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Answer 1 · 2015-03-13T18:25:50

The solution is: #y=ln(x+sqrt(x^2-1))#.

The exponential form of the function #y=arccoshx# is:

#y=(e^x+e^-x)/2#

and its graph is:

graph{coshx [-10, 10, -5, 5]}

So we have to restrict its domain to values in #(-oo,0]# or in #[0,+oo)#.

Usually the choice is #[0,+oo)#, the range of the variable #y# is #[1,+oo)#.

To find the inverse of a function usually we use this method:

#x=(e^y+e^-y)/2rArr2x=e^y+1/e^yrArre^(2y)-2xe^y+1=0rArr#

#e^y=x+-sqrt(x^2-1)#

Because the choice of the domain and the corresponding range of #y# we can take only the solution:

#e^y=x+sqrt(x^2-1)#,

and so:

#y=ln(x+sqrt(x^2-1))#.