Question
Find the domain of $f(x)=\sin ^{-1}\left(-x^2\right)$.
✓
Answer
The domain of $\sin ^{-1} x$ is $[-1,1]$.
Therefore, $f(x)=\sin ^{-1}\left(-x^2\right)$ is defined for all $x$ satisfying $-1 \leq-x^2 \leq 1$
$\Rightarrow 1 \geq x^2 \geq-1$
$\Rightarrow 0 \leq x^2 \leq 1$
$\Rightarrow x^2 \leq 1$
$\Rightarrow x^2-1 \leq 0$
$\Rightarrow(x-1)(x+1) \leq 0$
$\Rightarrow-1 \leq x \leq 1 $
Hence, the domain of $f(x)=\sin ^{-1}\left(-x^2\right)$ is $[-1,1]$.
Therefore, $f(x)=\sin ^{-1}\left(-x^2\right)$ is defined for all $x$ satisfying $-1 \leq-x^2 \leq 1$
$\Rightarrow 1 \geq x^2 \geq-1$
$\Rightarrow 0 \leq x^2 \leq 1$
$\Rightarrow x^2 \leq 1$
$\Rightarrow x^2-1 \leq 0$
$\Rightarrow(x-1)(x+1) \leq 0$
$\Rightarrow-1 \leq x \leq 1 $
Hence, the domain of $f(x)=\sin ^{-1}\left(-x^2\right)$ is $[-1,1]$.
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