$$ f(x)=e^{-x^{2}} $$\n(a) find the vertical asymptote(s). (enter your answers as a comma-separated…

$$ f(x)=e^{-x^{2}} $$\n(a) find the vertical asymptote(s). (enter your answers as a comma-separated list.)\n$$ x= $$\nfind the horizontal asymptote(s). (enter your answers as a comma-separated list.)\n$$ y= $$\n(b) find the interval(s) of increase. (enter your answer using interval notation.)\n$$ $$\nfind the interval(s) of decrease. (enter your answer using interval notation.)\n$$ $$\n(c) find the local maximum and minimum values.\nlocal maximum value\n$$ $$\nlocal minimum value\n$$ $$\n(d) find the interval(s) on which ( f ) is concave up. (enter your answer using interval notation.)\n$$ $$\nfind the interval(s) on which ( f ) is concave down. (enter your answer using interval notation.)\n$$ $$\nfind the inflection points.\nsmaller ( x )-value ( (x, y)=(quad) )\nlarger ( x )-value ( (x, y)=(quad) )

Answer

Explanation:

Step1: Analyze vertical asymptote

A vertical asymptote occurs where the function is undefined. For (y = e^{-x^{2}}), the exponential function (e^{u}) is defined for all real (u). So, there is no vertical asymptote.

Step2: Analyze horizontal asymptote

We find (\lim_{x\rightarrow\pm\infty}e^{-x^{2}}). Let (t = x^{2}), then (\lim_{x\rightarrow\pm\infty}e^{-x^{2}}=\lim_{t\rightarrow+\infty}e^{-t}). Using the property (\lim_{t\rightarrow+\infty}e^{-t}=\lim_{t\rightarrow+\infty}\frac{1}{e^{t}} = 0). So, (y = 0) is a horizontal asymptote.

Step3: Find the first - derivative

Use the chain rule. If (y = e^{-x^{2}}), let (u=-x^{2}), then (y = e^{u}). The derivative (y^\prime=\frac{dy}{du}\cdot\frac{du}{dx}). (\frac{dy}{du}=e^{u}) and (\frac{du}{dx}=-2x). So, (y^\prime=- 2xe^{-x^{2}}). Set (y^\prime = 0), then (-2xe^{-x^{2}}=0). Since (e^{-x^{2}}>0) for all (x\in R), (x = 0). Test intervals:

For (x\in(-\infty,0)), let (x=-1), then (y^\prime=-2\times(-1)\times e^{-(-1)^{2}} = 2e^{-1}>0).
For (x\in(0,+\infty)), let (x = 1), then (y^\prime=-2\times1\times e^{-1^{2}}=-2e^{-1}<0). So, the function is increasing on ((-\infty,0)) and decreasing on ((0,+\infty)).

Step4: Find local maximum and minimum

Since the function changes from increasing to decreasing at (x = 0). Substitute (x = 0) into (y = e^{-x^{2}}), (y(0)=e^{0}=1). There is a local maximum at (x = 0) and no local minimum (because the function only changes from increasing to decreasing).

Step5: Find the second - derivative

(y^\prime=-2xe^{-x^{2}}). Using the product rule ((uv)^\prime = u^\prime v+uv^\prime) where (u=-2x) and (v = e^{-x^{2}}). (u^\prime=-2) and (v^\prime=-2xe^{-x^{2}}). (y^{\prime\prime}=-2e^{-x^{2}}+(-2x)(-2xe^{-x^{2}})=e^{-x^{2}}(4x^{2}-2)). Set (y^{\prime\prime}=0), then (4x^{2}-2 = 0) (since (e^{-x^{2}}>0) for all (x)). (x^{2}=\frac{1}{2}), (x=\pm\frac{\sqrt{2}}{2}). Test intervals:

For (x\in(-\infty,-\frac{\sqrt{2}}{2})), let (x=-1), (y^{\prime\prime}=e^{-1}(4 - 2)=2e^{-1}>0).
For (x\in(-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})), let (x = 0), (y^{\prime\prime}=e^{0}(0 - 2)=-2<0).
For (x\in(\frac{\sqrt{2}}{2},+\infty)), let (x = 1), (y^{\prime\prime}=e^{-1}(4 - 2)=2e^{-1}>0).

The function is concave up on ((-\infty,-\frac{\sqrt{2}}{2})\cup(\frac{\sqrt{2}}{2},+\infty)) and concave down on ((-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})). Substitute (x=\pm\frac{\sqrt{2}}{2}) into (y = e^{-x^{2}}), (y = e^{-\frac{1}{2}}=\frac{1}{\sqrt{e}}).

Answer:

(a) (x=\text{None}), (y = 0) (b) Increasing interval: ((-\infty,0)), Decreasing interval: ((0,+\infty)) (c) Local maximum value: (1), Local minimum value: (\text{None}) (d) Concave up: ((-\infty,-\frac{\sqrt{2}}{2})\cup(\frac{\sqrt{2}}{2},+\infty)), Concave down: ((-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})), Inflection points: ((-\frac{\sqrt{2}}{2},\frac{1}{\sqrt{e}})), ((\frac{\sqrt{2}}{2},\frac{1}{\sqrt{e}}))