Let $f$ be a function defined by $$f(x)=\frac{e^x-e^{-x}}{2}$$ Find $f^{-1}(x)$. Domain for $x$ is $R$
My attempt,
Let $$f^{-1}(x)=a$$
$$x=f(a)$$
$$=\frac{e^a-e^{-a}}{2}$$
$$2x=e^a-e^{-a}$$
Let $$e^a=u$$
$$2x=u-u^{-1}$$
$$u^2-2xu-1=0$$
Solving $u$ by quadratic formula.
$$u=x\pm \sqrt{x^2+1}$$
$$e^a=x\pm \sqrt{x^2+1}$$
$$a=\ln(x\pm \sqrt{x^2+1})$$
$$f^{-1}(x)=\ln(x\pm \sqrt{x^2+1})$$
But my tutor said I'm wrong. Why?
On
Your tutor wants you to find the right root of your equation. Clearly, if $f$ is bijective, there can be only one answer. Note that $x^2+1>x^2$, so $\sqrt{x^2+1}>\left|x\right|$. So $x-\sqrt{x^2+1}<0$ whatever $x$ is (which can't be right because this is the value of $e^a$ which has to be positive). So the right answer is $$f^{-1}(x)=\ln(x+\sqrt{x^2+1})$$ sometimes (generally ?) noted ${\rm Argsh}(x)$.
Maybe because $x-\sqrt{x^2+1}<0$