Question

summarize the pertinent information obtained by applying the graphing strategy and sketch the graph of ( f(x)=7 x e^{-0.5 x} ).
find the intervals where ( f(x) ) is increasing or decreasing. select the correct choice below and fill in the answer box(es) to complete your choice.(type your answer in interval notation. use integers or decimals for any numbers in the expression. use a comma to separate answers as needed.)
a. the function is increasing on ( square ). it is never decreasing.
b. the function is increasing on ( square ). it is decreasing on ( square ).
c. the function is decreasing on ( square ). it is never increasing.
find the location of any local extrema of ( f(x) ). select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice.
a. there is a local maximum at ( x=square ). there is no local minimum.(type an integer or a decimal. use a comma to separate answers as needed.)
b. there is a local maximum at ( x=square ) and there is a local minimum at ( x=square ).(type integers or decimals. use a comma to separate answers as needed.)
c. there is a local minimum at ( x=square ). there is no local maximum.(type an integer or a decimal. use a comma to separate answers as needed.)
d. there are no local extrema.

Explanation:

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

Use the product rule \((uv)^\prime = u^\prime v+uv^\prime\), where \(u = 7x\) and \(v=e^{-0.5x}\).
\(u^\prime=7\) and \(v^\prime=- 0.5e^{-0.5x}\).
So \(f^\prime(x)=7e^{-0.5x}+7x(-0.5)e^{-0.5x}=7e^{-0.5x}(1 - 0.5x)\)

Step2: Find the critical points

Set \(f^\prime(x) = 0\).
Since \(e^{-0.5x}>0\) for all \(x\), we solve \(1-0.5x = 0\).
\(0.5x=1\), so \(x = 2\)

Step3: Determine the intervals of increase and decrease

Test intervals:

• For \(x<2\), let \(x = 0\). Then \(f^\prime(0)=7e^{0}(1-0)=7>0\), so \(f(x)\) is increasing on \((-\infty,2)\)
• For \(x>2\), let \(x = 3\). Then \(f^\prime(3)=7e^{-1.5}(1 - 1.5)=7e^{-1.5}(-0.5)<0\), so \(f(x)\) is decreasing on \((2,\infty)\)

Step4: Find local extrema

Since \(f(x)\) changes from increasing to decreasing at \(x = 2\), by the first - derivative test, there is a local maximum at \(x = 2\)

Answer:

For the increasing/decreasing part: B. The function is increasing on \((-\infty,2)\). It is decreasing on \((2,\infty)\)
For the local extrema part: A. There is a local maximum at \(x = 2\). There is no local minimum.