How do you find the domain and range for y = 5arcsin(2cos(3x))?

1 Answer
Jun 11, 2015

"Domain"={x|x in RR ^^( pi/9+2/3kpi<=x<=2/9pi+2/3kpi vv 4/9pi+2/3kpi<=x<=5/9pi+2/3kpi) " where "kinZZ}
"Range"={y|y in RR ^^ -5/2pi<=y<=5/2pi }

Explanation:

arcsin(x):[-1,1]->[-pi/2,pi/2]
So the argument of the function must be between 1 and -1.
-1<=2cos(3x)<=1
So we have to solve the system:
{(2cos(3x)>=-1),(2cos(3x)<=1):}
{(cos(3x)>=-1/2),(cos(3x)<=1/2):}
pi/3+2kpi<=3x<=2/3pi+2kpi vv 4/3pi+2kpi<=3x<=5/3pi+2kpi
So, isolating x, the domain would be:
pi/9+2/3kpi<=x<=2/9pi+2/3kpi vv 4/9pi+2/3kpi<=x<=5/9pi+2/3kpi.

Considering that 2cos(3x) in the domain would vary in [-1,1], the range of arcsin(x) willl vary in [-pi/2,pi/2]. But the function has a cohefficient "5" that extends the amplitude of our function to -5/2pi,5/2pi.

This strange function would be something like this:
http://rechneronline.de/http://rechneronline.de/