Having trouble inverting these functions, I get stuck when trying to solve for y,
$\displaystyle y=x+\sqrt{x}$
and
$\displaystyle y=\frac{1-\sqrt{x}}{1+\sqrt{x}}$
Hello !
$\displaystyle (\sqrt{x})^2+\sqrt{x}-y=0$
$\displaystyle u^2+u-y=0$
Now solve this quadratic equation for u. The root has to be positive. So following the values of y, you'll have two expressions for u.
Then substitute get back to $\displaystyle x=u^2$
And you're done =)
For the second one :
$\displaystyle y(1+\sqrt{x})=1-\sqrt{x}$
$\displaystyle y+y\sqrt{x}=1-\sqrt{x}$
$\displaystyle \sqrt{x}(y+1)=1-y$
$\displaystyle \sqrt{x}=\frac{1-y}{1+y}$
...