Injection and surjection of a function Let $f:(−1,\infty)\to (−1,\infty)$ be defined by $f(x)=x^2+2x $, study the injection and surjection of $f$ , then find the inverse function if exist . 
So i showed that the function is not $1-1$. my problem is am struggling with showing whether it’s surjective or not i know it’s surjective if the range of the function =the codomain of the function but i dont know how to .. also the inverse doesn’t exist since it’s not one-to-one ?? any help would be appreciated! 
 A: If we think of as a function from $\mathbb R$ to $\mathbb R$ it is neither injective nor surjective.
But, over a restricted domain and co-domain, it could be.  We have to find that domain and co-domain.  Since $f$ is continuous, it can only be 1-1 if is monotonic.  Where is $f$ strictly increasing, or strictly decreasing.  Either one of these will give us a suitable restriction for the domain.
And what is the max and min over this restricted domain?  This will give us the co-domain.
Now you can look for a suitable inverse.
A: To understand the function, complete the squares to write $f(x)=(x+1)^2-1$.
Now try to solve $(x+1)^2-1=c$ for some $c\in (-1, \infty)$.
We see that this has real solutions so long as $c≥-1$ and that if $c>-1$ it has two distinct solutions, given by $x=-1\pm \sqrt {1+c}$.  Since $\sqrt {1+c}>0$ we see that exactly one of these solutions is $>-1$, hence we get both surjectivity and injectivity and we have found an explicit inverse, namely $$f^{-1}(c)=-1+\sqrt {1+c}$$
A: For completeness, I'll prove that the function is bijective.
Injective: Note that for $x_1,x_2\in(-1,\infty)$, if $x_1^2+2x_1=x_2^2+2x_2$, then $x_1^2-x_2^2+2x_1-2x_2=0$, so $(x_1-x_2)(x_1+x_2+2)=0$. Assuming $x_1,x_2$ are distinct, $x_1-x_2\neq0$, so this implies that $x_1+x_2+2=0$, or in other words, $x_1+x_2=-2$. Now, since $x_1,x_2>-1$, $x_1+x_2>-2$, so this is a contradiction, showing injectivity.
Surjective: For any $y>-1$, consider $x=-1+\sqrt{1+y}$. Since $1+y>0$, $\sqrt{1+y}>0$, so we know that $x\in(-1,\infty)$. Moreover, $$(-1+\sqrt{1+y})^2+2(-1+\sqrt{1+y})=1-2\sqrt{1+y}+1+y-2+2\sqrt{1+y}=y$$
A: The function $f(x) = x^2 +2x$ is a bijection if and only if there exists a function $g: (-1, +\infty) \to (-1, +\infty)$ satisfying
$\tag 1 f \circ g = g \circ f = \text{Id}_{(-1, +\infty)}$
Solve for $x$ in terms of $y$:
$\quad y = x^2 + 2x \; \text{ iff } \; x^2 + 2x - y = 0 $
Using the quadratic formula,
$\quad \displaystyle x = \frac{-2 + \sqrt{4 + 4y}}{2} \text{ or } x = \frac{-2 - \sqrt{4 + 4y}}{2} $
Define $g(x): (-1, +\infty) \to (-1, +\infty)$ with
$\tag 2 g(x) =  \displaystyle \sqrt{x + 1} - 1$
checking that $g$ is indeed a transformational mapping of the interval $(-1, +\infty)$.
The function $g$ is a left inverse:
$\quad g \big(f (x) \big) =  \displaystyle \sqrt{(x^2 + 2x) + 1} - 1 = \sqrt{ (x+1)^2}  - 1 = x$
The function $g$ is a right inverse:
$\quad f \big( g (x) \big) =  \displaystyle (\sqrt{x + 1} - 1)^2 +2(\sqrt{x + 1} - 1) = (x+1) - 2\sqrt{x + 1} + 1 + 2\sqrt{x + 1} - 2 = x$
There is nothing left to do.
