What is the domain of $f (x) = arcsin [\sqrt{x}]$ $f(x)= arcsin[sqrt(x)]$ ?

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What is the domain of $f (x) = arcsin [\sqrt{x}]$ $f(x)= arcsin[sqrt(x)]$ ?

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Answer 1 · 2016-09-01T12:02:38

The domain of $f (x)$ $f(x)$ is $x \leq 1$ $x<=1$

Explanation:

As the range for $sin (x)$ $sin(x)$ is $[- 1, 1]$ $[-1,1]$ for all real $x$ $x$

So, if we simplify the above equation:

$f (x) = arcsin (\sqrt{x})$ $f(x)=arcsin(sqrt(x))$ ;or
$sin (f (x)) = \sqrt{x}$ $sin(f(x))=sqrt(x)$ ;or
So, we know that the range of above equation is $[- 1, 1]$ $[-1,1]$
thus, $\sqrt{x}$ $sqrt(x)$ is $[- 1, 1]$ $[-1,1]$ ;or
$x \leq 1$ $x<=1$ is the solution or the domain of the original $f (x)$ $f(x)$

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What is the domain of $f (x) = arcsin [\sqrt{x}]$ $f(x)= arcsin[sqrt(x)]$ ?

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What is the domain of $f (x) = arcsin [\sqrt{x}]$ $f(x)= arcsin[sqrt(x)]$ ?