Question 1106122
looking for g(x) such that  g(f(x)) = x   (i.e. if g(x) is the inverse function of f(x) then when it is applied to f(x) it should give us back x.

{{{ f(x) = x^2 + 2x }}}
{{{ f(x) + 1 = x^2 + 2x + 1 }}}
{{{ sqrt(f(x) + 1) = x+1 }}}
{{{ sqrt(f(x)+1) - 1 = x }}}

The LHS is the g(x) we were seeking, just replace "f(x)" with "x", and note that the domain for this x is {{{x>=0}}} because  f(x) >= 0 for x>=0:  
 {{{  highlight(g(x) = -1 + sqrt(x+1))}}}  {{{ x>=0}}}

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Check:  g(f(x)) = {{{ -1 + sqrt((x^2+2x)+1) = -1 + (x+1) = x }}}  (ok)
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