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I really need help on part of a proof that I just don't understand!
Clash Royale CLAN TAG#URR8PPP
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Problem:
a) Show that if $P(x)$ is a polynomial such that $P(a)=P'(a)=0$ then there exists a polynomial $Q(x)$ such that $[P(x)=(x-a)^2Q(x)$.
b) Show that if $P(x)$ is a quartic polynomial then there exists at most one line $ell$ that is tangent to the graph of $P(x)$ at two places.
I got an answer for the first part meaning that I proved that Q(x) exists. I don't understand how I can prove how there is only one tangent line to the polynomial P(x).
calculus derivatives proof-verification proof-writing proof-explanation
asked yesterday
mathperson1234
62
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up vote
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down vote
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Problem:
a) Show that if $P(x)$ is a polynomial such that $P(a)=P'(a)=0$ then there exists a polynomial $Q(x)$ such that $[P(x)=(x-a)^2Q(x)$.
b) Show that if $P(x)$ is a quartic polynomial then there exists at most one line $ell$ that is tangent to the graph of $P(x)$ at two places.
I got an answer for the first part meaning that I proved that Q(x) exists. I don't understand how I can prove how there is only one tangent line to the polynomial P(x).
calculus derivatives proof-verification proof-writing proof-explanation
asked yesterday
mathperson1234
62
Can you first show that the limit of a function (if exists) is unique ? Then can you also show that in $mathbbR^2$, if you know the slope of a line and a point that it passes, then you can determine that line uniquely ?
– onurcanbektas
yesterday
2
Welcome to MSE. Please read this text about how to ask a good question.
– José Carlos Santos
yesterday
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favorite
up vote
1
down vote
favorite
Problem:
a) Show that if $P(x)$ is a polynomial such that $P(a)=P'(a)=0$ then there exists a polynomial $Q(x)$ such that $[P(x)=(x-a)^2Q(x)$.
b) Show that if $P(x)$ is a quartic polynomial then there exists at most one line $ell$ that is tangent to the graph of $P(x)$ at two places.
I got an answer for the first part meaning that I proved that Q(x) exists. I don't understand how I can prove how there is only one tangent line to the polynomial P(x).
calculus derivatives proof-verification proof-writing proof-explanation
asked yesterday
mathperson1234
62
Problem:
a) Show that if $P(x)$ is a polynomial such that $P(a)=P'(a)=0$ then there exists a polynomial $Q(x)$ such that $[P(x)=(x-a)^2Q(x)$.
b) Show that if $P(x)$ is a quartic polynomial then there exists at most one line $ell$ that is tangent to the graph of $P(x)$ at two places.
I got an answer for the first part meaning that I proved that Q(x) exists. I don't understand how I can prove how there is only one tangent line to the polynomial P(x).
calculus derivatives proof-verification proof-writing proof-explanation
asked yesterday
mathperson1234
62
asked yesterday
mathperson1234
62
asked yesterday
mathperson1234
62
asked yesterday
mathperson1234
62
62
Can you first show that the limit of a function (if exists) is unique ? Then can you also show that in $mathbbR^2$, if you know the slope of a line and a point that it passes, then you can determine that line uniquely ?
– onurcanbektas
yesterday
2
Welcome to MSE. Please read this text about how to ask a good question.
– José Carlos Santos
yesterday
add a comment |Â
Can you first show that the limit of a function (if exists) is unique ? Then can you also show that in $mathbbR^2$, if you know the slope of a line and a point that it passes, then you can determine that line uniquely ?
– onurcanbektas
yesterday
2
Welcome to MSE. Please read this text about how to ask a good question.
– José Carlos Santos
yesterday
Can you first show that the limit of a function (if exists) is unique ? Then can you also show that in $mathbbR^2$, if you know the slope of a line and a point that it passes, then you can determine that line uniquely ?
– onurcanbektas
yesterday
Can you first show that the limit of a function (if exists) is unique ? Then can you also show that in $mathbbR^2$, if you know the slope of a line and a point that it passes, then you can determine that line uniquely ?
