Inverse of $(e^x - e^{-x})/2$

Question

What is the inverse of the function $f(x)=\frac{e^x - e^{-x}}2$? I tried replacing $e^x$ by a variable but I still can't get it.

Answer 1

$$\sinh x = \frac{e^x - e^{-x}}{2}$$

So $\sinh^{-1} x$ is the inverse.

Answer 2

Substituting is the right way to go. If we have $b = \frac{a - \frac{1}{a}}{2}$, then multiplying by 2 and then $a$ gives $2ab = a^2 - 1$. Now rearrange: $0 = a^2 - 2ab - 1$. Now it should look more familiar.

Answer 3

Let $y=f^{-1}(x)$. Then $f(y)=x$. So,

$(e^y-e^{-y})/2=x$. Now use your previous idea: Let $e^y=t$. Then we get the equation equation:

$t-1/t=2x$, or $t^2-2xt-1=0$. Solve for $t$ in terms of $x$. Then find $y$ in terms of $x$.

Answer 4

Let $f:\Bbb{R}\to{\Bbb{R}}$ be defined by $f(x)=(e^x-e^{-x})/2$.

(a) To show $f(x)$ is one-one: Let $x_1,x_2\in \Bbb{R}$ and let $f(x_1)=f(x_2)$.

$$\frac{(e^x_1-e^{-x_1})}{2}=\frac{(e^x_2-e^{-x_2})}{2}$$ $$\Rightarrow e^{x_1+x_2}(e^{x_1}-e^{x_2})=e^{x_2}-e^{x_1}$$ $\Rightarrow e^{x_1}-e^{x_2}=0$ or $e^{x_1+x_2}+1=0$

As $e^{x_1+x_2}$ cannot be negative as $x_1,x_2\in \Bbb{R}$ hence, $$e^{x_1}-e^{x_2}=0$$ $$\Rightarrow x_1=x_2$$

Q.E.D

(b) To show $f(x)$ is Onto: It can be easily shown that as $x\to\infty$,$f(x)\to\infty$.

Similarly as $x\to-\infty$,$f(x)\to-\infty$ i.e., $f(x)\in(-\infty,\infty)$ so long as $x\in(-\infty,\infty)$. Hence, the range of f is same as the set $\Bbb{R}$. Therefore $f(x)$ is Onto. Q.E.D.

And thus, $f(x)$ is invertive.

(c) To find $f^{-1}$.

$$f(x)=\frac{(e^x-e^{-x})}{2}$$ By identity, $$f(f^{-1}(x))=x$$

And,$$f(f^{-1}(x))=\frac{e^{f^{-1}(x)}-e^{-{f^{-1}(x)}}}{2}=x$$ $$e^{2f^{-1}(x)}-2xe^{f^{-1}(x)}-1=0$$ $$e^{f^{-1}(x)}=x\pm{\sqrt{1+x^2}}$$

Since $e^{f^{-1}(x)}>0$, negative sign is ruled out.

Hence, $$e^{f^{-1}(x)}=x+\sqrt{1+x^2}$$ $$\Rightarrow f^{-1}(x)=\ln(x+\sqrt{1+x^2})$$