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 $[0,sqrt(pi/2)]$ .

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