Is this a function and does F have an inverse? $F: 4x^2 + 8x + 7 \rightarrow 8x + 8$
is $F$ an function that is onto, one-to-one?
does $F$ have an inverse?
I don't know if it's a function, I guess the variable would be $x$, but it doesn't have an inverse because the integral of $8x + 8$ is: $4x^2 + 8x + C$
I think the function is not onto for $\mathbb{Z}$, but is onto for $\mathbb{R}$.
and not one to one because when $x=-2$ or $x = 2 y$ has the same value.
 A: $F$ fails to even be a function. To see this, note that $4x^2+8x+7=7$ for both $x=-2$ and $x=0$, but that $8x+8$ does not take the same value at these two $x$-values. If $F$ were a function, then we'd have $$-8=8(-2)+8=F\bigl(4(-2)^2+8(-2)+7\bigr)=F(7)=F\bigl(4(0)^2+8(0)+7\bigr)=8(0)+8=8.$$ But $-8\neq 8$, so that doesn't work.
Edit: When you write $F:4x^2+8x+7\to 8x+8$, that's very unclear notation.
Did you mean that it takes a point $(x,y)$ on the curve $y=4x^2+8x+7$ to the point on the curve $y=8x+8$ with the same $x$-coordinate? If so, it would be clearer to say "$F$ is the map that takes $(x,4x^2+8x+7)$ to $(x,8x+8)$." That would be a function. Your domain would then be $\bigl\{(x,4x^2+8x+7):x\in\Bbb R\bigr\}$ (or perhaps $\Bbb R$ would be replaced by something else). It would also be one-to-one, and might even be onto depending on your codomain (see "Important Points" below).
I was initially interpreting it as follows: given $y\in\Bbb R$ such that $y=4x^2+8x+7$ for some $x\in\Bbb R$, $F$ takes $y$ to $8x+8$. In that context, my previous discussion makes sense.
Upon rereading your post, it seems that you're treating $F$ as the function with domain $C^1(\Bbb R)$--that is, the functions on $\Bbb R$ with continuous first derivatives--given by $F(f)=f'$. That is a function (to see that, assume $f=g$ and show $F(f)=F(g)$). You're right that it isn't one-to-one, but I don't understand your argument for why this is the case. Instead, take $C$ to be any non-$0$ constant, $f$ a function on $\Bbb R$ with continuous first derivative, $g(x)=f(x)+C$ (so $f\neq g$), and show that $F(f)=F(g)$. Since it isn't one-to-one, then it can't have an inverse.

IMPORTANT POINTS:

(i) Typically, when we write $f:A\to B$ (in words, "$f$ is a function from $A$ to $B$"), we are taking $A$ and $B$ to be sets. $A$ is called the domain of $f$; $B$, the codomain of $f$. That's why what you wrote is so confusing.
(ii) In order to tell if a particular rule (method of calculation) gives us a function, we also need a specified domain. For example, let $h(\frac p q)=p$. If we take the domain to be the set of all $\frac p q$ such that $p$ is an integer and $q$ is a positive integer, then this fails to be a function--consider $\frac p q=\frac21$ and $\frac p q=\frac42$, which are equal, but taken to different places. On the other hand, if we take the domain to be the set of all $\frac p q$ with $p$ an integer and $q=3$, then it is a function.
(iii) In order to tell if a function is one-to-one, we need more than just a rule--we also need a specified domain. For example, consider the function $F(f)=f'$, but instead, let the domain be the set of all functions $f$ on $\Bbb R$ with continuous derivatives, such that $f(0)=0$. That one does turn out to be one-to-one, though without the requirement that $f(0)=0$ (or something similar), it wouldn't be.
(iv) In order to tell if a function is onto, we need a specified domain and a specified codomain. For example, consider the rule $g(x)=x^2$. If we take the domain and codomain both to be $\Bbb R$, then $g$ fails to be onto, since (for instance) there is no $x$ such that $g(x)=-1$. If we take the domain to be $\Bbb R$ and codomain to be the nonnegative real numbers, then $g$ is onto. If we take the domain to be $\Bbb Z$ and codomain to be the nonnegative real numbers, then $g$ isn't onto.
(v) Saying a function "has an inverse" can mean different things. Please specify what you mean by that.


The moral to the story is: please try to be as clear as possible how you write things, and give as much information as you can, to make it easier for us to help you. We're happy to help you out, but we'll be better equipped to do that if you make it clear what you're talking about.
Feel free to leave a comment on my answer asking for clarifications (or giving clarifications on your post so that I can give you a definitive answer).
