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

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Answer 1 · 2016-07-31T16:00:20

Domain is $[0, \frac{1}{2}]$ $[0,1/2]$ and, Range is $[0, \sqrt{\frac{π}{2}}]$ $[0,sqrt(pi/2)]$ .

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

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

is $[- \frac{π}{2}, \frac{π}{2}]$ $[-pi/2,pi/2]$ . So, for our $f$ $f$ ,

$2 x \in [0, 1] \Rightarrow - 1 \leq 2 x \leq 1 \Rightarrow - \frac{1}{2} \leq x \leq \frac{1}{2}$ $2x in [0,1] rArr -1<=2x<=1 rArr -1/2<=x<=1/2$

But, $- \frac{1}{2} \leq x < 0$ $-1/2<=x<0$ $\Rightarrow - 1 \leq 2 x < 0$ $rArr -1<=2x<0$ , and, since, ${sin}^{- 1}$ $sin^-1$ is $↑$ $uarr$ ,

${sin}^{- 1} (- 1) \leq {sin}^{- 1} (2 x) <, {sin}^{- 1} 0$ $sin^-1 (-1)<=sin^-1 (2x)<,sin^-1 0$ , i.e.,

$- \frac{π}{2} \leq {sin}^{- 1} 2 x < 0$ $-pi/2<=sin^-1 2x<0$ , and, hence $\sqrt{{sin}^{- 1} 2 x}$ $sqrt(sin^-1 2x)$ will be undefined.

Therefore, $x$ $x$ has to be restricted to $[0, \frac{1}{2}]$ $[0,1/2]$ , and as such,

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

Thus, Domain is $[0, \frac{1}{2}]$ $[0,1/2]$ and, Range is $[0, \sqrt{\frac{π}{2}}]$ $[0,sqrt(pi/2)]$ .

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