Find the inverse function $x + \sqrt {x}$ $ Y = x + \sqrt {x} $ 
Hello , I want to find the inverse function of this function , I know that it's injective How to prove the $f(x) = \sqrt{x + \sqrt{x}}$ is injective. but do not know how to make the inverse function step by step.

Given $y$, you need to find $x\ge0$ such that $y=x+\sqrt{x}$ or
$$
y-x=\sqrt{x}
$$
A necessary condition is $(y-x)^2=x$, which becomes
$$
y^2-2xy+x^2=x
$$
or
$$
x^2-(2y+1)x+y^2=0
$$
Can you find $x$ from this? Which one of the two solutions have you to consider?
Note that the condition $y-x=\sqrt{x}$ implies $y\ge x$, so the solution you must choose is the only one that satisfies $y\ge x$.

 Since you know that the function is injective, only one solution is acceptable; the two roots are $x_1=\dfrac{2y+1+\sqrt{4y+1}}{2}$ and $x_2=\dfrac{2y+1-\sqrt{4y+1}}{2}$. The first one is clearly greater than $y$, so we have $x=\dfrac{2y+1-\sqrt{4y+1}}{2}$, which is acceptable only for $y\ge0$. Indeed from $\dfrac{2y+1-\sqrt{4y+1}}{2}\le y$ we get $2y+1-\sqrt{4y+1}\le 2y$ or $1\le\sqrt{4y+1}$. So the inverse function is $y\mapsto \dfrac{2y+1-\sqrt{4y+1}}{2}$, defined for $y\ge0$.

