Using Rolle's theorem to show an equation has only one real root The Next CEO of Stack OverflowProving number of roots of a function using Rolle's theoremUsing 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.Proof polynomial has only one real root.prove to have at least one real root by Rolle's theoremProve that the equation $x + cos(x) + e^x = 0$ has *exactly* one rootProof using Rolle's theoremUsing Rolle's theorem and IVT, show that $x^4-7x^3+9=0$ has exactly $2$ roots.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



The Next CEO of Stack OverflowProving number of roots of a function using Rolle's theoremUsing 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.Proof polynomial has only one real root.prove to have at least one real root by Rolle's theoremProve that the equation $x + cos(x) + e^x = 0$ has *exactly* one rootProof using Rolle's theoremUsing Rolle's theorem and IVT, show that $x^4-7x^3+9=0$ has exactly $2$ roots.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
    9 mins 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
    9 mins 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 35 mins ago









Eevee Trainer

9,06731640




9,06731640










asked 43 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
    9 mins ago

















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
















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





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











1 Answer
1






active

oldest

votes


















5












$begingroup$

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_...
    34 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
    32 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_...
    27 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
    16 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_...
    13 mins ago











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









5












$begingroup$

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_...
    34 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
    32 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_...
    27 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
    16 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_...
    13 mins ago















5












$begingroup$

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_...
    34 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
    32 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_...
    27 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
    16 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_...
    13 mins ago













5












5








5





$begingroup$

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 15 mins ago

























answered 37 mins ago









Eevee TrainerEevee Trainer

9,06731640




9,06731640











  • $begingroup$
    I don't understand the second para.
    $endgroup$
    – blue_eyed_...
    34 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
    32 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_...
    27 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
    16 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_...
    13 mins ago
















  • $begingroup$
    I don't understand the second para.
    $endgroup$
    – blue_eyed_...
    34 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
    32 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_...
    27 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
    16 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_...
    13 mins ago















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




$begingroup$
I don't understand the second para.
$endgroup$
– blue_eyed_...
34 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
32 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
32 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_...
27 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_...
27 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
16 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
16 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_...
13 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_...
13 mins ago

















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