Question

summarize the pertinent information obtained by applying the graphing strategy and sketch the graph of ( f(x)=9 e^{-0.5 x^{2}} )

b. there is a local minimum at ( x= ). there is no local maximum.
(type an integer or a decimal. use a comma to separate answers as needed.)
c. there is a local maximum at ( x=0 ). there is no local minimum.
(type an integer or a decimal. use a comma to separate answers as needed.)
d. there are no local extrema.
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.
a. the function is concave upward on. it is never concave downward.
(type your answer in interval notation. type integers or decimals. use a comma to separate answers as needed.)
b. the function is concave upward on. it is concave downward on.
(type your answers in interval notation. type integers or decimals. use a comma to separate answers as needed.)
c. the function is concave downward on. it is never concave upward.
(type your answer in interval notation. type integers or decimals. use a comma to separate answers as needed.)

Explanation:

Step1: Find the first - derivative

Use the chain rule. If \(y = 9e^{-0.5x^{2}}\), let \(u=-0.5x^{2}\), then \(y = 9e^{u}\).
The derivative of \(y\) with respect to \(u\) is \(y_{u}=9e^{u}\), and the derivative of \(u\) with respect to \(x\) is \(u_{x}=-x\).
By the chain rule \(y_{x}=y_{u}\cdot u_{x}\), so \(f^{\prime}(x)=9e^{-0.5x^{2}}\cdot(-x)=-9xe^{-0.5x^{2}}\).
Set \(f^{\prime}(x) = 0\), then \(-9xe^{-0.5x^{2}}=0\). Since \(e^{-0.5x^{2}}>0\) for all \(x\in R\), the critical point is \(x = 0\).
Take the second - derivative:
\(f^{\prime}(x)=-9xe^{-0.5x^{2}}\). Using the product rule \((uv)^\prime=u^\prime v + uv^\prime\) where \(u=-9x\) and \(v = e^{-0.5x^{2}}\).
\(u^\prime=-9\) and \(v^\prime=-9xe^{-0.5x^{2}}\) (from the first - derivative calculation).
\(f^{\prime\prime}(x)=-9e^{-0.5x^{2}}+(-9x)(-x)e^{-0.5x^{2}}=9e^{-0.5x^{2}}(x^{2}-1)\).

Step2: Analyze concavity

Set \(f^{\prime\prime}(x)=0\), then \(9e^{-0.5x^{2}}(x^{2}-1)=0\). Since \(e^{-0.5x^{2}}>0\) for all \(x\in R\), we solve \(x^{2}-1 = 0\), so \(x=\pm1\).

• When \(x\in(-\infty,-1)\), let \(x=-2\), then \(f^{\prime\prime}(-2)=9e^{-0.5\times(-2)^{2}}((-2)^{2}-1)=9e^{-2}\times3>0\).
• When \(x\in(-1,1)\), let \(x = 0\), then \(f^{\prime\prime}(0)=9e^{0}(0 - 1)=-9<0\).
• When \(x\in(1,\infty)\), let \(x = 2\), then \(f^{\prime\prime}(2)=9e^{-0.5\times2^{2}}(2^{2}-1)=9e^{-2}\times3>0\).

Answer:

B. The function is concave upward on \((-\infty,-1)\cup(1,\infty)\). It is concave downward on \((-1,1)\).