Finding range of Domain where there is only one solution

Question

Given $$f(x) = \left(2^{x}-1-\frac{1}{x}\right)$$ how to find all values of $x$ such that there is only one $y$ such that $f(x)=y $ $$$$ or in other terms when $f^{-1}(y)$ has only 1 solution.

without using the lambert function.

Answer 1

Per definition $\operatorname f(x)$ exists only for $x$ belonging to $R$ \ ${0}$.

So why do you want to avoid using the Lambert function? You mean $W(x) = xe^x$. Right? Just for my understanding...

Further it seems, without having proven this, from the function graph that the solution interval such as $\operatorname f(x) = y$ is unique in not the wohle definition interval because you can see counter-examples in the graph.

Calculate $f'(x)$ first and see whether f is strictly increasing. Then you would be able to proove my assumptions.

If the number derived from a function is positive over an interval, this function will increase over that same interval. Conversely, if it is negative, it will be decreasing. When the derived number is zero at one point, the curve admits a horizontal tangent in this.

Let's continue. Here $f'(x) = 2^x.\operatorname ln(2) + 1/x²$

which means that $f'(x) > 0$ for all $x$ belonging to $R$ \ ${0}$.

Therefore the function $f$ is strictly increasing respectively over $R^+$ \ ${0}$ and $R^-$ \ ${0}$