Problem: Find domain of $ \sin ^ {-1} [\log_2(\frac{x}{2})]$
Solution: $\log_2(\frac{x}{2})$ is defined for $\frac{x}{2} > 0$
$\log_2(\frac{x}{2})$ is defined for $x > 0$
Also domain of $\sin ^ {-1}x$ is $[-1,1]$
When $x=1$ ,then $\log_2(\frac{x}{2})$ becomes $-1$
When $x=4$ ,then $\log_2(\frac{x}{2})$ becomes $1$
So domain is $[1,4]$
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$\begingroup$ Is $[.]$ the floor function? $\endgroup$– MikasaJul 16, 2013 at 12:57
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$\begingroup$ No , I used it for square bracket $\endgroup$– rstJul 16, 2013 at 12:58
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$\begingroup$ Check the graph. I think it is $\pi/2$ to positive infinity. $\endgroup$– Torsten Hĕrculĕ CärlemänJul 16, 2013 at 13:04
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3 Answers
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Hint:
You can search the proper $x$ in which we can have $$\log_2(x/2)\in[-1,+1], ~~\text{and}~~~x/2>0$$ simultaneously. Note that the logarithm function is a one-one function here.
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1$\begingroup$ It is also monotone increasing, which facilitates handling inequalities. $\endgroup$ Jul 16, 2013 at 13:12
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$\begingroup$ @rst: When $-1\le\log_2(x/2)\le 1 $ then according to comment below my answer you can write $2^{-1}\le x/2\le 2^1$ $\endgroup$– MikasaJul 16, 2013 at 13:20
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here log(x/2) should be between -1&1 because if arc sin a=b then
sin (b)=a or a is between -1 &1
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2$\begingroup$ You could make your previous answer fixed. :) $\endgroup$– MikasaJul 16, 2013 at 13:21
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the domain is [1,4] as sin (y) is between -1 &1