Single Variable Calculus, Chapter 4, 4.4, Section 4.4, Problem 44

Determine the horizontal asymptotes of the curve $\displaystyle y = \frac{1 + 2x^2}{1 + x^2}$ and use them together with concavity and intervals of increase and decrease, to sketch the curve.

$\displaystyle y = \frac{1 + 2x^2}{1 + x^2}$ has a domain $(- \infty, \infty)$

So there are no vertical asymptote

Solving for the horizontal asymptote

$\displaystyle \lim_{x \to \pm \infty} \frac{1 + 2x^2}{1 + x^2} = \lim_{x \to \pm \infty} \frac{2\cancel{x^2}}{\cancel{x^2}} = \frac{2}{1} = 2 $

So the horizontal asymptote is $y = 2$

If we take the derivative of $f(x) = y$


$
\begin{equation}
\begin{aligned}

y' =& \frac{d}{dx} \left( \frac{1 + 2x^2}{1 + x^2} \right)
\\
\\
y' =& \frac{\displaystyle (1 + x^2) \frac{d}{dx} (1 + 2x^2) - (1 + 2x^2) \frac{d}{dx} (1 + x^2) }{(1 + x^2)^2}
\\
\\
y' =& \frac{(1 + x^2) (4x) - (1 + 2x^2)(2x) }{(1 + x^2)^2}
\\
\\
y' =& \frac{4x + \cancel{4x^3} - 2x - \cancel{4x^3}}{(1 + x^2)^2}
\\
\\
y' =& \frac{2x}{(1 + x^2)^2}

\end{aligned}
\end{equation}
$


when $y' = 0$


$
\begin{equation}
\begin{aligned}

0 =& \frac{2x}{(1 + x^2)^2}
\\
\\
2x =& 0
\\
\\
\frac{\cancel{2}x}{\cancel{2}} =& \frac{0}{2}
\\
\\
x =& 0

\end{aligned}
\end{equation}
$


The critical number is $x = 0$

Hence, the intervals of increase or decrease are

$
\begin{array}{|c|c|c|}
\hline\\
\text{Interval} & f'(x)/y' & f \\
x < 0 & - & \text{decreasing on } (- \infty, 0) \\
x > 0 & + & \text{increasing on } (0, \infty)\\
\hline
\end{array}
$

Since $f'(x)$ changes from negative to positive at $x = 0, f(0) = 1$ is a local minimum.

Solving for concavity and inflection points

If $\displaystyle y' = \frac{2x}{(1 + x^2)^2}$, then


$
\begin{equation}
\begin{aligned}

y'' =& \frac{d}{dx} \left[ \frac{2x}{(1 + x^2)^2} \right]
\\
\\
y'' =& \frac{\displaystyle (1 + x^2)^2 \frac{d}{dx} (2x) - (2x) \frac{d}{dx} (1 + x^2) }{[(1 + x^2)^2]^2}
\\
\\
y'' =& \frac{\displaystyle (1 + x^2)^2 (2) = (2x)(2) (1 + x^2) \frac{d}{dx} (1 + x^2) }{(1 + x^2)^4}
\\
\\
y'' =& \frac{(1 + 2x^2 + x^4)(2) - (4x)(1 + x^2)(2x) }{(1 + x^2)^4}
\\
\\
y'' =& \frac{2 + 4x^2 + 2x^4 - (8x^2 + 8x^4)}{(1 + x^2)^4}
\\
\\
y'' =& \frac{2 + 4x^2 + 2x^4 - 8x^2 - 8x^4}{(1 + x^2)^4}
\\
\\
y'' =& \frac{2 - 4x^2 - 6x^4}{(1 + x^2)^4}
\\
\\
y'' =& \frac{(2 - 6x^2)(1 + x^2)}{(1 + x^2)^4}
\\
\\
y'' =& \frac{2 - 6x^2}{(1 + x^2)^3}



\end{aligned}
\end{equation}
$


when $y'' = 0$


$
\begin{equation}
\begin{aligned}

0 =& \frac{2 - 6x^2}{(1 + x^2)^2}
\\
\\
0 =& 2 - 6x^2
\\
\\
6x^2 =& 2
\\
\\
x^2 =& \frac{2}{6}
\\
\\
x =& \pm \sqrt{\frac{1}{3}}

\end{aligned}
\end{equation}
$


Therefore the inflection points are

$\displaystyle f \left( \sqrt{\frac{1}{3}} \right) = \frac{5}{4}$ and $\displaystyle f \left(- \sqrt{\frac{1}{3}} \right) = \frac{5}{4}$

Thus, concavity is...

$
\begin{array}{|c|c|c|}
\hline\\
\text{Interval} & f''(x) & \text{Concavity} \\
\hline\\
\displaystyle x < - \sqrt{\frac{1}{3}} & - & \text{Downward} \\
\hline\\
\displaystyle - \sqrt{\frac{1}{3}} < x < \sqrt{\frac{1}{3}}& + & \text{Upward} \\
\hline\\
\displaystyle x > \sqrt{\frac{1}{3}} & - & \text{Downward}\\
\hline
\end{array}
$