Using Rolle's theorem to show an equation has only one real root The Next CEO of Stack OverflowUsing the Intermediate Value Theorem and Rolle's theorem to determine number of rootsProve using Rolle's Theorem that an equation has exactly one real solution.Rolle's theorem prove polynomial has only 1 rootprove to have at least one real root by Rolle's theoremProve that $tan (x)=sin (x)+1$ has only one solution in $left(−frac π2,frac π2right)$Prove that the equation $x + cos(x) + e^x = 0$ has *exactly* one rootUsing Rolle's theorem and IVT, show that $x^4-7x^3+9=0$ has exactly $2$ roots.How to show that an equation has exactly two solutions?Proving the equation $4x^3+6x^2+5x=-7$ has has only one solution using Rolle's or Lagrange's theoremProve, without using Rolle's theorem, that a polynomial $f$ with $f'(a) = 0 = f'(b)$ for some $a < b$, has at most one root

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Using Rolle's theorem to show an equation has only one real root

Where do students learn to solve polynomial equations these days?



Using Rolle's theorem to show an equation has only one real root



The Next CEO of Stack OverflowUsing the Intermediate Value Theorem and Rolle's theorem to determine number of rootsProve using Rolle's Theorem that an equation has exactly one real solution.Rolle's theorem prove polynomial has only 1 rootprove to have at least one real root by Rolle's theoremProve that $tan (x)=sin (x)+1$ has only one solution in $left(−frac π2,frac π2right)$Prove that the equation $x + cos(x) + e^x = 0$ has *exactly* one rootUsing Rolle's theorem and IVT, show that $x^4-7x^3+9=0$ has exactly $2$ roots.How to show that an equation has exactly two solutions?Proving the equation $4x^3+6x^2+5x=-7$ has has only one solution using Rolle's or Lagrange's theoremProve, without using Rolle's theorem, that a polynomial $f$ with $f'(a) = 0 = f'(b)$ for some $a < b$, has at most one root










2












$begingroup$



Applying Rolle's Theorem, prove that the given equation has only one root:
$$e^x=1+x$$




By inspection, we can say that $x=0$ is one root of the equation. But how can we use Rolle's theorem to prove this root is unique?










share|cite|improve this question











$endgroup$











  • $begingroup$
    It is $$exp(x)geq 1+x$$ for all real $x$
    $endgroup$
    – Dr. Sonnhard Graubner
    55 secs ago
















2












$begingroup$



Applying Rolle's Theorem, prove that the given equation has only one root:
$$e^x=1+x$$




By inspection, we can say that $x=0$ is one root of the equation. But how can we use Rolle's theorem to prove this root is unique?










share|cite|improve this question











$endgroup$











  • $begingroup$
    It is $$exp(x)geq 1+x$$ for all real $x$
    $endgroup$
    – Dr. Sonnhard Graubner
    55 secs ago














2












2








2





$begingroup$



Applying Rolle's Theorem, prove that the given equation has only one root:
$$e^x=1+x$$




By inspection, we can say that $x=0$ is one root of the equation. But how can we use Rolle's theorem to prove this root is unique?










share|cite|improve this question











$endgroup$





Applying Rolle's Theorem, prove that the given equation has only one root:
$$e^x=1+x$$




By inspection, we can say that $x=0$ is one root of the equation. But how can we use Rolle's theorem to prove this root is unique?







calculus applications rolles-theorem






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 26 mins ago









Eevee Trainer

9,05731640




9,05731640










asked 34 mins ago









blue_eyed_...blue_eyed_...

3,30221755




3,30221755











  • $begingroup$
    It is $$exp(x)geq 1+x$$ for all real $x$
    $endgroup$
    – Dr. Sonnhard Graubner
    55 secs ago

















  • $begingroup$
    It is $$exp(x)geq 1+x$$ for all real $x$
    $endgroup$
    – Dr. Sonnhard Graubner
    55 secs ago
















$begingroup$
It is $$exp(x)geq 1+x$$ for all real $x$
$endgroup$
– Dr. Sonnhard Graubner
55 secs ago





$begingroup$
It is $$exp(x)geq 1+x$$ for all real $x$
$endgroup$
– Dr. Sonnhard Graubner
55 secs ago











1 Answer
1






active

oldest

votes


















4












$begingroup$

Let $f(x) = e^x - 1 - x$, and we observe that $f(0)=0$. $f$ is also obviously continuous and differentiable over the real numbers (if you wish to verify that in detail, you can do that separately).



Suppose there exists a second root $b neq 0$ such that $f(0) = f(b) = 0$. Then there exists some $c in (0,b)$ (or $(b,0)$ if $b<0$) such that $f'(c) = 0$ by Rolle's theorem.



