y = \dfrac{1}{x^2-1}

The Rational Function $y = \dfrac{1}{x^2-1}$

$y = \dfrac{1}{x^2 - 1}$ is a rational function whose denominator factors as $(x-1)(x+1)$ . Because the denominator vanishes at two points, the graph is cut into three pieces by two vertical asymptotes¹.

The contrast with $y = \dfrac{1}{x^2+1}$ , the Witch of Agnesi, is striking: changing the constant term from $-1$ to $+1$ turns this broken curve into a single smooth bell defined on the whole real line.

Domain and range

Range of $x$	Sign of $x^2 - 1$	Values taken
$\|x\| > 1$	positive	$y > 0$
$\|x\| < 1$	negative	$y \leq -1$

The domain is every real number except $x = \pm 1$ . No value in $-1 < y \leq 0$ is ever taken.

Symmetry

Since $f(-x) = f(x)$ , the function is even and the graph is symmetric about the $y$ -axis.

Monotonicity and extrema

The derivative is $f'(x) = -\dfrac{2x}{(x^2-1)^2}$ . The denominator is positive throughout the domain, so the sign comes from $-2x$ alone: the function increases for $x < 0$ and decreases for $x > 0$ , giving a local maximum of $-1$ at the origin.

That point is the summit of the middle branch, but it is not the maximum of the whole graph, since the outer branches grow without bound. This is a clear picture of the fact that a local maximum need not be a global one.

Asymptotes

Approach	Behaviour
$x \to 1^{+}$	$y \to +\infty$
$x \to 1^{-}$	$y \to -\infty$
$x \to \pm\infty$	$y \to 0$

The branches separate above and below the asymptote $x = 1$ , and the same happens on either side of $x = -1$ .

Concavity

The second derivative is $f''(x) = \dfrac{6x^2 + 2}{(x^2-1)^3}$ . The numerator is always positive, so the sign matches that of $x^2 - 1$ . The two outer branches are concave up and the middle branch is concave down. The sign changes at $x = \pm 1$ , but those points lie outside the domain, so there are no inflection points.

Partial fractions

\frac{1}{x^2-1} = \frac{1}{2}\left(\frac{1}{x-1} - \frac{1}{x+1}\right)

Viewed as a difference of two reciprocals², it is clear that near $x = 1$ and near $x = -1$ only one term at a time blows up. Integrating gives the following.

\int \frac{dx}{x^2-1} = \frac{1}{2}\ln\left|\frac{x-1}{x+1}\right| + C

That equals $-\operatorname{artanh} x$ for $|x| < 1$ and $-\operatorname{arcoth} x$ for $|x| > 1$ .

Applications

In undamped forced oscillation the amplitude is proportional to $\dfrac{1}{\omega_0^2 - \omega^2}$ . As the driving frequency approaches the natural frequency the amplitude grows without limit, and that resonance is exactly the vertical asymptote of this function.

Asymptote, Wikipedia
Partial fraction decomposition, Wikipedia

The Rational Function y=1x2−1y = \dfrac{1}{x^2-1}y=x2−11​