema occur.
g(x)=3x^{3}e^{-x}, - 1≤x≤4
find the absolute maximum. select the correct choice below and, if necessary, fill in the answer box choice.
a. the absolute maximum value occurs at x =
(use a comma to separate answers as needed. round to three decimal places as needed.)
b. there is no absolute maximum.

Explanation:

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

Use the product - rule \((uv)^\prime = u^\prime v+uv^\prime\), where \(u = 3x^{3}\) and \(v = e^{-x}\). \(u^\prime=9x^{2}\) and \(v^\prime=-e^{-x}\). So \(g^\prime(x)=9x^{2}e^{-x}-3x^{3}e^{-x}=3x^{2}e^{-x}(3 - x)\).

Step2: Find the critical points

Set \(g^\prime(x) = 0\). Since \(3x^{2}e^{-x}(3 - x)=0\), and \(e^{-x}\gt0\) for all real \(x\), then \(3x^{2}(3 - x)=0\). Solving gives \(x = 0\) and \(x = 3\). Both \(x = 0\) and \(x = 3\) are in the interval \([-1,4]\).

Step3: Evaluate the function at critical points and endpoints

Evaluate \(g(x)\) at \(x=-1\), \(x = 0\), \(x = 3\), and \(x = 4\).

• When \(x=-1\), \(g(-1)=3(-1)^{3}e^{-(-1)}=-3e\approx - 8.155\).
• When \(x = 0\), \(g(0)=3(0)^{3}e^{-0}=0\).
• When \(x = 3\), \(g(3)=3(3)^{3}e^{-3}=81e^{-3}\approx3.347\).
• When \(x = 4\), \(g(4)=3(4)^{3}e^{-4}=192e^{-4}\approx3.547\).

Answer:

A. The absolute maximum value \(3.547\) occurs at \(x = 4\).