$f'(x) = e^x - 1$, however, which satisfies $f'(x) = 0$ only when $x=0$, which is not in any interval $(0,b)$ (or $(b,0)$).



Thus, since no satisfactory $c$ exists, we conclude the equation only has one real root.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    I don't understand the second para.
    $endgroup$
    – blue_eyed_...
    25 mins ago










  • $begingroup$
    We want to show that there exists no second (unique) root, so we seek a contradiction by supposing it exists. Okay, so if the second root is not unique, it is some real number $b$ that is not equal to our first root, $0$. If $b$ is a root, then we are ensured $f(b) =0$. Coincidentally, $f(b) = f(0)$, which gives us a situation in which Rolle's theorem applies. Then, there exists some point $c$ between $b$ and $0$ such that the derivative of $f$ is equal to zero.
    $endgroup$
    – Eevee Trainer
    23 mins ago







  • 1




    $begingroup$
    Do we not need to check for continuity and differentiability of $f(x)$ in $[0,b]$ and $(0,b)$ respectively before applying Rolle's Theorem?
    $endgroup$
    – blue_eyed_...
    18 mins ago










  • $begingroup$
    Yeah, technically you do if you want to be rigorous (and that's a fair point to bring up). Though in this case it's one of those cases where it's "obvious" in the sense that $f$ is obviously continuous and differentiable over $Bbb R$. I suppose whether you want to prove that, or just state it as an obvious thing, depends on the rigor expected of you in your course.
    $endgroup$
    – Eevee Trainer
    7 mins ago










  • $begingroup$
    With regard to my course, we need to prove those conditions of Rolle's Theorem everytime we are willing to use it.
    $endgroup$
    – blue_eyed_...
    4 mins ago











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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

Let $f(x) = e^x - 1 - x$, and we observe that $f(0)=0$. $f$ is also obviously continuous and differentiable over the real numbers (if you wish to verify that in detail, you can do that separately).



Suppose there exists a second root $b neq 0$ such that $f(0) = f(b) = 0$. Then there exists some $c in (0,b)$ (or $(b,0)$ if $b<0$) such that $f'(c) = 0$ by Rolle's theorem.



$f'(x) = e^x - 1$, however, which satisfies $f'(x) = 0$ only when $x=0$, which is not in any interval $(0,b)$ (or $(b,0)$).



Thus, since no satisfactory $c$ exists, we conclude the equation only has one real root.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    I don't understand the second para.
    $endgroup$
    – blue_eyed_...
    25 mins ago










  • $begingroup$
    We want to show that there exists no second (unique) root, so we seek a contradiction by supposing it exists. Okay, so if the second root is not unique, it is some real number $b$ that is not equal to our first root, $0$. If $b$ is a root, then we are ensured $f(b) =0$. Coincidentally, $f(b) = f(0)$, which gives us a situation in which Rolle's theorem applies. Then, there exists some point $c$ between $b$ and $0$ such that the derivative of $f$ is equal to zero.
    $endgroup$
    – Eevee Trainer
    23 mins ago







  • 1




    $begingroup$
    Do we not need to check for continuity and differentiability of $f(x)$ in $[0,b]$ and $(0,b)$ respectively before applying Rolle's Theorem?
    $endgroup$
    – blue_eyed_...
    18 mins ago










  • $begingroup$
    Yeah, technically you do if you want to be rigorous (and that's a fair point to bring up). Though in this case it's one of those cases where it's "obvious" in the sense that $f$ is obviously continuous and differentiable over $Bbb R$. I suppose whether you want to prove that, or just state it as an obvious thing, depends on the rigor expected of you in your course.
    $endgroup$
    – Eevee Trainer
    7 mins ago










  • $begingroup$
    With regard to my course, we need to prove those conditions of Rolle's Theorem everytime we are willing to use it.
    $endgroup$
    – blue_eyed_...
    4 mins ago















4












$begingroup$

Let $f(x) = e^x - 1 - x$, and we observe that $f(0)=0$. $f$ is also obviously continuous and differentiable over the real numbers (if you wish to verify that in detail, you can do that separately).



Suppose there exists a second root $b neq 0$ such that $f(0) = f(b) = 0$. Then there exists some $c in (0,b)$ (or $(b,0)$ if $b<0$) such that $f'(c) = 0$ by Rolle's theorem.



$f'(x) = e^x - 1$, however, which satisfies $f'(x) = 0$ only when $x=0$, which is not in any interval $(0,b)$ (or $(b,0)$).



