How do you find the domain and range of #sqrt(sin^-1(2x))#?

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Answer 1 · 2016-07-31T16:00:20

Domain is #[0,1/2]# and, Range is #[0,sqrt(pi/2)]#.

Let #f(x)=sqrt{sin^-1(2x)}#.

We recall that the Domain of #sin^-1# function is [-1,1], and, its Range

is #[-pi/2,pi/2]#. So, for our #f#,

#2x in [0,1] rArr -1<=2x<=1 rArr -1/2<=x<=1/2#

But, #-1/2<=x<0##rArr -1<=2x<0#, and, since, #sin^-1# is #uarr#,

#sin^-1 (-1)<=sin^-1 (2x)<,sin^-1 0#, i.e.,

#-pi/2<=sin^-1 2x<0#, and, hence #sqrt(sin^-1 2x)# will be undefined.

Therefore, #x# has to be restricted to #[0,1/2]#, and as such,

#sin^-1 2x# will be in {0,1], so that, #f(x)=sin^-1 2x in [0,sqrt(pi/2)]#.

Thus, Domain is #[0,1/2]# and, Range is #[0,sqrt(pi/2)]#.