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 concave upward or downward. 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 concave upward on ( square ). it is never concave downward.
b. the function is concave upward on ( square ). it is concave downward on ( square ).
c. the function is concave downward on ( square ). it is never concave upward.
find the location of any inflection points of ( f(x) ). select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. there is an inflection point at ( x= ) (type an integer or a decimal. use a comma to separate answers as needed.)
b. there are no inflection points.

Explanation:

Step1: Find the first derivative

Use the product rule \((uv)^\prime = u^\prime v+uv^\prime\), where \(u = 7x\) and \(v=e^{-0.5x}\).
\(u^\prime=7\), \(v^\prime=- 0.5e^{-0.5x}\)
\(f^\prime(x)=7e^{-0.5x}+7x(-0.5)e^{-0.5x}=(7 - 3.5x)e^{-0.5x}\)

Step2: Find the second derivative

Again use the product rule. Let \(u = 7 - 3.5x\) and \(v = e^{-0.5x}\)
\(u^\prime=-3.5\), \(v^\prime=-0.5e^{-0.5x}\)
\(f^{\prime\prime}(x)=-3.5e^{-0.5x}+(7 - 3.5x)(-0.5)e^{-0.5x}\)
\(=(-3.5-3.5 + 1.75x)e^{-0.5x}=(1.75x - 7)e^{-0.5x}\)

Step3: Find where \(f^{\prime\prime}(x)=0\)

Set \(f^{\prime\prime}(x) = 0\), since \(e^{-0.5x}>0\) for all \(x\), we solve \(1.75x-7 = 0\)
\(1.75x=7\), \(x = 4\)

Step4: Test intervals

• For \(x<4\), let \(x = 0\), then \(f^{\prime\prime}(0)=(1.75\times0 - 7)e^{0}=-7<0\)
• For \(x > 4\), let \(x=5\), then \(f^{\prime\prime}(5)=(1.75\times5 - 7)e^{-2.5}=(8.75 - 7)e^{-2.5}=1.75e^{-2.5}>0\)

Answer:

• For concavity: B. The function is concave upward on \((4,\infty)\). It is concave downward on \((-\infty,4)\)
• For inflection point: A. There is an inflection point at \(x = 4\)