Y=arccos(2x-3)/pi +1

Question

How to find the domain and range.  

Answer 1

Hi Rebecah.

 

arccos(x) is a bit of a tricky function. It's strange in the sense that it's the inverse of cos(x), and cos(x) should not be allowed to have an inverse, since it fails the horizontal line test miserably. In order for arccos(x) to exist, we have to restrict cos(x) to some domain that allows it to pass the horizontal line test.

 

The best we can do for cos(x) is to restrict it to values between 0 and pi, which is half of its normal period. cos(0) is 1, and cos(pi) is -1, so in this restricted domain of 0 to pi, the range of cos(x) covers all of the values between -1 and 1 exactly once.

 

The domain of a function is the range of its inverse, and the range of a function is the domain of its inverse.

 

So,

 

the domain of arccos(x) is -1 to 1

 

and

 

the range of arccos(x) is 0 to pi

 

We don't have arccos(x) though. We have arccos(2x-3)/pi + 1.

 

So we need to take what we know about arccos(x) and apply it to this function.

 

We know for the domain that

 

-1 < x < 1

 

But because we don't have x, we have 2x - 3, we have to do the following:

 

-1 < 2x - 3 < 1

 

2 < 2x < 4

 

1 < x < 2

 

So the domain is all real numbers from 1 to 2.

 

For the range, we know that:

 

0 < arccos(x) < pi

 

but we actually have arccos(x)/pi + 1

 

So

 

0 < arccos(x)/pi + 1 < pi

 

-1 < arccos(x)/pi < pi - 1

 

-pi < arccos(x) < pi^2 - pi

 

So the range is all real numbers from (-pi) to (pi^2 - pi).

 

Hope this helps.