Find the Implied Domain+Range Q

[VCE1996]

Hi,

Can someone please explain how to do this question and provide solutions?

arcsin (2/pi arctan(x) + 1)

[rhinwarr]

You can solve it using composite functions.
Let f(x)=g(h(x)), where g(x)=arcsin(x) and h(x)=2/pi arctan(x) + 1
You should know that for f(x) to exist, the range of h(x) must be a subset or equal to the domain of g(x). Also, the domain of f(x) is the domain of h(x).
dom g = [-1,1]
range g = [-pi/2, pi/2]
dom h = R
range h = (-pi/2 +1, pi/2 +1)
You can see that in this case the range of h does not fit into the domain of g so we must restrict the domain of h to change the range of h.
To do this, we can change the range of h to (-pi/2 + 1, 1].
To find the new domain of h we need to find when h(x) = 1.
2/pi arctan(x) + 1 = 1
x = 0
Therefore, we need to change the domain of h to (-infinity, 0]
Since the domain of f(x) is equal to the domain of h(x), the implied domain is (-infinity, 0].
To find the range you'll have to draw the graph. The range should be (0,pi/2]

[kinslayer]

The domain of is and the range is

First, implies that which just means that .

So the implied domain is .

When , we know that which means that , so 

So if then the implied range is .

[VCE1996]

thanks guys!