How do you find the domain and range of ln (inverse sinx)?

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How do you find the domain and range of ln (inverse sinx)?

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Answer 1 · 2017-08-31T18:11:58

Domain: $(0, 1]$ $(0,1]$
Range: $(- \infty, ln (\frac{π}{2})]$ $(-∞,ln (pi/2)]$

Explanation:

Let $f (x) = ln ({sin}^{- 1} x)$ $f (x)=ln (sin^-1x)$
Now as $ln y$ $lny$ is defined only if $y > 0$ $y>0$
$:.$ For $ln (sin^-1x)$ to be defined we must have
$sin^-1x>0$
Also $sin^-1x$ is only defined in the interval $[-1,1]$
and $sin^-1x >0$ for $x in (0,1]$
$:.$ The domain of $f (x)$ is $x in (0,1]$
And as $sin^-1y$ and $lny$ are increasing functions
$:.$ The range of $f (x)$ will be $(-∞,ln(pi/2)]$

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How do you find the domain and range of ln (inverse sinx)?

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