Finding the domain of arcsin function

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I have to find the domain of:$ f(x) = 10 \frac{2}{3\arcsin x^2}$

Assumption 1:

$-1 \leq 3\arcsin x^2 \leq 1$

$-1 \leq x^2 \leq 1$

$x \in [-1;1]$

Assumption 2:

$3\arcsin x^2 \neq 0$

$x^2 \neq \sin 0$

$x^2 \neq 0$

$x \neq 0$

Answer: $D_{f} = [-1;0) \cup (0;1] $

However, the correct answer should be $D_{f} = (0, 1]$

Anyone sees mistake? :) Thanks~!

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2 Answers

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No. Your answer is correct since the function is even i.e. $$f(x)=f(-x)$$ therefore the domain must also be symmetric

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There is nothing wrong with your answer, but your first case is just incorrect.

$$\arcsin x \implies -1 \leq x \leq 1$$

Obviously, $x^2$ is non-negative.

$$\arcsin x^2 \implies x^2 \leq 1 \implies x \leq \vert1\vert \implies -1\leq x \leq 1$$

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