Thus, since no satisfactory $c$ exists, we conclude the equation only has one real root.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    I don't understand the second para.
    $endgroup$
    – blue_eyed_...
    25 mins ago










  • $begingroup$
    We want to show that there exists no second (unique) root, so we seek a contradiction by supposing it exists. Okay, so if the second root is not unique, it is some real number $b$ that is not equal to our first root, $0$. If $b$ is a root, then we are ensured $f(b) =0$. Coincidentally, $f(b) = f(0)$, which gives us a situation in which Rolle's theorem applies. Then, there exists some point $c$ between $b$ and $0$ such that the derivative of $f$ is equal to zero.
    $endgroup$
    – Eevee Trainer
    23 mins ago







  • 1




    $begingroup$
    Do we not need to check for continuity and differentiability of $f(x)$ in $[0,b]$ and $(0,b)$ respectively before applying Rolle's Theorem?
    $endgroup$
    – blue_eyed_...
    18 mins ago










  • $begingroup$
    Yeah, technically you do if you want to be rigorous (and that's a fair point to bring up). Though in this case it's one of those cases where it's "obvious" in the sense that $f$ is obviously continuous and differentiable over $Bbb R$. I suppose whether you want to prove that, or just state it as an obvious thing, depends on the rigor expected of you in your course.
    $endgroup$
    – Eevee Trainer
    7 mins ago










  • $begingroup$
    With regard to my course, we need to prove those conditions of Rolle's Theorem everytime we are willing to use it.
    $endgroup$
    – blue_eyed_...
    4 mins ago













4












4








4





$begingroup$

Let $f(x) = e^x - 1 - x$, and we observe that $f(0)=0$. $f$ is also obviously continuous and differentiable over the real numbers (if you wish to verify that in detail, you can do that separately).



Suppose there exists a second root $b neq 0$ such that $f(0) = f(b) = 0$. Then there exists some $c in (0,b)$ (or $(b,0)$ if $b<0$) such that $f'(c) = 0$ by Rolle's theorem.



$f'(x) = e^x - 1$, however, which satisfies $f'(x) = 0$ only when $x=0$, which is not in any interval $(0,b)$ (or $(b,0)$).



Thus, since no satisfactory $c$ exists, we conclude the equation only has one real root.






share|cite|improve this answer











$endgroup$



Let $f(x) = e^x - 1 - x$, and we observe that $f(0)=0$. $f$ is also obviously continuous and differentiable over the real numbers (if you wish to verify that in detail, you can do that separately).



Suppose there exists a second root $b neq 0$ such that $f(0) = f(b) = 0$. Then there exists some $c in (0,b)$ (or $(b,0)$ if $b<0$) such that $f'(c) = 0$ by Rolle's theorem.



$f'(x) = e^x - 1$, however, which satisfies $f'(x) = 0$ only when $x=0$, which is not in any interval $(0,b)$ (or $(b,0)$).



Thus, since no satisfactory $c$ exists, we conclude the equation only has one real root.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 6 mins ago

























answered 28 mins ago









Eevee TrainerEevee Trainer

9,05731640




9,05731640











  • $begingroup$
    I don't understand the second para.
    $endgroup$
    – blue_eyed_...
    25 mins ago










  • $begingroup$
    We want to show that there exists no second (unique) root, so we seek a contradiction by supposing it exists. Okay, so if the second root is not unique, it is some real number $b$ that is not equal to our first root, $0$. If $b$ is a root, then we are ensured $f(b) =0$. Coincidentally, $f(b) = f(0)$, which gives us a situation in which Rolle's theorem applies. Then, there exists some point $c$ between $b$ and $0$ such that the derivative of $f$ is equal to zero.
    $endgroup$
    – Eevee Trainer
    23 mins ago







  • 1




    $begingroup$
    Do we not need to check for continuity and differentiability of $f(x)$ in $[0,b]$ and $(0,b)$ respectively before applying Rolle's Theorem?
    $endgroup$
    – blue_eyed_...
    18 mins ago










  • $begingroup$
    Yeah, technically you do if you want to be rigorous (and that's a fair point to bring up). Though in this case it's one of those cases where it's "obvious" in the sense that $f$ is obviously continuous and differentiable over $Bbb R$. I suppose whether you want to prove that, or just state it as an obvious thing, depends on the rigor expected of you in your course.
    $endgroup$
    – Eevee Trainer
    7 mins ago










  • $begingroup$
    With regard to my course, we need to prove those conditions of Rolle's Theorem everytime we are willing to use it.
    $endgroup$
    – blue_eyed_...
    4 mins ago
















  • $begingroup$
    I don't understand the second para.
    $endgroup$
    – blue_eyed_...
    25 mins ago










