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};-1,5
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 minimum is at x =, but there is no absolute maximum.
b. the absolute maximum is at x = and the absolute minimum is at x =.
c. the absolute maximum is at x =, but there is no absolute minimum.
d. there are no absolute extreme values for f(x) on -1,5.

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 \(5x^{2}e^{-x}(3 - x)=0\) and \(e^{-x}>0\) for all \(x\), then \(x = 0\) or \(x=3\) (because \(5x^{2}=0\) gives \(x = 0\) and \(3 - x=0\) gives \(x = 3\))

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

• For \(x=-1\): \(f(-1)=5\times(-1)^{3}\times e^{-(-1)}=-5e\approx - 13.59\)
• For \(x = 0\): \(f(0)=5\times0^{3}\times e^{-0}=0\)
• For \(x = 3\): \(f(3)=5\times3^{3}\times e^{-3}=\frac{135}{e^{3}}\approx 6.79\)
• For \(x = 5\): \(f(5)=5\times5^{3}\times e^{-5}=\frac{625}{e^{5}}\approx 4.19\)

Answer:

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