Show that for $f(x)=x^2+x$ the mapping $f: A \rightarrow B$ is a bijection. Here is a problem I am working on and a few parts are stumping me.
Problem
Assume $A=[-\frac{1}{2},\infty), B=[-\frac{1}{4},\infty)$, where, for $a \in \mathbb{R}, [a,\infty)=\{x\in\mathbb{R}|x\geq a\}.$ Consider $f:A \rightarrow B$ defined by $f(x)=x^2+x, x\in A.$


*

*Show that $f$ is indeed into $B$, i.e. show that if $x \in A$ then $f(x) \in B$.

*Show that $f$ is one to one.

*Show that $f$ maps onto $B$, i.e. show that for $y\in B$ there is $x \in A$ such that $f(x)=y$. For your choice of $x$, verify directly that $f(x)=y$.
My attempt


*

*I originally thought I would have to do something complicated to do this but at this point I am thinking it is fairly simple. Is it not true that $f:A \rightarrow B$ by definition implies that any element from $A$ is mapped to an element of $B$? If so I would imagine there is no need for an explanation other than citing the definition, but at the same time I am a bit hesitant as it seems like perhaps there is a way to verify this further.

*I start by letting $x_1,x_2 \in [-\frac{1}{2},\infty)$ such that $f(x_1)=f(x_2)$. Then we can say that $x_1^2+x_1=x_2^2+x_2$. The goal is to show that $x_1=x_2$, but I am stuck on the algebra on how to show that. I tried factoring in multiple ways, such as
$$x_1(x_1+1)=x_2(x_2+1)$$
and
$$(x_1-x_2)(x_1+x_2)+x_1-x_2=0$$
but nothing is jumping out at me to simplify to $x_1=x_2$.

*Solving $y=x^2+x$ for $x$ we get $x=\pm \sqrt{y+\frac{1}{4}}-\frac{1}{2}.$ If we then let $y\in B$, we see that $y\geq-\frac{1}{4}$. If we apply our previous function of $y$ to our inequality we get that $\sqrt{y+\frac{1}{4}}-\frac{1}{2}\geq-\frac{1}{2}$, telling us that $x=\sqrt{y+\frac{1}{4}}-\frac{1}{2}\in A$. I verify this by showing if $x=-\frac{1}{2}$ then $y=-\frac{1}{4}$ and that if $y=-\frac{1}{4}$ then $x=-\frac{1}{2}$. Is this sufficient?
 A: An elementary way.
(1) To prove $f$ is injective, let $x,y\in A$ such that $x \neq y$. If $x^{2}+x = y^{2}+y$, then $(x-y)(x+y) = y-x$, that is, we have $x+y = -1$. But this is impossible since $x\neq y$ and $x,y \geq -1/2$. Thus $f: A \to B$ is injective.
(2) To show $f$ is surjective, let $z \in B$. I claim that there is some $x \in A$ such that $f(x) = x^{2}+x = z$. But, since $x^{2}+x = z$ for some $x \in A$ if and only if $x = \frac{-1}{2} \pm \sqrt{1+4z}$, and since $z \geq -1/4$, so $x \geq -1/2$, and we are done.
A: For part A,  they want you to confirm the range is accurate, that whatever you put in from $A$ will actually take you to $B$ and not to some other part of $\mathbb R$.   For that,  you can just use the derivative:  $f'(x)=2x+1$,  hence on $A$,  $f'(x)\ge 0$,  hence it is increasing, and the smallest value is at $f(\frac {-1} 2)=\frac {-1} {4}$
For B: Again use the derivative, $f'(x)>0$  for all but the starting point says that the function is strictly increasing, and strictly increasing functions are 1-1.
C:  Yes, that's pretty much it, you show that for any number $y\in B$ there is a preimage in $A$
Non calculus versions;  A:  This is the right half of the parabola,  the vertex is at $-\frac 1 2$,  so your range is the vertex to infinity (Same for C, honestly)
B:  This is the right half of a parabola,  so it passes the "horizontal line" test of a 1-1 function.
A: $x^2+x=y^2+y$ then $(x+\frac{1}{2})^2=(y+\frac{1}{2})^2$ thus $(x+\frac{1}{2})=(y+\frac{1}{2})$ that will give $x=y$ or $(x+\frac{1}{2})=-(y+\frac{1}{2})$ which gives $x+y=-1$ because of domain it gives $x=y=\dfrac{-1}{2}$
A: *

*You have to check that if you take something from $A$, you'll get in $B$, a priori, this application ends in $\mathbb{R}$, but you can narrow it down to $B$.

*Good attempt, but it doesn't seem to work. Compute the derivative (it works).

*I don't really understand what you are doing. Take $y \in B$. For $y\in B-\{\frac{1}{4}\}$ there is two real solutions to the equation $x^2 + x - y = 0$ (check it!). Only one of them is in $B$. 
You've got to solve the previous equation, take the one in $B$, then apply $f$ to that solution and check if it works.
Then Work on the case $y=\frac{1}{4}$ separately as you've done in your attempt.
EDIT : Oh alright I now understand what you've done in 3... Yea that's it, I just find the way you wrote it a bit weird.
