Finding the inverse of an exponential function $h(x)$.

Question

A function is defined by

$$h(x) = 2 \exp(x) - \dfrac{1}{\exp(x)}, x \in \mathbb R.$$

How to find an expression for $h^{-1}(x)$. Any leads appreciated.

Answer 1

In $$y=h(x)=2e^x-\dfrac{1}{e^x}\tag{1}$$

we first observe that $h'(x)=2e^x+e^{-x}>0$ ; therefore $h$ is an increasing function ; as such, it is a bijection thus has a reciprocal function.

In order to obtain it, we are going to express $x$ as a function of $y$. Let

$$e^x=X\tag{2}$$

Using (2) in (1) gives the following quadratic equation with unknown $X$, $y$ being considered as a parameter :

$$2X^2-yX-1=0$$

Its solutions are :

$$X=\tfrac14(y\color{red}{\pm}\sqrt{y^2+8}).$$

Taking (2) into account gives the result :

$$x=\ln\left(\tfrac14(y\color{red}{+}\sqrt{y^2+8})\right)$$

You have noticed that the $\pm$ sign has been replaced by a $+$ sign ; this is compulsory in order to have a positive fraction (unless, choosing the minus sign, one would have a negative fraction and one couldn't take its logarithm because $\sqrt{y^2+8}>|y|$).

Fig. 1 : The curve of $h$ is colored in blue ; the curve of $h^{-1}$ is colored in magenta ($h^{-1}(x)=\ln\left(\tfrac14(x+\sqrt{x^2+8})\right)$. These curves are symmetrical with respect to line $y=x$.