Show that $f$ is bijective and also to study the injectivity of the function $g:[1,\infty)\rightarrow \mathbb{R}$, $g(x)=x^{2}+(f(x)-4)x-2f(x)+7$, Let $f:[1,\infty)\rightarrow [1,\infty)$ be a function such that for every $x\in [1,\infty)$, $f(f(x))=2x^{2}-3x+2$. I am required to show that $f$ is bijective and also to study the injectivity of the function $g:[1,\infty)\rightarrow \mathbb{R}$, $g(x)=x^{2}+(f(x)-4)x-2f(x)+7$, for every $x\in\mathbb{R}$.
For the first task I selected  $x,y \in [1,\infty)$ such that $x\neq y$. Then, $f(f(x))=f(f(y))$ iff $2x^{2}-3x=2y^{2}-3y$, meaning that $2x-3=2y-3 \iff x=y$, which is not true. Thus, the function is not injective.
For every $x \in [1,\infty)$, we want to show that there is a $z$  in $[1,\infty)$ such that $z=2x^{2}-3x+2$; because $2x^2-3x+2=2x(x-1)-(x-1)+1=(2x-1)(x-1)+1$ and  $x\geq 1$, then $2x\geq 1$ and $z \geq 1$, so there exists $z \in [1,\infty)$ such that $z=2x^{2}-3x+2$. Thus the function is surjective.
I am quite clueless on how to study the injectivity of the other function, not knowing who $f$ is and what properties does it have.
 A: Your proof that $f$ is (not) injective has a flaw, since $2x^{2}-3x=2y^{2}-3y$ does not imply $2x-3=2y-3$. One can proceed as follows:
If $x, y \in [1, \infty)$ with $f(x) = f(y)$ then
$$
 2x^2-3x+2 = f(f(x)) = f(f(y)) = 2y^2-3y+2 \\
\implies 0 = (2x^2-3x+2) - (2y^2-3y+2) = (x-y)(2x+2y-3) \\
\implies x=y \, .
$$
The last conclusion holds because $2x+2y-3 \ge 1$ so that the second factor is not zero. This shows that $f$ is injective.
Your proof that $f:[1, \infty) \to [1, \infty)$ is surjective is correct, I would write it as follows:
$p(x) =  2x^2-3x+2$ maps $[1, \infty)$ onto $[1, \infty)$. For $z \in [1, \infty)$ there is a $x \in [1, \infty)$ such that $p(x) = z$. Then $y=f(x)$ satisfies
$$
 f(y) = f(f(x)) = p(x) = z \, .
$$
This shows that $f$ is surjective.
A: If $a:=f(f(x))=f(f(y))$ with $x\ne y$, then $x,y$ are two roots of the quadratic polynomial $2X^2-3X+2-a$. By Vieta, their sum is $x+y=\frac 32$, so they cannot both be $\ge 1$. This shows that $f\circ f$ is injective on $[1,\infty)$, hence so is $f$.
Or note that $$f(f(1+h))=2(1+h)^2-3(1+h)+2=1+h+2h^2$$
and read off that $f\circ f$ is a bijection of $[1,\infty)$ to itself.
A: For the second part, notice that $g(x)=(x-2)(f(x)+x-2)+3$.
We have $g(2)=3$.
If one can show that $f(1)=1$, then $g(1)=3$, and therefore $g$ is not injective.
From $f(f(1))=1$ we get $f(f(f(1)))=f(1)$, and then $2f(1)^2-3f(1)+2=f(1)$ which shows that $f(1)=1$.
