Domains, ranges and reciprocal hyperbolic functionsAQA A-Level Further Maths: Revision notes

The domain of a function is the set of allowed inputs; the range is the set of outputs.

• $\sinh x$: domain all real $x$; range all real $y$ (odd, increasing, unbounded).
• $\cosh x$: domain all real $x$; range $y\ge1$, because $\frac12(e^x+e^{-x})\ge1$ with equality only at $x=0$.
• $\tanh x$: domain all real $x$; range $-1<y<1$, because the graph lies strictly between its asymptotes $y=\pm1$. The range can be read from the graph: look at the lowest and highest $y$-values reached, and whether they are attained or only approached.

Three further functions are defined as reciprocals: $\operatorname{sech}x=\frac{1}{\cosh x}=\frac{2}{e^x+e^{-x}},\quad\operatorname{cosech}x=\frac{1}{\sinh x}=\frac{2}{e^x-e^{-x}},\quad\coth x=\frac{1}{\tanh x}=\frac{\cosh x}{\sinh x}=\frac{e^x+e^{-x}}{e^x-e^{-x}}.$ Read them as 'sheck', 'cosheck' and 'coth'. They are defined wherever the denominator is not zero. Example: $\coth(\ln3)=\frac{3+\frac13}{3-\frac13}=\frac{10/3}{8/3}=\frac54$.

• $\operatorname{sech}x$: since $\cosh x\ge1$ is never zero, the domain is all real $x$. The range is $0<y\le1$, with the maximum value $1$ at $x=0$ and $y\to0$ as $|x|\to\infty$.
• $\operatorname{cosech}x$: $\sinh x=0$ only at $x=0$, so the domain is $x\ne0$. Since $\sinh x$ takes every non-zero real value, the range is all real $y\ne0$.
• $\coth x$: $\tanh x=0$ only at $x=0$, so the domain is $x\ne0$. Since $-1<\tanh x<1$ and $\tanh x\ne0$, the range is $y<-1$ or $y>1$. As $x\to\infty$, $\coth x\to1$ from above; as $x\to-\infty$, $\coth x\to-1$ from below. The reciprocal of a number in $(0,1]$ is at least 1, so reciprocals turn bounded ranges into unbounded ones and vice versa.

The domain of an inverse is the range of the original function, and its range is the domain of the original.

• $\sinh^{-1}x$: $\sinh$ is one-to-one, so the domain is all real $x$ and the range is all real $y$.
• $\cosh^{-1}x$: $\cosh x$ is not one-to-one on the whole line, so it is restricted to $x\ge0$. Then the domain of $\cosh^{-1}$ is $x\ge1$ and the range is $y\ge0$.
• $\tanh^{-1}x$: $\tanh$ is one-to-one with range $-1<y<1$, so the domain is $-1<x<1$ and the range is all real $y$. This matches the logarithmic forms: $\cosh^{-1}x=\ln\left(x+\sqrt{x^2-1}\right)$ needs $x\ge1$, and $\tanh^{-1}x=\frac12\ln\frac{1+x}{1-x}$ needs $|x|<1$.

Domain and range tell you at once whether an equation has solutions: $\cosh x=\frac12$ has none because $\cosh x\ge1$, and $\operatorname{sech}x=2$ has none because $\operatorname{sech}x\le1$. Equations such as $\operatorname{sech}x=\frac12$ become $\cosh x=2$, giving $x=\pm\ln\left(2+\sqrt3\right)$, two solutions since $\cosh$ is even. Always state domains when you define a new function, and check inputs to $\cosh^{-1}$ and $\tanh^{-1}$ lie in their domains.