– onurcanbektas
yesterday
2
2
Welcome to MSE. Please read this text about how to ask a good question.
– José Carlos Santos
yesterday
Welcome to MSE. Please read this text about how to ask a good question.
– José Carlos Santos
yesterday
add a comment |Â
1 Answer
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Suppose $l$ and $k$ are lines, tangent to $p$ at $a,b$ and $c,d$, respectively.
By part (a) of the problem,
beginalign*
p(x)-k(x)=(x-a)^2(x-b)^2 \
p(x)-l(x)=(x-c)^2(x-d)^2.
endalign*
Subtracting one equation from the other we eliminate $p$ and obtain
beginequation*
l(x)-k(x) = (x-a)^2(x-b)^2 - (x-c)^2(x-d)^2
endequation*
But the difference of two degree 1 polynomials has degree $leq$ 1. The degree of the RHS is $leq 1$ only if $a,b = c,d$, i.e., only if $l(x)equiv k(x)$.
answered yesterday
user581726
12
thank you very much for your answer! However, since this is my first year taking calculus, would you mind explaining it in less complicated terms? for example, what is RHS? And I assume when you wrote that l(x) ≡ k(x), you meant that they are congruent to one another?
– mathperson1234
yesterday
Do you understand that if $l(x)$ and $k(x)$ are equations of lines, then the difference $l(x)-k(x)$ is also the equation of a line? The right hand side (RHS) of the equation “$l(x)-k(x) = ... $†must be the equation of a line, namely, $l(x)-k(x) = mx + n$ for some constants $m$ and $n$. If you expand the RHS and set the coefficients of the quadratic and cubic terms equal to zero, you should be able to see that these restrictions imply that $k$ and $l$ intersect $p$ at precisely the same points, namely, $a,b=c,d$. Now, what can you say about two lines that share two points in common?
– user581726
yesterday
Right, I understood the first part, and thanks for clarifying what RHS means (can't believe I missed that). Regarding your last question, I believe that would simply mean that the lines intersect, am I correct? Note also in the question I'm trying to prove that there can at most be ONE line tangent to the quartic. I thought the best way to do this was to first assume that such a line exists, and use that to prove that there can't be any other line that fits the criteria. In other words, by completely proving one exists, I can eliminate all other possibilities.
– mathperson1234
yesterday
If $l$ and $k$ are lines that intersect twice, then they are in fact the same line. Can you see why this is the case? Your idea of proof is essentially what is given in the answer above: given two lines, $k$ and $l$ that intersect $p$ in two places, show that they are in fact the same line, $k(x) equiv l(x)$. This is achieved by showing that they must be tangent to $p$ at the same two points.
– user581726
yesterday
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Suppose $l$ and $k$ are lines, tangent to $p$ at $a,b$ and $c,d$, respectively.
By part (a) of the problem,
beginalign*
p(x)-k(x)=(x-a)^2(x-b)^2 \
p(x)-l(x)=(x-c)^2(x-d)^2.
endalign*
Subtracting one equation from the other we eliminate $p$ and obtain
beginequation*
l(x)-k(x) = (x-a)^2(x-b)^2 - (x-c)^2(x-d)^2
endequation*
But the difference of two degree 1 polynomials has degree $leq$ 1. The degree of the RHS is $leq 1$ only if $a,b = c,d$, i.e., only if $l(x)equiv k(x)$.
answered yesterday
user581726
12
thank you very much for your answer! However, since this is my first year taking calculus, would you mind explaining it in less complicated terms? for example, what is RHS? And I assume when you wrote that l(x) ≡ k(x), you meant that they are congruent to one another?
– mathperson1234
yesterday
Do you understand that if $l(x)$ and $k(x)$ are equations of lines, then the difference $l(x)-k(x)$ is also the equation of a line? The right hand side (RHS) of the equation “$l(x)-k(x) = ... $†must be the equation of a line, namely, $l(x)-k(x) = mx + n$ for some constants $m$ and $n$. If you expand the RHS and set the coefficients of the quadratic and cubic terms equal to zero, you should be able to see that these restrictions imply that $k$ and $l$ intersect $p$ at precisely the same points, namely, $a,b=c,d$. Now, what can you say about two lines that share two points in common?
