Question #e8d28

1 Answer
John D.
Jan 24, 2018

I'm not sure what this question is asking, but see below for some information about #cos^-1(1/x)#

Explanation:

The domain of #cos^-1(x)# is from -1 to 1. Its graph looks like this:
graph{arccos(x) [-2, 2, -0.82, 4.18]}

Therefore, the domain of #cos^-1(1/x)# will be wherever #-1 le 1/x le 1#.

This is #x < -1 " " or " " x > 1#

Because #1/x# essentially takes the function and flips it inside out, putting #x=oo# at #x=0# and vice versa, we can imagine taking the left side of the function (where x is negative), flipping it around, and stretching it out all the way to negative infinity. We will also do the same thing to the right side.

If that description was confusing, this is what #cos^-1(1/x)# looks like graphically:

graph{arccos(1/x) [-3.634, 4.366, -3.44, 6.56]}

Hope this helps!

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