Question 1170306
For (a), when you find the candidate inverse, you get {{{f^-1(x) = (1/2)(x +- sqrt(x^2+4)) }}} <br> 

It is that +- that tells you this candidate inverse is not a function (there are many places on its graph where there are two "y" values for a single x input).  When this happens, the relationship does not pass the vertical line test (it is not one-to-one) so the candidate inverse is NOT a function.  Therefore, f(x) has no inverse function. 
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For (b), limiting the domain to {{{0<x<infinity}}}, results in {{{g^-1(x) =  (1/2)(x+sqrt(x^2+4)) }}} which indeed is a function, therefore g(x) has an inverse function.   
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Check (b):  if you apply {{{g^-1(x) }}} to {{{g(x)}}} you should get back x:

  {{{g^-1(g(x)) = (1/2)((x-1/x) + sqrt((x-1/x)^2 + 4)) }}}<br>
           = {{{ (1/2)(((x^2-1)/x) + sqrt((x^2-1)^2 + 4x^2) / x) }}}<br>
           = {{{ (1/2x) ((x^2 - 1) + sqrt(x^4+2x^2 + 1)) }}}<br>
           = {{{ (1/2x)((x^2 - 1) + sqrt((x^2+1)^2)) }}}<br>
           = {{{ (1/2x)((x^2 - 1) + (x^2+1)) }}}<br>
           = {{{ (1/2x)(2x^2) }}}<br>
           =  x    
so g(x) has an inverse function


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Here are two pictures to help illustrate this.  The first picture shows f(x) (red) and the required "inverse" (green and blue).   The 2nd picture shows g(x) (red) and its inverse (blue).

*[illustration f_of_x.png]


*[illustration g_of_x.png]