Finding number of roots using Rolle's Theorem, and depending on parameter The Next CEO of Stack OverflowProve there are 3 real roots to this equation using Rolle's TheoremTwo factored polynomials and a parameterFinding the scope of a parameter where a polynomial can have rootsFinding all complex roots of an equation with exponentials.A stronger form of Rolle's Theorem in the direction of number of roots of $f'(x)$Finding the Roots of this CubicFind number of real roots using Sturm's method.The number of real roots of continuous $f(x)$Using Rolle's Theorem to prove roots.Finding the real roots and complex roots to a polynomial
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Finding number of roots using Rolle's Theorem, and depending on parameter
The Next CEO of Stack OverflowProve there are 3 real roots to this equation using Rolle's TheoremTwo factored polynomials and a parameterFinding the scope of a parameter where a polynomial can have rootsFinding all complex roots of an equation with exponentials.A stronger form of Rolle's Theorem in the direction of number of roots of $f'(x)$Finding the Roots of this CubicFind number of real roots using Sturm's method.The number of real roots of continuous $f(x)$Using Rolle's Theorem to prove roots.Finding the real roots and complex roots to a polynomial
$begingroup$
I need to count the number of real solutions for $ f(x) = 0 $ but I have an $m$ in there.
$$ f(x) = x^3+3x^2-mx+5 $$
I know I need to study $m$ to get the number of roots, but I don't know where to begin. Any suggestions?
calculus polynomials roots
$endgroup$
add a comment |
$begingroup$
I need to count the number of real solutions for $ f(x) = 0 $ but I have an $m$ in there.
$$ f(x) = x^3+3x^2-mx+5 $$
I know I need to study $m$ to get the number of roots, but I don't know where to begin. Any suggestions?
calculus polynomials roots
$endgroup$
add a comment |
$begingroup$
I need to count the number of real solutions for $ f(x) = 0 $ but I have an $m$ in there.
$$ f(x) = x^3+3x^2-mx+5 $$
I know I need to study $m$ to get the number of roots, but I don't know where to begin. Any suggestions?
calculus polynomials roots
$endgroup$
I need to count the number of real solutions for $ f(x) = 0 $ but I have an $m$ in there.
$$ f(x) = x^3+3x^2-mx+5 $$
I know I need to study $m$ to get the number of roots, but I don't know where to begin. Any suggestions?
calculus polynomials roots
calculus polynomials roots
edited Mar 22 '16 at 19:18
Rory Daulton
29.5k63355
29.5k63355
asked Mar 22 '16 at 18:58
Vlad PotraVlad Potra
373
373
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Taking the derivative of the function: $$f'(x)=3x^2+6x-m$$
If the function $f$ has two roots with $x=a$ and $x=b$, then $$f(a)=0=f(b)$$
Since $f(x)$ is continuous and differentiable, by Rolle's theorem, we have that there exists at least one $c in (a,b)$ such that $$f'(c)=0$$
Namely, $$f'(c)=3c^2+6c-m=0$$
And here is where you begin to consider how the roots exist depend on the parameter.
New contributor
$endgroup$
add a comment |
$begingroup$
The theme is that between any two roots, the derivative has a root. This comes directly from Rolle's Theorem.
So one way to get a first understanding of the number of roots is to find the extrema. This leads you to consider
$$ f'(x) = 3x^2 + 6x - m = 0.$$
As this is a quadratic, you can very quickly write down the two roots,
$$ x = frac-6 pm sqrt36 + 12m6.$$
When $m < -3$, there are no roots of the derivative, and correspondingly there can be at most one root of $f(x)$. As $f to -infty$ as $x to -infty$ and $f to infty$ as $x to infty$, we know that $f$ always has at least one root. So for $m < -3$, we know that $f$ has exactly one root.
More generally, you can determine the number of roots by classifying the extrema. Denote the two roots by $r_1$ and $r_2$ with $r_1 < r_2$ (which both depend on $m$). When $m > -3$, these are two distinct real numbers. One way to understand more roots is to see when $f(r_1) > 0$ and $f(r_2) < 0$.
