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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?

mauna
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  • The domain is $\mathbb{R}^{+}$, so you have to verify that the formula (1) always returns an answer in $\mathbb{R}^{+}$ – Alex Zorn Jul 11 '14 at 17:46
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    I think the solution is just demonstrating the existence of at least one solution $x$ to $f(x) = y$ for a given $y\geq 1$. That's all you need for surjectivity -- it isn't necessary to find all solutions, just to show that at least one exists. – MPW Jul 11 '14 at 17:58
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    @MPW Why is one solution value sufficient to claim exhaustion of the whole target set? – mvw Jul 11 '14 at 18:29
  • Because instead of a specific number, a generic $y$ was used that stood for an arbitrary choice in the target set, and it was shown that a domain element mapped to it. That shows that every such element is actually an image of something, right? – MPW Jul 11 '14 at 19:04

3 Answers3

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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.

Lubin
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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.

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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.

Fly by Night
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  • Thanks for the detailed explanation. In this case, shouldn't we say $f(0.5-\epsilon) = f(0.5+\epsilon)$ for all $0<\epsilon<0.5$? And also we should say that in this case, there are two $x$ values that give the same $y$ value when $0<y<2$? – mauna Jul 11 '14 at 22:57