Question

Question #9ef51

Answer 1

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))`.