Question

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

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 \(\lim_{x
ightarrow\pm\infty}e^{-x^{2}}\). Let \(t = x^{2}\), then \(\lim_{x
ightarrow\pm\infty}e^{-x^{2}}=\lim_{t
ightarrow+\infty}e^{-t}\). Using the property \(\lim_{t
ightarrow+\infty}e^{-t}=\lim_{t
ightarrow+\infty}\frac{1}{e^{t}} = 0\). 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: \((-\frac{\sqrt{2}}{2},\frac{1}{\sqrt{e}})\), \((\frac{\sqrt{2}}{2},\frac{1}{\sqrt{e}})\)