– user581726
yesterday
Right, I understood the first part, and thanks for clarifying what RHS means (can't believe I missed that). Regarding your last question, I believe that would simply mean that the lines intersect, am I correct? Note also in the question I'm trying to prove that there can at most be ONE line tangent to the quartic. I thought the best way to do this was to first assume that such a line exists, and use that to prove that there can't be any other line that fits the criteria. In other words, by completely proving one exists, I can eliminate all other possibilities.
– mathperson1234
yesterday
If $l$ and $k$ are lines that intersect twice, then they are in fact the same line. Can you see why this is the case? Your idea of proof is essentially what is given in the answer above: given two lines, $k$ and $l$ that intersect $p$ in two places, show that they are in fact the same line, $k(x) equiv l(x)$. This is achieved by showing that they must be tangent to $p$ at the same two points.
– user581726
yesterday
add a comment |Â
up vote
0
down vote
Suppose $l$ and $k$ are lines, tangent to $p$ at $a,b$ and $c,d$, respectively.
By part (a) of the problem,
beginalign*
p(x)-k(x)=(x-a)^2(x-b)^2 \
p(x)-l(x)=(x-c)^2(x-d)^2.
endalign*
Subtracting one equation from the other we eliminate $p$ and obtain
beginequation*
l(x)-k(x) = (x-a)^2(x-b)^2 - (x-c)^2(x-d)^2
endequation*
But the difference of two degree 1 polynomials has degree $leq$ 1. The degree of the RHS is $leq 1$ only if $a,b = c,d$, i.e., only if $l(x)equiv k(x)$.
answered yesterday
user581726
12
thank you very much for your answer! However, since this is my first year taking calculus, would you mind explaining it in less complicated terms? for example, what is RHS? And I assume when you wrote that l(x) ≡ k(x), you meant that they are congruent to one another?
– mathperson1234
yesterday
Do you understand that if $l(x)$ and $k(x)$ are equations of lines, then the difference $l(x)-k(x)$ is also the equation of a line? The right hand side (RHS) of the equation “$l(x)-k(x) = ... $†must be the equation of a line, namely, $l(x)-k(x) = mx + n$ for some constants $m$ and $n$. If you expand the RHS and set the coefficients of the quadratic and cubic terms equal to zero, you should be able to see that these restrictions imply that $k$ and $l$ intersect $p$ at precisely the same points, namely, $a,b=c,d$. Now, what can you say about two lines that share two points in common?
– user581726
yesterday
Right, I understood the first part, and thanks for clarifying what RHS means (can't believe I missed that). Regarding your last question, I believe that would simply mean that the lines intersect, am I correct? Note also in the question I'm trying to prove that there can at most be ONE line tangent to the quartic. I thought the best way to do this was to first assume that such a line exists, and use that to prove that there can't be any other line that fits the criteria. In other words, by completely proving one exists, I can eliminate all other possibilities.
– mathperson1234
yesterday
If $l$ and $k$ are lines that intersect twice, then they are in fact the same line. Can you see why this is the case? Your idea of proof is essentially what is given in the answer above: given two lines, $k$ and $l$ that intersect $p$ in two places, show that they are in fact the same line, $k(x) equiv l(x)$. This is achieved by showing that they must be tangent to $p$ at the same two points.
– user581726
yesterday
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Suppose $l$ and $k$ are lines, tangent to $p$ at $a,b$ and $c,d$, respectively.
By part (a) of the problem,
beginalign*
p(x)-k(x)=(x-a)^2(x-b)^2 \
p(x)-l(x)=(x-c)^2(x-d)^2.
endalign*
Subtracting one equation from the other we eliminate $p$ and obtain
beginequation*
l(x)-k(x) = (x-a)^2(x-b)^2 - (x-c)^2(x-d)^2
endequation*
But the difference of two degree 1 polynomials has degree $leq$ 1. The degree of the RHS is $leq 1$ only if $a,b = c,d$, i.e., only if $l(x)equiv k(x)$.
answered yesterday
user581726
12
Suppose $l$ and $k$ are lines, tangent to $p$ at $a,b$ and $c,d$, respectively.
