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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

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?

calculus applications rolles-theorem

share|cite|improve this question

edited 26 mins ago

Eevee Trainer

9,05731640

asked 34 mins ago

blue_eyed_...blue_eyed_...

3,30221755

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

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?

calculus applications rolles-theorem

share|cite|improve this question

edited 26 mins ago

Eevee Trainer

9,05731640

asked 34 mins ago

blue_eyed_...blue_eyed_...

3,30221755

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

$begingroup$

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?

calculus applications rolles-theorem

share|cite|improve this question

edited 26 mins ago

Eevee Trainer

9,05731640

asked 34 mins ago

blue_eyed_...blue_eyed_...

3,30221755

$endgroup$

share|cite|improve this question

edited 26 mins ago

Eevee Trainer

9,05731640

asked 34 mins ago

blue_eyed_...blue_eyed_...

3,30221755

share|cite|improve this question

edited 26 mins ago

Eevee Trainer

9,05731640

asked 34 mins ago

blue_eyed_...blue_eyed_...

3,30221755

edited 26 mins ago

Eevee Trainer

9,05731640

edited 26 mins ago

Eevee Trainer

9,05731640

asked 34 mins ago

blue_eyed_...blue_eyed_...

3,30221755

asked 34 mins ago

blue_eyed_...blue_eyed_...

3,30221755

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

edited 6 mins ago

answered 28 mins ago

Eevee TrainerEevee Trainer

9,05731640

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

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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

edited 6 mins ago

answered 28 mins ago

Eevee TrainerEevee Trainer

9,05731640

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

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

edited 6 mins ago

answered 28 mins ago

Eevee TrainerEevee Trainer

9,05731640

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

$begingroup$

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

edited 6 mins ago

answered 28 mins ago

Eevee TrainerEevee Trainer

9,05731640

$endgroup$

share|cite|improve this answer

edited 6 mins ago

answered 28 mins ago

Eevee TrainerEevee Trainer

9,05731640

answered 28 mins ago

Eevee TrainerEevee Trainer

9,05731640

answered 28 mins ago

Eevee TrainerEevee Trainer

9,05731640