$endgroup$
add a comment |
$begingroup$
You can use the discriminant of the cubic function to find the nature of the roots. The general cubic equation is
$$ax^3+bx^2+cx+d=0$$
so in your case
$$a=1, b=3, c=-m, d=5$$
In general, the discriminant is
$$Delta=18abcd-4b^3d+b^2c^2-4ac^3-27a^2d^2$$
You can substitute in your values and get a cubic expression in $m$. Then your equation has
- three distinct real roots, if $Delta>0$.
- one or two distinct real roots, if $Delta=0$.
- one real roots (and two complex roots), if $Delta<0$.
To really finish your answer, you would need to distinguish one or two real roots in the second case. You may also want so solve for $m$ in each (in)equality, to give simpler conditions on $m$. Since you asked for "suggestions" and "where to begin," I'll leave it at that.
Note that this did not need Rolle's theorem. Do you really need to use it?
$endgroup$
1
$begingroup$
OP is asking for something completely different.
$endgroup$
– KonKan
Mar 22 '16 at 19:23
$begingroup$
Except for the phrase "using Rolle's Theorem" I have answered precisely his question, leaving out a few things that I mentioned at the end of my answer. Then I ask if that phrase is critical to him. It is very common at this site to ask for a questioner to request a solution using one method and accept a solution using a different method. Since the OP has not yet said Rolle's Theorem is a requirement, why do you say the question is "something completely different"?
$endgroup$
– Rory Daulton
Mar 22 '16 at 19:28
$begingroup$
Because the requirement for the use of Rolle's theorem lies in the title of the question.
$endgroup$
– KonKan
Mar 22 '16 at 19:32
$begingroup$
As I said, I have seem many other questions here that state a method, including in the title, but a solution using another method was accepted.
$endgroup$
– Rory Daulton
Mar 22 '16 at 19:34
1
$begingroup$
Reading through the answer, I can't help noticing, that what it actually does, is to translate the problem of solving a $3rd$ degree polynomial equation in terms of $x$, to the solution of another $3rd$ degree polynomial equation in terms of $m$....
$endgroup$
– KonKan
Mar 22 '16 at 22:37
|
show 1 more comment
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Taking the derivative of the function: $$f'(x)=3x^2+6x-m$$
If the function $f$ has two roots with $x=a$ and $x=b$, then $$f(a)=0=f(b)$$
Since $f(x)$ is continuous and differentiable, by Rolle's theorem, we have that there exists at least one $c in (a,b)$ such that $$f'(c)=0$$
Namely, $$f'(c)=3c^2+6c-m=0$$
And here is where you begin to consider how the roots exist depend on the parameter.
New contributor
$endgroup$
add a comment |
$begingroup$
Taking the derivative of the function: $$f'(x)=3x^2+6x-m$$
If the function $f$ has two roots with $x=a$ and $x=b$, then $$f(a)=0=f(b)$$
Since $f(x)$ is continuous and differentiable, by Rolle's theorem, we have that there exists at least one $c in (a,b)$ such that $$f'(c)=0$$
Namely, $$f'(c)=3c^2+6c-m=0$$
And here is where you begin to consider how the roots exist depend on the parameter.
New contributor
$endgroup$
add a comment |
$begingroup$
Taking the derivative of the function: $$f'(x)=3x^2+6x-m$$
If the function $f$ has two roots with $x=a$ and $x=b$, then $$f(a)=0=f(b)$$
Since $f(x)$ is continuous and differentiable, by Rolle's theorem, we have that there exists at least one $c in (a,b)$ such that $$f'(c)=0$$
Namely, $$f'(c)=3c^2+6c-m=0$$
And here is where you begin to consider how the roots exist depend on the parameter.