By part (a) of the problem,
beginalign*
p(x)-k(x)=(x-a)^2(x-b)^2 \
p(x)-l(x)=(x-c)^2(x-d)^2.
endalign*
Subtracting one equation from the other we eliminate $p$ and obtain
beginequation*
l(x)-k(x) = (x-a)^2(x-b)^2 - (x-c)^2(x-d)^2
endequation*
But the difference of two degree 1 polynomials has degree $leq$ 1. The degree of the RHS is $leq 1$ only if $a,b = c,d$, i.e., only if $l(x)equiv k(x)$.
answered yesterday
user581726
12
answered yesterday
user581726
12
answered yesterday
user581726
12
answered yesterday
user581726
12
12
thank you very much for your answer! However, since this is my first year taking calculus, would you mind explaining it in less complicated terms? for example, what is RHS? And I assume when you wrote that l(x) ≡ k(x), you meant that they are congruent to one another?
– mathperson1234
yesterday
Do you understand that if $l(x)$ and $k(x)$ are equations of lines, then the difference $l(x)-k(x)$ is also the equation of a line? The right hand side (RHS) of the equation “$l(x)-k(x) = ... $†must be the equation of a line, namely, $l(x)-k(x) = mx + n$ for some constants $m$ and $n$. If you expand the RHS and set the coefficients of the quadratic and cubic terms equal to zero, you should be able to see that these restrictions imply that $k$ and $l$ intersect $p$ at precisely the same points, namely, $a,b=c,d$. Now, what can you say about two lines that share two points in common?
– user581726
yesterday
Right, I understood the first part, and thanks for clarifying what RHS means (can't believe I missed that). Regarding your last question, I believe that would simply mean that the lines intersect, am I correct? Note also in the question I'm trying to prove that there can at most be ONE line tangent to the quartic. I thought the best way to do this was to first assume that such a line exists, and use that to prove that there can't be any other line that fits the criteria. In other words, by completely proving one exists, I can eliminate all other possibilities.
– mathperson1234
yesterday
If $l$ and $k$ are lines that intersect twice, then they are in fact the same line. Can you see why this is the case? Your idea of proof is essentially what is given in the answer above: given two lines, $k$ and $l$ that intersect $p$ in two places, show that they are in fact the same line, $k(x) equiv l(x)$. This is achieved by showing that they must be tangent to $p$ at the same two points.
– user581726
yesterday
add a comment |Â
thank you very much for your answer! However, since this is my first year taking calculus, would you mind explaining it in less complicated terms? for example, what is RHS? And I assume when you wrote that l(x) ≡ k(x), you meant that they are congruent to one another?
– mathperson1234
yesterday
Do you understand that if $l(x)$ and $k(x)$ are equations of lines, then the difference $l(x)-k(x)$ is also the equation of a line? The right hand side (RHS) of the equation “$l(x)-k(x) = ... $†must be the equation of a line, namely, $l(x)-k(x) = mx + n$ for some constants $m$ and $n$. If you expand the RHS and set the coefficients of the quadratic and cubic terms equal to zero, you should be able to see that these restrictions imply that $k$ and $l$ intersect $p$ at precisely the same points, namely, $a,b=c,d$. Now, what can you say about two lines that share two points in common?
– user581726
yesterday
Right, I understood the first part, and thanks for clarifying what RHS means (can't believe I missed that). Regarding your last question, I believe that would simply mean that the lines intersect, am I correct? Note also in the question I'm trying to prove that there can at most be ONE line tangent to the quartic. I thought the best way to do this was to first assume that such a line exists, and use that to prove that there can't be any other line that fits the criteria. In other words, by completely proving one exists, I can eliminate all other possibilities.
