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If $f: [0,2] \to [0,2]$ defined by $f(x)=ax^2+bx+c$ is a bijective function, find $f(2)$.

I don’t know how to start solving it. Please help

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closed as off-topic by Namaste, Xander Henderson, Arthur, José Carlos Santos, Parcly Taxel Jul 23 '18 at 1:46

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    $\begingroup$ When you say you have no idea, I don't believe you. I believe you know more than you think. $\endgroup$ – Arthur Jul 22 '18 at 12:18
  • $\begingroup$ Do you know what a bijective function is? $\endgroup$ – Joel Reyes Noche Jul 22 '18 at 12:23
  • $\begingroup$ If $f(x)$ were not monotonic on $[0,2]$ then there would be three points $a<b<c$ in $[0,2]$ such that either $f(a)>f(b)<f(c)$ or $f(a)<f(b)>f(c)$. In either case, by the intermediate value theorem, the value $(f(b)+\min(f(a),f(c)))/2$, or the value $(f(b)+\max(f(a),f(c)))/2$ is taken both in $(a,b)$ and in $b,c$. Therefore, the function wouldn't be injective. This means that $f$ must be monotonic. Therefore, $f(2)=2$ if it is increasing, or $f(2)=0$ if it is decreasing. $\endgroup$ – user574889 Jul 22 '18 at 12:34
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Since you have the algebra-precalculus tag, you may appreciate a less "high-powered" answer.

The graph of a quadratic function is a parabola; its vertex is the extreme value of the function, the function is symmetric about the vertex, and the function is strictly increasing or decreasing on either side of the vertex.

If the vertex of the parabola were not at $x = 0$ or $x = 2$, then $f$ could not be injective, since moving slightly to either side of it would produce equal values. If the vertex were not at $y = 0$ or $y = 2$, then $f$ could not be surjective, since the values above (or below, depending on the direction of $f$) could not be reached.

It is not hard to picture the possibilities now. The parabola can start in the "top-left" corner and go down, the "bottom-left" corner and go up, and so on. This requires $f(2) = 0$ in some cases, and $f(2) = 2$ in other cases.

As a graphical reference, all of the (segments of) parabolas below satisfy the hypotheses of your question.

Drawing of valid parabolas

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$f(2)$ is either $0$ or $2$.

Indeed, $f|_{[0,2)}$ is a continuous bijection of $[0,2)$ and $[0,2] \setminus \{f(2)\}$. The former is a connected set so the latter also must be connected, which isn't true unless $f(2) \in \{0, 2\}$.

Both cases are possible:

For $f(2) = 2$ consider $f(x) = \frac12x^2$ and for $f(2) = 0$ consider $f(x) = 2-\frac12 x^2$.

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