New contributor
$endgroup$
Taking the derivative of the function: $$f'(x)=3x^2+6x-m$$
If the function $f$ has two roots with $x=a$ and $x=b$, then $$f(a)=0=f(b)$$
Since $f(x)$ is continuous and differentiable, by Rolle's theorem, we have that there exists at least one $c in (a,b)$ such that $$f'(c)=0$$
Namely, $$f'(c)=3c^2+6c-m=0$$
And here is where you begin to consider how the roots exist depend on the parameter.
New contributor
New contributor
answered Mar 28 at 3:19
UnbelieveTableUnbelieveTable
965
965
New contributor
New contributor
add a comment |
add a comment |
$begingroup$
The theme is that between any two roots, the derivative has a root. This comes directly from Rolle's Theorem.
So one way to get a first understanding of the number of roots is to find the extrema. This leads you to consider
$$ f'(x) = 3x^2 + 6x - m = 0.$$
As this is a quadratic, you can very quickly write down the two roots,
$$ x = frac-6 pm sqrt36 + 12m6.$$
When $m < -3$, there are no roots of the derivative, and correspondingly there can be at most one root of $f(x)$. As $f to -infty$ as $x to -infty$ and $f to infty$ as $x to infty$, we know that $f$ always has at least one root. So for $m < -3$, we know that $f$ has exactly one root.
More generally, you can determine the number of roots by classifying the extrema. Denote the two roots by $r_1$ and $r_2$ with $r_1 < r_2$ (which both depend on $m$). When $m > -3$, these are two distinct real numbers. One way to understand more roots is to see when $f(r_1) > 0$ and $f(r_2) < 0$.
$endgroup$
add a comment |
$begingroup$
The theme is that between any two roots, the derivative has a root. This comes directly from Rolle's Theorem.
So one way to get a first understanding of the number of roots is to find the extrema. This leads you to consider
$$ f'(x) = 3x^2 + 6x - m = 0.$$
As this is a quadratic, you can very quickly write down the two roots,
$$ x = frac-6 pm sqrt36 + 12m6.$$
When $m < -3$, there are no roots of the derivative, and correspondingly there can be at most one root of $f(x)$. As $f to -infty$ as $x to -infty$ and $f to infty$ as $x to infty$, we know that $f$ always has at least one root. So for $m < -3$, we know that $f$ has exactly one root.
More generally, you can determine the number of roots by classifying the extrema. Denote the two roots by $r_1$ and $r_2$ with $r_1 < r_2$ (which both depend on $m$). When $m > -3$, these are two distinct real numbers. One way to understand more roots is to see when $f(r_1) > 0$ and $f(r_2) < 0$.
$endgroup$
add a comment |
$begingroup$
The theme is that between any two roots, the derivative has a root. This comes directly from Rolle's Theorem.
So one way to get a first understanding of the number of roots is to find the extrema. This leads you to consider
$$ f'(x) = 3x^2 + 6x - m = 0.$$
As this is a quadratic, you can very quickly write down the two roots,
$$ x = frac-6 pm sqrt36 + 12m6.$$
When $m < -3$, there are no roots of the derivative, and correspondingly there can be at most one root of $f(x)$. As $f to -infty$ as $x to -infty$ and $f to infty$ as $x to infty$, we know that $f$ always has at least one root. So for $m < -3$, we know that $f$ has exactly one root.
More generally, you can determine the number of roots by classifying the extrema. Denote the two roots by $r_1$ and $r_2$ with $r_1 < r_2$ (which both depend on $m$). When $m > -3$, these are two distinct real numbers. One way to understand more roots is to see when $f(r_1) > 0$ and $f(r_2) < 0$.
$endgroup$
The theme is that between any two roots, the derivative has a root. This comes directly from Rolle's Theorem.
