Question

determine the location and value of the absolute extreme values of f on the given interval, if they exist.
$f(x)=5x^{3}e^{-x};-1,6$
select the correct choice below and, if necessary, fill in the answer boxes to complete your choice.
(type exact answers. use a comma to separate answers as needed.)
a. the absolute maximum is at ( x = ) and the absolute minimum is at ( x = ).
b. the absolute minimum is at ( x = ), but there is no absolute maximum.
c. the absolute maximum is at ( x = ), but there is no absolute minimum.
d. there are no absolute extreme values for ( f(x) ) on ( -1,6 ).

Explanation:

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

Use the product rule \((uv)^\prime = u^\prime v+uv^\prime\), where \(u = 5x^{3}\) and \(v=e^{-x}\).
\(u^\prime=15x^{2}\), \(v^\prime=-e^{-x}\)
\(f^\prime(x)=15x^{2}e^{-x}-5x^{3}e^{-x}=5x^{2}e^{-x}(3 - x)\)

Step2: Find the critical points

Set \(f^\prime(x) = 0\).
Since \(5x^{2}e^{-x}\gt0\) for \(x\in[-1,6]\) (except \(x = 0\), and \(f^\prime(0)=0\)), then \(3 - x=0\) gives \(x = 3\). Also, \(x = 0\) is a critical point (where \(f^\prime(x)=0\) due to \(x^{2}\) factor).

Step3: Evaluate \(f(x)\) at critical points and endpoints

• For \(x=-1\): \(f(-1)=5(-1)^{3}e^{-(-1)}=-5e\approx - 13.59\)
• For \(x = 0\): \(f(0)=5(0)^{3}e^{-0}=0\)
• For \(x = 3\): \(f(3)=5(3)^{3}e^{-3}=\frac{135}{e^{3}}\approx5.49\)
• For \(x = 6\): \(f(6)=5(6)^{3}e^{-6}=\frac{1080}{e^{6}}\approx2.19\)

Answer:

A. The absolute maximum is \(\frac{135}{e^{3}}\) at \(x = 3\) and the absolute minimum is \(-5e\) at \(x=-1\)