Domain, Range and Graphs of Hyperbolic Functions

Because each hyperbolic function is built from $e^x$ and $e^{-x}$, their domains and ranges follow from the behaviour of the exponential function rather than from any periodicity — hyperbolic functions are not periodic, unlike their circular namesakes.

$\sinh x$ is defined for all real $x$; as $x \to \infty$, $e^{-x} \to 0$ so $\sinh x$ grows without bound like $\tfrac12 e^x$, and as $x \to -\infty$ it falls without bound. Since $\sinh x$ is continuous, odd, and strictly increasing (its derivative $\cosh x$ is always positive), its range is all of $\mathbb{R}$. Its graph passes through the origin and looks like a stretched, smoothed-out version of $y = x^3$ near $0$, flaring out steeply on both sides.

$\cosh x$ is also defined for all real $x$, but since $\cosh x = \frac{e^x+e^{-x}}{2} \geq \sqrt{e^x \cdot e^{-x}} = 1$ by AM–GM (with equality exactly at $x=0$), its range is $[1, \infty)$. Its graph is the classic catenary — the shape of a heavy chain or cable hanging under its own weight between two supports — symmetric about the $y$-axis with minimum value $1$ at $x = 0$.

$\tanh x$ is defined for all real $x$. Dividing the exponential form of $\sinh x$ and $\cosh x$ by $e^x$ gives $\tanh x = \frac{1 - e^{-2x}}{1+e^{-2x}}$, which shows that as $x \to \infty$, $\tanh x \to 1$, and as $x \to -\infty$, $\tanh x \to -1$; since it is continuous and strictly increasing, its range is the open interval $(-1, 1)$. The graph is a smooth S-shaped (sigmoid) curve with horizontal asymptotes $y = \pm 1$, rising steeply through the origin. …