# How do you find the domain & range for f(x)=sqrt(sinx-1)?

Oct 22, 2015

$\mathrm{do} m f = \left\{x | x = \frac{\pi}{2} + k \pi , k \in 2 \mathbb{Z}\right\}$
$r a n f = \left\{0\right\}$

#### Explanation:

Domain
The number inside the radical must be greater than or equal to $0$.

$\left[1\right] \textcolor{w h i t e}{X X} \sin x - 1 \ge 0$

$\left[2\right] \textcolor{w h i t e}{X X} \sin x \ge 1$

However, the possible values of $\sin$ are only $\left[- 1 , 1\right]$. Therefore, for the inequality to be true, $\sin x$ must be equal to $1$. For $\sin x$ to be equal to 1, $x$ must be $\frac{\pi}{2}$ or an angle that is coterminal with $\frac{\pi}{2}$.

color(red)(domf={x|x=pi/2+kpi, kin2ZZ}

Range
Since the only value $\sin x$ can have is $1$, the only value in the range is $0$:

$\left[1\right] \textcolor{w h i t e}{X X} y = \sqrt{\sin x - 1}$

$\left[2\right] \textcolor{w h i t e}{X X} y = \sqrt{1 - 1}$

$\left[3\right] \textcolor{w h i t e}{X X} y = \sqrt{0}$

$\left[4\right] \textcolor{w h i t e}{X X} y = 0$

$\textcolor{b l u e}{r a n f = \left\{0\right\}}$