So one way to get a first understanding of the number of roots is to find the extrema. This leads you to consider
$$ f'(x) = 3x^2 + 6x - m = 0.$$
As this is a quadratic, you can very quickly write down the two roots,
$$ x = frac-6 pm sqrt36 + 12m6.$$
When $m < -3$, there are no roots of the derivative, and correspondingly there can be at most one root of $f(x)$. As $f to -infty$ as $x to -infty$ and $f to infty$ as $x to infty$, we know that $f$ always has at least one root. So for $m < -3$, we know that $f$ has exactly one root.
More generally, you can determine the number of roots by classifying the extrema. Denote the two roots by $r_1$ and $r_2$ with $r_1 < r_2$ (which both depend on $m$). When $m > -3$, these are two distinct real numbers. One way to understand more roots is to see when $f(r_1) > 0$ and $f(r_2) < 0$.
answered Mar 22 '16 at 19:40
davidlowryduda♦davidlowryduda
75.1k7121256
75.1k7121256
add a comment |
add a comment |
$begingroup$
You can use the discriminant of the cubic function to find the nature of the roots. The general cubic equation is
$$ax^3+bx^2+cx+d=0$$
so in your case
$$a=1, b=3, c=-m, d=5$$
In general, the discriminant is
$$Delta=18abcd-4b^3d+b^2c^2-4ac^3-27a^2d^2$$
You can substitute in your values and get a cubic expression in $m$. Then your equation has
- three distinct real roots, if $Delta>0$.
- one or two distinct real roots, if $Delta=0$.
- one real roots (and two complex roots), if $Delta<0$.
To really finish your answer, you would need to distinguish one or two real roots in the second case. You may also want so solve for $m$ in each (in)equality, to give simpler conditions on $m$. Since you asked for "suggestions" and "where to begin," I'll leave it at that.
Note that this did not need Rolle's theorem. Do you really need to use it?
$endgroup$
1
$begingroup$
OP is asking for something completely different.
$endgroup$
– KonKan
Mar 22 '16 at 19:23
$begingroup$
Except for the phrase "using Rolle's Theorem" I have answered precisely his question, leaving out a few things that I mentioned at the end of my answer. Then I ask if that phrase is critical to him. It is very common at this site to ask for a questioner to request a solution using one method and accept a solution using a different method. Since the OP has not yet said Rolle's Theorem is a requirement, why do you say the question is "something completely different"?
$endgroup$
– Rory Daulton
Mar 22 '16 at 19:28
$begingroup$
Because the requirement for the use of Rolle's theorem lies in the title of the question.
$endgroup$
– KonKan
Mar 22 '16 at 19:32
$begingroup$
As I said, I have seem many other questions here that state a method, including in the title, but a solution using another method was accepted.
$endgroup$
– Rory Daulton
Mar 22 '16 at 19:34
1
$begingroup$
Reading through the answer, I can't help noticing, that what it actually does, is to translate the problem of solving a $3rd$ degree polynomial equation in terms of $x$, to the solution of another $3rd$ degree polynomial equation in terms of $m$....
$endgroup$
– KonKan
Mar 22 '16 at 22:37
|
show 1 more comment
$begingroup$
You can use the discriminant of the cubic function to find the nature of the roots. The general cubic equation is
$$ax^3+bx^2+cx+d=0$$
so in your case
$$a=1, b=3, c=-m, d=5$$
In general, the discriminant is
$$Delta=18abcd-4b^3d+b^2c^2-4ac^3-27a^2d^2$$
You can substitute in your values and get a cubic expression in $m$. Then your equation has
- three distinct real roots, if $Delta>0$.
- one or two distinct real roots, if $Delta=0$.
- one real roots (and two complex roots), if $Delta<0$.
To really finish your answer, you would need to distinguish one or two real roots in the second case. You may also want so solve for $m$ in each (in)equality, to give simpler conditions on $m$. Since you asked for "suggestions" and "where to begin," I'll leave it at that.
Note that this did not need Rolle's theorem. Do you really need to use it?
$endgroup$
1
$begingroup$
OP is asking for something completely different.
