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If a quadratic function $[0,2] to [0,2]$ is bijective, find $f(2)$ [closed]
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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
algebra-precalculus
edited Jul 22 at 12:24
mechanodroid
22.2k52041
asked Jul 22 at 12:15
learner_avid
687313
closed as off-topic by amWhy, Xander Henderson, Arthur, José Carlos Santos, Parcly Taxel Jul 23 at 1:46
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РamWhy, Xander Henderson, Arthur, Jos̩ Carlos Santos, Parcly Taxel
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up vote
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down vote
favorite
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
algebra-precalculus
edited Jul 22 at 12:24
mechanodroid
22.2k52041
asked Jul 22 at 12:15
learner_avid
687313
closed as off-topic by amWhy, Xander Henderson, Arthur, José Carlos Santos, Parcly Taxel Jul 23 at 1:46
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РamWhy, Xander Henderson, Arthur, Jos̩ Carlos Santos, Parcly Taxel
1
When you say you have no idea, I don't believe you. I believe you know more than you think.
– Arthur
Jul 22 at 12:18
Do you know what a bijective function is?
– Joel Reyes Noche
Jul 22 at 12:23
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.
– user574889
Jul 22 at 12:34
add a comment |Â
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
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
algebra-precalculus
edited Jul 22 at 12:24
mechanodroid
22.2k52041
asked Jul 22 at 12:15
learner_avid
687313
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
algebra-precalculus
edited Jul 22 at 12:24
mechanodroid
22.2k52041
asked Jul 22 at 12:15
learner_avid
687313
edited Jul 22 at 12:24
mechanodroid
22.2k52041
edited Jul 22 at 12:24
mechanodroid
22.2k52041
edited Jul 22 at 12:24
mechanodroid
22.2k52041
22.2k52041
asked Jul 22 at 12:15
learner_avid
687313
asked Jul 22 at 12:15
learner_avid
687313
asked Jul 22 at 12:15
learner_avid
687313
687313
closed as off-topic by amWhy, Xander Henderson, Arthur, José Carlos Santos, Parcly Taxel Jul 23 at 1:46
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РamWhy, Xander Henderson, Arthur, Jos̩ Carlos Santos, Parcly Taxel
closed as off-topic by amWhy, Xander Henderson, Arthur, José Carlos Santos, Parcly Taxel Jul 23 at 1:46
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РamWhy, Xander Henderson, Arthur, Jos̩ Carlos Santos, Parcly Taxel
1
When you say you have no idea, I don't believe you. I believe you know more than you think.
– Arthur
Jul 22 at 12:18
Do you know what a bijective function is?
– Joel Reyes Noche
Jul 22 at 12:23
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.
– user574889
Jul 22 at 12:34
add a comment |Â
1
When you say you have no idea, I don't believe you. I believe you know more than you think.
– Arthur
Jul 22 at 12:18
Do you know what a bijective function is?
– Joel Reyes Noche
Jul 22 at 12:23
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.
– user574889
Jul 22 at 12:34
1
1
When you say you have no idea, I don't believe you. I believe you know more than you think.
– Arthur
Jul 22 at 12:18
When you say you have no idea, I don't believe you. I believe you know more than you think.
– Arthur
Jul 22 at 12:18
Do you know what a bijective function is?
– Joel Reyes Noche
Jul 22 at 12:23
Do you know what a bijective function is?
– Joel Reyes Noche
Jul 22 at 12:23
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.
– user574889
Jul 22 at 12:34
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.
– user574889
Jul 22 at 12:34
add a comment |Â
2 Answers
2
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oldest
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up vote
1
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accepted
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.
answered Jul 22 at 15:22
rwbogl
716415
add a comment |Â
up vote
5
down vote
$f(2)$ 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 $f(x) = 2-frac12 x^2$.
answered Jul 22 at 12:22
mechanodroid
22.2k52041
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
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.
answered Jul 22 at 15:22
rwbogl
716415
add a comment |Â
up vote
1
down vote
accepted
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.
answered Jul 22 at 15:22
rwbogl
716415
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
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.
answered Jul 22 at 15:22
rwbogl
716415
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.
answered Jul 22 at 15:22
rwbogl
716415
edited Jul 22 at 17:08
answered Jul 22 at 15:22
rwbogl
716415
answered Jul 22 at 15:22
rwbogl
716415
answered Jul 22 at 15:22
rwbogl
716415
716415
add a comment |Â
add a comment |Â
up vote
5
down vote
$f(2)$ 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 $f(x) = 2-frac12 x^2$.
answered Jul 22 at 12:22
mechanodroid
22.2k52041
add a comment |Â
up vote
5
down vote
$f(2)$ 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 $f(x) = 2-frac12 x^2$.
answered Jul 22 at 12:22
mechanodroid
22.2k52041
add a comment |Â
up vote
5
down vote
up vote
5
down vote
$f(2)$ 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 $f(x) = 2-frac12 x^2$.
answered Jul 22 at 12:22
mechanodroid
22.2k52041
$f(2)$ 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 $f(x) = 2-frac12 x^2$.
answered Jul 22 at 12:22
mechanodroid
22.2k52041
answered Jul 22 at 12:22
mechanodroid
22.2k52041
answered Jul 22 at 12:22
mechanodroid
22.2k52041
answered Jul 22 at 12:22
mechanodroid
22.2k52041
22.2k52041
add a comment |Â
add a comment |Â
1
When you say you have no idea, I don't believe you. I believe you know more than you think.
– Arthur
Jul 22 at 12:18
Do you know what a bijective function is?
– Joel Reyes Noche
Jul 22 at 12:23
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.
– user574889
Jul 22 at 12:34