Is $4x^2-4x+2$ surjective? Determine whether the function $f_4:\mathbb{R^+}\rightarrow \{x \in \mathbb{R^+} x \ge 1\}$ given by $f_4(x)=4x^2-4x+2$ is injective, surjective or bijective.
I will just show parts of the solution I don't understand.
The formula for the preimage is:
$x = (1 \pm \sqrt{y-1})/2 , \tag{1}$

...if $y$ is in the codomain then $y\ge 1$ and so the square root in the formula for a pre-image element does determine a real number. Furthermore $(1+\sqrt{y-1}/2)>\frac{1}{2}$ and so this gives one pre-image for $y$ showing that $f_4$ is surjective.

I don't understand why we need to specifically consider $(1+\sqrt{y-1}/2)>\frac{1}{2}$ to determine whether $f_4$ is surjective. I would have just looked at equation (1) and note that it is defined for all $y \in \mathbb{R^+}$. Hence, all the elements in the codomain is assigned a value in the domain. Therefore, $f_4$ is surjective. Is this answer acceptable?
 A: My approach would be to draw the picture, even if only in my mind. Ignoring temporarily the shrunken domain and codomain, we have a parabola opening upwards, with minimum at $x=1/2$ (by locating the axis as the vertical line $x=-\frac12b/a$). The minimum value you see to be $1$. Now considering the specific domain and codomain, we see that our whole graph is to lie in the quarter-space $x\ge0$, $y\ge1$. No problem, none of the graph we drew lies below the lower boundary. All values in the codomain clearly occur as function values, voilà, surjective.
A: We have $f'_4(x)=8x-4=0\iff x=\frac12$ hence the function $f_4$ is monotonically decreasing on $\left[0,\frac12\right]$ and monotonically increasing on $\left[\frac12,\infty\right)$ so this function isn't injective. Moreover we have $f_4\left(\frac12\right)=1$ and $\lim\limits_{x\to\infty}f_4(x)=+\infty$ so by the continuity we have
$$f\left(\left[\frac12,\infty\right)\right)=[1,+\infty)$$
and then the function is surjective.
A: The easiest way to find the range of a quadratic polynomial is to complete the square:
$$\begin{eqnarray*}
4x^2-4x+2 &\equiv& 4\left(x^2 - x + \tfrac{1}{2}\right) \\ \\
&\equiv& 4\left(\left[x-\tfrac{1}{2}\right]^{\! 2}+\tfrac{1}{4}\right) \\ \\
&\equiv& 4\left(x-\tfrac{1}{2}\right)^{\! 2}+1
\end{eqnarray*}$$
The graph $y=4x^2-4x+2$ has a turning point at $(x,y)=\left(\tfrac{1}{2},1\right)$. Moreover, since the $x^2$-coefficient is positive, the graph is a $\cup$-shape and the turning point is a global minimum. The line of symmetry of the parabola has equation $x=\tfrac{1}{2}$.
In your case, you have $x>0$. The line of symmetry is $x=\tfrac{1}{2}$ and so the function $\mathrm{f}(x)=4x^2-4x+2$ is not injective. We have $\mathrm{f}(0.5-\varepsilon)=\mathrm{f}(0.5+\varepsilon)$ for all $\varepsilon$. For example:
$$\mathrm{f}(0.5-0.1)=\mathrm{f}(0.4)=1.04$$
$$\mathrm{f}(0.5+0.1)=\mathrm{f}(0.6)=1.04$$
The function is surjective onto $y\ge 1$. To see this, try to solve $y=4x^2-4x+2$ for $x$:
$$x=\frac{1}{2}\pm\frac{1}{2}\sqrt{y-1}$$
This has real solutions for all $y \ge 1$. Moreover, it again show that there are two $x$ values that give the same $y$ value when $y > 0$, i.e. $\mathrm{f}(x)=4x^2-4x+2$ is not injective.
It is not bijective because it is not injective.