$endgroup$
– KonKan
Mar 22 '16 at 19:23
$begingroup$
Except for the phrase "using Rolle's Theorem" I have answered precisely his question, leaving out a few things that I mentioned at the end of my answer. Then I ask if that phrase is critical to him. It is very common at this site to ask for a questioner to request a solution using one method and accept a solution using a different method. Since the OP has not yet said Rolle's Theorem is a requirement, why do you say the question is "something completely different"?
$endgroup$
– Rory Daulton
Mar 22 '16 at 19:28
$begingroup$
Because the requirement for the use of Rolle's theorem lies in the title of the question.
$endgroup$
– KonKan
Mar 22 '16 at 19:32
$begingroup$
As I said, I have seem many other questions here that state a method, including in the title, but a solution using another method was accepted.
$endgroup$
– Rory Daulton
Mar 22 '16 at 19:34
1
$begingroup$
Reading through the answer, I can't help noticing, that what it actually does, is to translate the problem of solving a $3rd$ degree polynomial equation in terms of $x$, to the solution of another $3rd$ degree polynomial equation in terms of $m$....
$endgroup$
– KonKan
Mar 22 '16 at 22:37
|
show 1 more comment
$begingroup$
You can use the discriminant of the cubic function to find the nature of the roots. The general cubic equation is
$$ax^3+bx^2+cx+d=0$$
so in your case
$$a=1, b=3, c=-m, d=5$$
In general, the discriminant is
$$Delta=18abcd-4b^3d+b^2c^2-4ac^3-27a^2d^2$$
You can substitute in your values and get a cubic expression in $m$. Then your equation has
- three distinct real roots, if $Delta>0$.
- one or two distinct real roots, if $Delta=0$.
- one real roots (and two complex roots), if $Delta<0$.
To really finish your answer, you would need to distinguish one or two real roots in the second case. You may also want so solve for $m$ in each (in)equality, to give simpler conditions on $m$. Since you asked for "suggestions" and "where to begin," I'll leave it at that.
Note that this did not need Rolle's theorem. Do you really need to use it?
$endgroup$
You can use the discriminant of the cubic function to find the nature of the roots. The general cubic equation is
$$ax^3+bx^2+cx+d=0$$
so in your case
$$a=1, b=3, c=-m, d=5$$
In general, the discriminant is
$$Delta=18abcd-4b^3d+b^2c^2-4ac^3-27a^2d^2$$
You can substitute in your values and get a cubic expression in $m$. Then your equation has
- three distinct real roots, if $Delta>0$.
- one or two distinct real roots, if $Delta=0$.
- one real roots (and two complex roots), if $Delta<0$.
To really finish your answer, you would need to distinguish one or two real roots in the second case. You may also want so solve for $m$ in each (in)equality, to give simpler conditions on $m$. Since you asked for "suggestions" and "where to begin," I'll leave it at that.
Note that this did not need Rolle's theorem. Do you really need to use it?
answered Mar 22 '16 at 19:15
Rory DaultonRory Daulton
29.5k63355
29.5k63355
1
$begingroup$
OP is asking for something completely different.
$endgroup$
– KonKan
Mar 22 '16 at 19:23
$begingroup$
Except for the phrase "using Rolle's Theorem" I have answered precisely his question, leaving out a few things that I mentioned at the end of my answer. Then I ask if that phrase is critical to him. It is very common at this site to ask for a questioner to request a solution using one method and accept a solution using a different method. Since the OP has not yet said Rolle's Theorem is a requirement, why do you say the question is "something completely different"?
$endgroup$
– Rory Daulton
Mar 22 '16 at 19:28
$begingroup$
Because the requirement for the use of Rolle's theorem lies in the title of the question.
$endgroup$
– KonKan
Mar 22 '16 at 19:32
$begingroup$
As I said, I have seem many other questions here that state a method, including in the title, but a solution using another method was accepted.
