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I am having problem with proper math-fashioned style of solving this task. It's task from my exam and apparently I didn't do well enough to get 2.5 out of 5 points.

Find inverse function $f^{-1}$to function $f(x)= \arcsin{x^3}+\frac{\pi}{2}$. Find the domain of $f$ and $f^{-1}$.

Step 1: domain of $f$:

$-1 \leq x^3 \leq 1$

$\Rightarrow -1 \leq x^3 \land x^3\leq1$

$-1 \leq x \land x \leq 1$

$\Longrightarrow D_{f}=[-1;1]$

Step 2: finding the inverse function.

$$f(x)=\arcsin{x^3} + \frac{\pi}{2}$$

$$y=\arcsin{x^3} + \frac{\pi}{2}$$

$$x \longleftrightarrow y$$

$$x=\arcsin{y^3} + \frac{\pi}{2}$$

$$x-\frac{\pi}{2}=\arcsin{y^3} \quad \quad /\sin(...) \quad \quad \quad \quad **HERE**$$

$$\sin{(x-\frac{\pi}{2})} = y^3 \quad \quad / ^{\sqrt[3]{}}\quad \quad \quad \quad \quad \quad **HERE**$$

$$\sqrt[3]{\sin{(x-\frac{\pi}{2})}} = y$$

$$\sqrt[3]{\sin{(x-\frac{\pi}{2})}} = f^{-1}(x)$$

Question 1: in the parts where I wrote HERE the teacher on my paper wrote "assumptions!". (URL to my paper: https://i.imgur.com/zkJxAOD.jpg ) What does it mean, what kind of assumptions I should have made?

Question 2: What if I was multiplying by $\log{(...)}$, $\ln{(...)}$ or taking the even root like $^{\sqrt[4]{}}$? What kind of assumptions should I make? Are there any other cases different than $\arcsin$, $\arccos$, $\log$, even root?

Step 3: Finding the domain of $f^{-1}(x)$:

Let's ignore it, but I have some questions here.

Because the initial function was $\arcsin{x^3}$ and $\arcsin{\text{(...something...)}}$ is based on $[-\frac{\pi}{2}; \frac{\pi}{2}]$ from domain of $\sin{\text{(...something...)}}$ (that's the only way, only in that interval sine function is injective, right?), then this is directly related to my $Question 1$?

Assumptions from that part are needed and directly related to the domain of $f^{-1}(x)$?

Check my calculations and paper please and tell me all mistakes I did, I need someone unforgiving to check what I wrote. Thanks

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