What is the inverse of the function $\tan^{-1}\sqrt{x^3-1}$?

Question

I have the following function $$\tan^{-1}\sqrt{x^3-1}$$ But I don't really know how would the inverse function look like, it is a one-to-one function so I know that it has an inverse, but I've tried many times and I can't get any good result, can anyone help me? -if you have any reference that can help, I would be happy to give a look-

Answer 1

Let $y = f(x)$ $$y=\tan^{-1}\sqrt{x^3-1}$$

Taking the tangent function for both sides you get:

$$\tan y=\sqrt{x^3-1}$$ Then you square both sides:

$$\tan^2y=x^3-1$$

Add $1$ to both sides:

$$\tan^2y+1=x^3$$

Take cube root for both sides

$$\sqrt[3]{\tan^2y+1}=x$$

And you can name that function $g(y)$ for example

The original domain for $f(x)$ is $\{x \mid x \in [1, \infty)\}$, and range $\{y \mid y \in [0, π/2)\}$.

The domain for $g(y)$ is $\{y \mid y \in [0, π/2)\}$, and the domain is $\{x \mid x \in [1, \infty)\}$.

Answer 2

$$y=\tan^{-1}\sqrt{x^3-1}$$

$$\tan y=\sqrt{x^3-1}$$

$$\tan^2y=x^3-1$$

$$\tan^2y+1=x^3$$

$$\sqrt[3]{\tan^2y+1}=x$$

Now you have to discuss the domain, range, signs...

The function $f$ is defined for $x\ge0$ and yields values in $[0,\frac\pi2)$.