Question

summarize the pertinent information obtained by applying the graphing strategy and sketch the graph of ( f(x)=9e^{-0.3x^{2}} ).

b. the function is increasing on ( (-infty,0) ). it is decreasing on ( (0,infty) ).
(type your answers in interval notation. type integers or decimals. use a comma to separate answers as needed.)
c. the function is decreasing on. it is never increasing.
(type your answer in interval notation. type integers or decimals. use a comma to separate answers as needed.)
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= ) and there is a local minimum at ( x= ).
(type integers or decimals. use a comma to separate answers as needed.)
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= ). 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.

Explanation:

Step1: Find the first derivative

Given \(y = 9e^{-0.3x^{2}}\), use the chain - rule. If \(y = 9e^{u}\) and \(u=-0.3x^{2}\), then \(\frac{dy}{du}=9e^{u}\) and \(\frac{du}{dx}=- 0.6x\). By the chain - rule \(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=9e^{-0.3x^{2}}\cdot(-0.6x)=-5.4xe^{-0.3x^{2}}\).

Step2: Determine where the function is increasing or decreasing

Set \(\frac{dy}{dx}>0\) to find where the function is increasing.
\(-5.4xe^{-0.3x^{2}}>0\). Since \(e^{-0.3x^{2}}>0\) for all \(x\in R\), then \(-5.4x>0\), which gives \(x < 0\).
Set \(\frac{dy}{dx}<0\) to find where the function is decreasing.
\(-5.4xe^{-0.3x^{2}}<0\). Since \(e^{-0.3x^{2}}>0\) for all \(x\in R\), then \(-5.4x<0\), which gives \(x>0\).

Step3: Find the local extrema

Set \(\frac{dy}{dx} = 0\), so \(-5.4xe^{-0.3x^{2}}=0\). Since \(e^{-0.3x^{2}}
eq0\) for all \(x\in R\), then \(x = 0\).
Find the second derivative using the product - rule. If \(y=-5.4xe^{-0.3x^{2}}\), let \(u=-5.4x\) and \(v = e^{-0.3x^{2}}\). Then \(u^\prime=-5.4\) and \(v^\prime=-0.6xe^{-0.3x^{2}}\).
By the product - rule \(y^\prime=u^\prime v+uv^\prime=-5.4e^{-0.3x^{2}}+(-5.4x)(-0.6xe^{-0.3x^{2}})=e^{-0.3x^{2}}(-5.4 + 3.24x^{2})\).
Evaluate the second derivative at \(x = 0\): \(y^{\prime\prime}(0)=e^{0}(-5.4+0)=-5.4<0\).

Answer:

For the increasing and decreasing intervals: The function is increasing on \((-\infty,0)\) and decreasing on \((0,\infty)\).
For the local extrema: There is a local maximum at \(x = 0\). So the answer for the local - extrema part is \(C\).