Question

summarize the pertinent information obtained by applying the graphing strategy and sketch the graph of ( f(x)=9 e^{-0.5 x^{2}} ).
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 ( (-infty,-1),(1, infty) ). it is concave downward on ( (-1,1) ).
(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.)
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 chain rule. If \(y = 9e^{-0.5x^{2}}\), let \(u=-0.5x^{2}\), then \(y = 9e^{u}\).
\(y^\prime=\frac{dy}{du}\cdot\frac{du}{dx}\)
\(\frac{dy}{du}=9e^{u}\), \(\frac{du}{dx}=-x\)
\(y^\prime=-9xe^{-0.5x^{2}}\)

Step2: Find the second derivative

Use the product rule \((uv)^\prime = u^\prime v+uv^\prime\), where \(u=-9x\), \(v = e^{-0.5x^{2}}\)
\(u^\prime=-9\), \(v^\prime=-9xe^{-0.5x^{2}}\cdot(-x)= - 9xe^{-0.5x^{2}}\) (using chain - rule again for \(v = e^{-0.5x^{2}}\))
\(y^{\prime\prime}=-9e^{-0.5x^{2}}+(-9x)\cdot(-x)e^{-0.5x^{2}}=9e^{-0.5x^{2}}(x^{2}-1)\)

Step3: Find concavity intervals

Set \(y^{\prime\prime}=0\), \(9e^{-0.5x^{2}}(x^{2}-1)=0\)
Since \(e^{-0.5x^{2}}>0\) for all \(x\in R\), then \(x^{2}-1 = 0\), \(x=\pm1\)
Test intervals:

• For \(x\in(-\infty,-1)\), let \(x=-2\), \(y^{\prime\prime}=9e^{-2}(4 - 1)>0\), so concave upward on \((-\infty,-1)\)
• For \(x\in(-1,1)\), let \(x = 0\), \(y^{\prime\prime}=9e^{0}(0 - 1)<0\), so concave downward on \((-1,1)\)
• For \(x\in(1,\infty)\), let \(x = 2\), \(y^{\prime\prime}=9e^{-2}(4 - 1)>0\), so concave upward on \((1,\infty)\)

Step4: Find inflection points

Inflection points occur where \(y^{\prime\prime}\) changes sign.
Since \(y^{\prime\prime}=0\) at \(x=-1\) and \(x = 1\) and \(y^{\prime\prime}\) changes sign at these points.

Answer:

For concavity: B. The function is concave upward on \((-\infty,-1),(1,\infty)\). It is concave downward on \((-1,1)\)
For inflection points: A. There is an inflection point at \(x=-1,1\)