# Question #9ef51

Mar 13, 2015

The solution is: $y = \ln \left(x + \sqrt{{x}^{2} - 1}\right)$.

The exponential form of the function $y = \arccos h x$ is:

$y = \frac{{e}^{x} + {e}^{-} x}{2}$

and its graph is:

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

So we have to restrict its domain to values in $\left(- \infty , 0\right]$ or in $\left[0 , + \infty\right)$.

Usually the choice is $\left[0 , + \infty\right)$, the range of the variable $y$ is $\left[1 , + \infty\right)$.

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

$x = \frac{{e}^{y} + {e}^{-} y}{2} \Rightarrow 2 x = {e}^{y} + \frac{1}{e} ^ y \Rightarrow {e}^{2 y} - 2 x {e}^{y} + 1 = 0 \Rightarrow$

${e}^{y} = x \pm \sqrt{{x}^{2} - 1}$

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:

$y = \ln \left(x + \sqrt{{x}^{2} - 1}\right)$.