  • $begingroup$
    We want to show that there exists no second (unique) root, so we seek a contradiction by supposing it exists. Okay, so if the second root is not unique, it is some real number $b$ that is not equal to our first root, $0$. If $b$ is a root, then we are ensured $f(b) =0$. Coincidentally, $f(b) = f(0)$, which gives us a situation in which Rolle's theorem applies. Then, there exists some point $c$ between $b$ and $0$ such that the derivative of $f$ is equal to zero.
    $endgroup$
    – Eevee Trainer
    23 mins ago







  • 1




    $begingroup$
    Do we not need to check for continuity and differentiability of $f(x)$ in $[0,b]$ and $(0,b)$ respectively before applying Rolle's Theorem?
    $endgroup$
    – blue_eyed_...
    18 mins ago










  • $begingroup$
    Yeah, technically you do if you want to be rigorous (and that's a fair point to bring up). Though in this case it's one of those cases where it's "obvious" in the sense that $f$ is obviously continuous and differentiable over $Bbb R$. I suppose whether you want to prove that, or just state it as an obvious thing, depends on the rigor expected of you in your course.
    $endgroup$
    – Eevee Trainer
    7 mins ago










  • $begingroup$
    With regard to my course, we need to prove those conditions of Rolle's Theorem everytime we are willing to use it.
    $endgroup$
    – blue_eyed_...
    4 mins ago















$begingroup$
I don't understand the second para.
$endgroup$
– blue_eyed_...
25 mins ago




$begingroup$
I don't understand the second para.
$endgroup$
– blue_eyed_...
25 mins ago












$begingroup$
We want to show that there exists no second (unique) root, so we seek a contradiction by supposing it exists. Okay, so if the second root is not unique, it is some real number $b$ that is not equal to our first root, $0$. If $b$ is a root, then we are ensured $f(b) =0$. Coincidentally, $f(b) = f(0)$, which gives us a situation in which Rolle's theorem applies. Then, there exists some point $c$ between $b$ and $0$ such that the derivative of $f$ is equal to zero.
$endgroup$
– Eevee Trainer
23 mins ago





$begingroup$
We want to show that there exists no second (unique) root, so we seek a contradiction by supposing it exists. Okay, so if the second root is not unique, it is some real number $b$ that is not equal to our first root, $0$. If $b$ is a root, then we are ensured $f(b) =0$. Coincidentally, $f(b) = f(0)$, which gives us a situation in which Rolle's theorem applies. Then, there exists some point $c$ between $b$ and $0$ such that the derivative of $f$ is equal to zero.
$endgroup$
– Eevee Trainer
23 mins ago





1




1




$begingroup$
Do we not need to check for continuity and differentiability of $f(x)$ in $[0,b]$ and $(0,b)$ respectively before applying Rolle's Theorem?
$endgroup$
– blue_eyed_...
18 mins ago




$begingroup$
Do we not need to check for continuity and differentiability of $f(x)$ in $[0,b]$ and $(0,b)$ respectively before applying Rolle's Theorem?
$endgroup$
– blue_eyed_...
18 mins ago












$begingroup$
Yeah, technically you do if you want to be rigorous (and that's a fair point to bring up). Though in this case it's one of those cases where it's "obvious" in the sense that $f$ is obviously continuous and differentiable over $Bbb R$. I suppose whether you want to prove that, or just state it as an obvious thing, depends on the rigor expected of you in your course.
$endgroup$
– Eevee Trainer
7 mins ago




$begingroup$
Yeah, technically you do if you want to be rigorous (and that's a fair point to bring up). Though in this case it's one of those cases where it's "obvious" in the sense that $f$ is obviously continuous and differentiable over $Bbb R$. I suppose whether you want to prove that, or just state it as an obvious thing, depends on the rigor expected of you in your course.
$endgroup$
– Eevee Trainer
7 mins ago












$begingroup$
With regard to my course, we need to prove those conditions of Rolle's Theorem everytime we are willing to use it.
$endgroup$
– blue_eyed_...
4 mins ago




$begingroup$
With regard to my course, we need to prove those conditions of Rolle's Theorem everytime we are willing to use it.
$endgroup$
– blue_eyed_...
4 mins ago

















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ValueError: Expected n_neighbors <= n_samples, but n_samples = 1, n_neighbors = 6 (SMOTE) The 2019 Stack Overflow Developer Survey Results Are InCan SMOTE be applied over sequence of words (sentences)?ValueError when doing validation with random forestsSMOTE and multi class oversamplingLogic behind SMOTE-NC?ValueError: Error when checking target: expected dense_1 to have shape (7,) but got array with shape (1,)SmoteBoost: Should SMOTE be ran individually for each iteration/tree in the boosting?solving multi-class imbalance classification using smote and OSSUsing SMOTE for Synthetic Data generation to improve performance on unbalanced dataproblem of entry format for a simple model in KerasSVM SMOTE fit_resample() function runs forever with no result