$endgroup$
– Rory Daulton
Mar 22 '16 at 19:34
1
$begingroup$
Reading through the answer, I can't help noticing, that what it actually does, is to translate the problem of solving a $3rd$ degree polynomial equation in terms of $x$, to the solution of another $3rd$ degree polynomial equation in terms of $m$....
$endgroup$
– KonKan
Mar 22 '16 at 22:37
|
show 1 more comment
1
$begingroup$
OP is asking for something completely different.
$endgroup$
– KonKan
Mar 22 '16 at 19:23
$begingroup$
Except for the phrase "using Rolle's Theorem" I have answered precisely his question, leaving out a few things that I mentioned at the end of my answer. Then I ask if that phrase is critical to him. It is very common at this site to ask for a questioner to request a solution using one method and accept a solution using a different method. Since the OP has not yet said Rolle's Theorem is a requirement, why do you say the question is "something completely different"?
$endgroup$
– Rory Daulton
Mar 22 '16 at 19:28
$begingroup$
Because the requirement for the use of Rolle's theorem lies in the title of the question.
$endgroup$
– KonKan
Mar 22 '16 at 19:32
$begingroup$
As I said, I have seem many other questions here that state a method, including in the title, but a solution using another method was accepted.
$endgroup$
– Rory Daulton
Mar 22 '16 at 19:34
1
$begingroup$
Reading through the answer, I can't help noticing, that what it actually does, is to translate the problem of solving a $3rd$ degree polynomial equation in terms of $x$, to the solution of another $3rd$ degree polynomial equation in terms of $m$....
$endgroup$
– KonKan
Mar 22 '16 at 22:37
1
1
$begingroup$
OP is asking for something completely different.
$endgroup$
– KonKan
Mar 22 '16 at 19:23
$begingroup$
OP is asking for something completely different.
$endgroup$
– KonKan
Mar 22 '16 at 19:23
$begingroup$
Except for the phrase "using Rolle's Theorem" I have answered precisely his question, leaving out a few things that I mentioned at the end of my answer. Then I ask if that phrase is critical to him. It is very common at this site to ask for a questioner to request a solution using one method and accept a solution using a different method. Since the OP has not yet said Rolle's Theorem is a requirement, why do you say the question is "something completely different"?
$endgroup$
– Rory Daulton
Mar 22 '16 at 19:28
$begingroup$
Except for the phrase "using Rolle's Theorem" I have answered precisely his question, leaving out a few things that I mentioned at the end of my answer. Then I ask if that phrase is critical to him. It is very common at this site to ask for a questioner to request a solution using one method and accept a solution using a different method. Since the OP has not yet said Rolle's Theorem is a requirement, why do you say the question is "something completely different"?
$endgroup$
– Rory Daulton
Mar 22 '16 at 19:28
$begingroup$
Because the requirement for the use of Rolle's theorem lies in the title of the question.
$endgroup$
– KonKan
Mar 22 '16 at 19:32
$begingroup$
Because the requirement for the use of Rolle's theorem lies in the title of the question.
$endgroup$
– KonKan
Mar 22 '16 at 19:32
$begingroup$
As I said, I have seem many other questions here that state a method, including in the title, but a solution using another method was accepted.
$endgroup$
– Rory Daulton
Mar 22 '16 at 19:34
$begingroup$
As I said, I have seem many other questions here that state a method, including in the title, but a solution using another method was accepted.
$endgroup$
– Rory Daulton
Mar 22 '16 at 19:34
1
1
$begingroup$
Reading through the answer, I can't help noticing, that what it actually does, is to translate the problem of solving a $3rd$ degree polynomial equation in terms of $x$, to the solution of another $3rd$ degree polynomial equation in terms of $m$....
$endgroup$
– KonKan
Mar 22 '16 at 22:37
$begingroup$
Reading through the answer, I can't help noticing, that what it actually does, is to translate the problem of solving a $3rd$ degree polynomial equation in terms of $x$, to the solution of another $3rd$ degree polynomial equation in terms of $m$....
$endgroup$
– KonKan
Mar 22 '16 at 22:37
|
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