For problems like this, we should work inside out.
First, what values of xx can go into the squareroot? We can't take a squareroot of a negative, so x > 0x>0. Let's also realize that the squareroot can go from 0 to infinity.
Therefore, the number that goes into cosecant is within (-infty, 0](−∞,0]. Cosecant repeats for every 2 pi2π so we know this will go through its entire range by 4pi^2 approx 39.54π2≈39.5. Therefore, the range of the cosecant (before multiplying by 3) is the usual: (-infty, -1] cup [1,infty) (−∞,−1]∪[1,∞). The 3 gets multiplied through, meaning that in the end,
Domain: [0, infty)[0,∞)
Range: (-infty, -3] cup [3, infty)(−∞,−3]∪[3,∞)
So it crosses the yy axis and exists in the 1st and 4th quadrants. Putting it into a plotting program confirms everything we just said:
graph{1/sin(0-x^(1/2)) [-5.55, 96.65, -20.24, 30.9]}
