Single Variable Calculus, Chapter 4, 4.4, Section 4.4, Problem 46

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

$\displaystyle y = \frac{x}{\sqrt{x^2 + 1}}$ has a domain $(- \infty, \infty)$, So there are no vertical asymptote

Solving for the horizontal asymptote

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

So the horizontal asymptotes are $y = 1$ and $y = -1$

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


$
\begin{equation}
\begin{aligned}

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

\end{aligned}
\end{equation}
$


when $y' = 0$


$
\begin{equation}
\begin{aligned}

0 =& \frac{1}{(x^2 + 1)^{\frac{3}{2}}}

\end{aligned}
\end{equation}
$


Since $y' = 0$ does not exist, the function has no critical number

Hence, the intervals of increase or decrease are

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

Solving for concavity and inflection points

If $\displaystyle y' = \frac{1}{(x^2 + 1)^{\frac{3}{2}}}$ or $y' = (x^2 + 1)^{\frac{-3}{2}}$, then


$
\begin{equation}
\begin{aligned}

y'' =& \frac{d}{dx} (x^2 + 1)^{\frac{-3}{2}}
\\
\\
y'' =& \frac{-3}{2} (x^2 + 1)^{\frac{-5}{2}} \frac{d}{dx} (x^2 + 1)
\\
\\
y'' =& \frac{-3}{\cancel{2}} (x^2 + 1)^{\frac{-5}{2}} (\cancel{2}x)
\\
\\
y'' =& -3x(x^2 + 1)^{\frac{-5}{2}}
\\
\\
y'' =& \frac{-3x}{(x^2 + 1)^{\frac{5}{2}}}

\end{aligned}
\end{equation}
$


when $y'' = 0$


$
\begin{equation}
\begin{aligned}

0 =& \frac{-3x}{(x^2 + 1) ^{\frac{5}{2}}}
\\
\\
0 =& -3x
\\
\\
x =& \frac{0}{-3}
\\
\\
x =& 0

\end{aligned}
\end{equation}
$


Therefore, the inflection point is

$f(0) = 0$

Thus, the concavity is..

$
\begin{array}{|c|c|c|}
\hline\\
\text{Interval} & f''(x)/y'' & \text{Concavity} \\
x < 0 & + & \text{Upward} \\
x > 0 & - & \text{Downward}\\
\hline
\end{array}
$