Please help.

AsheB:

Given that p(x)= sin^-1 x and q(x)= x+7, what is the domain of p(q(x))?

p(q(x)) basically means put the q(x) equation into the p(x) equation.

p(q(x)) = sin^-1 (x+7)

Now you can either try to figure out the domain by looking at the equation or you can input the equation into the Desmos graphing calculator, which is what I do. Just google 'desmos' to find it.

^_^ Hope it helps!

Thanks so much!!

Given that p(x)= sin^-1 x and q(x)= x+7, what is the domain of p(q(x))?

The instruction Special has given you is good but what if you didn't have access to the Demos calculator?

By the way, the Demos calculator is fabulous and I use it all the time. But mostly it should be used to check your answer or give you guidance, not to answer the question for you.

With your question you must first look at whether q(x) has put any restrictions on the domain. And no, it has not, q(x) is defined for all real x.

Now look at p(q(x))

p(q(x)) = sin-1(x+7) also written as p(q(x)) = asin(x+7)

Now, I think, AshB, that you are supposed to know what y=sin ^-1x looks like and therefore that it is restricted to -1 < x < 1

So, looking at your problem

-1 <= x+7 <= 1
-1 -7 <= x <= 1 - 7
So the domain of p(q(x)) will be -8 <= x <= -6 ( the errors that I speak about in the next post have been fixed)

I might have a play with demos and see if I can get you a graph or 2

This is the graph of y=sin ^-1x
It looks like it finishes early, I don't know why, but the end points are (-1,-pi/2) and (1, pi/2)
http://gyazo.com/7f6851adf451e3505295f474e55dfd33

I am going to add more but there are mistakes in my last post so I will fix them first and I will tell you when they are fixed
It was actually the graphing on desmos that showed me my error, which is exactly what Demos is great for.

this is a graph of both on the same grid, I am short of time so I must get this post finished.

http://gyazo.com/a8e1bcc4426b67d5c773a77bd39fc6ca