Examine the sign of the second derivative of y = x/(x²+1)

Examine the sign of the second derivative of $y = \dfrac{x}{x^{2}+1}$.

Step 1: The first derivative

\[ y' = \frac{\bigl(x^{2}+1\bigr)-x\cdot2x}{\bigl(x^{2}+1\bigr)^{2}} = \frac{1-x^{2}}{\bigl(x^{2}+1\bigr)^{2}}. \]

Step 2: The second derivative

With $P = 1-x^{2}$, $Q = \bigl(x^{2}+1\bigr)^{2}$, so $P' = -2x$ and $Q' = 4x\bigl(x^{2}+1\bigr)$, one factor $\bigl(x^{2}+1\bigr)$ cancels: \[y'' = \frac{2x\left[-\bigl(x^{2}+1\bigr)-2\bigl(1-x^{2}\bigr)\right]}{\bigl(x^{2}+1\bigr)^{3}} = \frac{2x\bigl(x^{2}-3\bigr)}{\bigl(x^{2}+1\bigr)^{3}}. \tag{1}\]

Step 3: Sign

The denominator is positive for every $x$, so the sign of $y''$ is that of $2x\bigl(x^{2}-3\bigr)$, which vanishes at $x = -\sqrt3,\ 0,\ \sqrt3$: \[\begin{array}{c|c|c|c} \text{interval} & x & x^{2}-3 & y'' \\ \hline x\lt -\sqrt3 & - & + & -\\ -\sqrt3\lt x\lt 0 & - & - & +\\ 0\lt x\lt \sqrt3 & + & - & -\\ x\gt \sqrt3 & + & + & + \end{array}\]

All three roots are simple, so the concavity reverses at each: three inflection points.

Step 4: The points

\[ y\bigl(-\sqrt3\bigr) = -\frac{\sqrt3}{4}, \qquad y(0) = 0, \qquad y\bigl(\sqrt3\bigr) = \frac{\sqrt3}{4}. \]

Step 5: Verification

\[y''(-1) = \frac{-2\cdot(-2)}{8} = 0.5\gt 0; \qquad y''(1) = \frac{2\cdot(-2)}{8} = -0.5\lt 0 \;\;\checkmark\]

\[y''(2) = \frac{4\cdot1}{125} = 0.032\gt 0; \qquad y''(-2) = -0.032\lt 0 \;\;\checkmark\]

\[\text{The function is odd, so its inflections come in symmetric pairs about the origin} \;\;\checkmark\]