– mathperson1234
yesterday
If $l$ and $k$ are lines that intersect twice, then they are in fact the same line. Can you see why this is the case? Your idea of proof is essentially what is given in the answer above: given two lines, $k$ and $l$ that intersect $p$ in two places, show that they are in fact the same line, $k(x) equiv l(x)$. This is achieved by showing that they must be tangent to $p$ at the same two points.
– user581726
yesterday
thank you very much for your answer! However, since this is my first year taking calculus, would you mind explaining it in less complicated terms? for example, what is RHS? And I assume when you wrote that l(x) ≡ k(x), you meant that they are congruent to one another?
– mathperson1234
yesterday
thank you very much for your answer! However, since this is my first year taking calculus, would you mind explaining it in less complicated terms? for example, what is RHS? And I assume when you wrote that l(x) ≡ k(x), you meant that they are congruent to one another?
– mathperson1234
yesterday
Do you understand that if $l(x)$ and $k(x)$ are equations of lines, then the difference $l(x)-k(x)$ is also the equation of a line? The right hand side (RHS) of the equation “$l(x)-k(x) = ... $†must be the equation of a line, namely, $l(x)-k(x) = mx + n$ for some constants $m$ and $n$. If you expand the RHS and set the coefficients of the quadratic and cubic terms equal to zero, you should be able to see that these restrictions imply that $k$ and $l$ intersect $p$ at precisely the same points, namely, $a,b=c,d$. Now, what can you say about two lines that share two points in common?
– user581726
yesterday
Do you understand that if $l(x)$ and $k(x)$ are equations of lines, then the difference $l(x)-k(x)$ is also the equation of a line? The right hand side (RHS) of the equation “$l(x)-k(x) = ... $†must be the equation of a line, namely, $l(x)-k(x) = mx + n$ for some constants $m$ and $n$. If you expand the RHS and set the coefficients of the quadratic and cubic terms equal to zero, you should be able to see that these restrictions imply that $k$ and $l$ intersect $p$ at precisely the same points, namely, $a,b=c,d$. Now, what can you say about two lines that share two points in common?
– user581726
yesterday
Right, I understood the first part, and thanks for clarifying what RHS means (can't believe I missed that). Regarding your last question, I believe that would simply mean that the lines intersect, am I correct? Note also in the question I'm trying to prove that there can at most be ONE line tangent to the quartic. I thought the best way to do this was to first assume that such a line exists, and use that to prove that there can't be any other line that fits the criteria. In other words, by completely proving one exists, I can eliminate all other possibilities.
– mathperson1234
yesterday
Right, I understood the first part, and thanks for clarifying what RHS means (can't believe I missed that). Regarding your last question, I believe that would simply mean that the lines intersect, am I correct? Note also in the question I'm trying to prove that there can at most be ONE line tangent to the quartic. I thought the best way to do this was to first assume that such a line exists, and use that to prove that there can't be any other line that fits the criteria. In other words, by completely proving one exists, I can eliminate all other possibilities.
– mathperson1234
yesterday
If $l$ and $k$ are lines that intersect twice, then they are in fact the same line. Can you see why this is the case? Your idea of proof is essentially what is given in the answer above: given two lines, $k$ and $l$ that intersect $p$ in two places, show that they are in fact the same line, $k(x) equiv l(x)$. This is achieved by showing that they must be tangent to $p$ at the same two points.
– user581726
yesterday
If $l$ and $k$ are lines that intersect twice, then they are in fact the same line. Can you see why this is the case? Your idea of proof is essentially what is given in the answer above: given two lines, $k$ and $l$ that intersect $p$ in two places, show that they are in fact the same line, $k(x) equiv l(x)$. This is achieved by showing that they must be tangent to $p$ at the same two points.
– user581726
yesterday
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Can you first show that the limit of a function (if exists) is unique ? Then can you also show that in $mathbbR^2$, if you know the slope of a line and a point that it passes, then you can determine that line uniquely ?
– onurcanbektas
yesterday
2
Welcome to MSE. Please read this text about how to ask a good question.
– José Carlos Santos
yesterday