Question

For the function $y=\frac{1}{1+x^2}$, which of the following is/are correct:

• A. It has a horizontal asymptote.
• B. It is symmetric about $x=0$
• C. It is increasing $\forall x\in R$
• D. It has maxima at $x=0$

Answer 1

Answer: (A), (B), (D)

It has maxima at $x=0$

The function is defined for all real values of $x,$ Consequently, it has no vertical asymptotes.
Since
$\underset{x\to \pm \infty }{lim}y\left(x\right)=\underset{x\to \pm \infty }{lim}\frac{1}{1+x^2}=0,$
then the graph of the function has horizontal asymptote $y=0,$ that is the $x-$axis is the horizontal asymptote.
This function is even.

$y\left(–x\right)=\frac{1}{1+{\left(–x\right)}^2}=\frac{1}{1+x^2}=y\left(x\right).$
It is obvious that the function has no roots and positive for all $x.$
At the point $x=0,$ its value is
$y\left(0\right)=\frac{1}{1+0^2}=1$
The first derivative:
$y^{\prime }\left(x\right)={\left(\frac{1}{1+x^2}\right)}^{\prime }$
$=–\frac{1}{{\left(1+x^2\right)}^2}\cdot {\left(1+x^2\right)}^{\prime }=–\frac{2x}{{\left(1+x^2\right)}^2}.$

This shows that $x=0$ is a stationary point. When passing through this point the derivative changes sign from plus to minus. Therefore, we have a maximum at $x=0.$ Its value is $y\left(0\right)=1.$
Calculate the second derivative:

$y^{\prime \prime }\left(x\right)={\left(–\frac{2x}{{\left(1+x^2\right)}^2}\right)}^{\prime }$
$=–\frac{2{\left(1+x^2\right)}^2–2\cdot 2x\cdot 2x\left(1+x^2\right)}{{\left(1+x^2\right)}^4}$
$=\frac{8x^2–2–2x^2}{{\left(1+x^2\right)}^3}=\frac{6x^2–2}{{\left(1+x^2\right)}^3}.$
It is equal to zero at the following points:

$y^{\prime \prime }\left(x\right)=0\Rightarrow \frac{6x^2–2}{{\left(1+x^2\right)}^3}=0$
$\Rightarrow \frac{2\left(\sqrt{3}x–1\right)\left(\sqrt{3}x+1\right)}{{\left(1+x^2\right)}^3}=0,\Rightarrow x_1=–\frac{1}{\sqrt{3}},x_2=\frac{1}{\sqrt{3}}.$
When passing through these points the second derivative changes its sign. Therefore, both points are inflection points. The function is strictly convex downward in the intervals $(-\infty ,-\frac{1}{\sqrt{3}})$ and $(\frac{1}{\sqrt{3}},+\infty )$ and accordingly, strictly convex upward in the interval $(-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})$ Since the function is even, the found inflection points have the same values of $y:$

$y\left(\pm \frac{1}{\sqrt{3}}\right)=\frac{1}{1+{\left(\pm \frac{1}{\sqrt{3}}\right)}^2}=\frac{1}{1+\frac{1}{3}}=\frac{3}{4}.$
Figure presents a schematic graph of the function

Hence, we can conclude clearly from the graph that it has a horizontal asymtote and it is symmetrical about $x=0$ or the $y$ axis. Also it has maxima at $x=0$