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Using Rolle's theorem to show $e^x=1+x$ has only one real root

3

1

$begingroup$

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 Mar 31 at 7:56

YuiTo Cheng

2,4064937

asked Mar 31 at 5:58

pi-πpi-π

3,35021755

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is $$exp(x)geq 1+x$$ for all real $x$
  $endgroup$
  – Dr. Sonnhard Graubner
  Mar 31 at 6:31
  
  

  
  
  
  

add a comment | 

3

1

$begingroup$

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 Mar 31 at 7:56

YuiTo Cheng

2,4064937

asked Mar 31 at 5:58

pi-πpi-π

3,35021755

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is $$exp(x)geq 1+x$$ for all real $x$
  $endgroup$
  – Dr. Sonnhard Graubner
  Mar 31 at 6:31
  
  

  
  
  
  

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 Mar 31 at 7:56

YuiTo Cheng

2,4064937

asked Mar 31 at 5:58

pi-πpi-π

3,35021755

$endgroup$

share|cite|improve this question

edited Mar 31 at 7:56

YuiTo Cheng

2,4064937

asked Mar 31 at 5:58

pi-πpi-π

3,35021755

share|cite|improve this question

edited Mar 31 at 7:56

YuiTo Cheng

2,4064937

asked Mar 31 at 5:58

pi-πpi-π

3,35021755

edited Mar 31 at 7:56

YuiTo Cheng

2,4064937

edited Mar 31 at 7:56

YuiTo Cheng

2,4064937

asked Mar 31 at 5:58

pi-πpi-π

3,35021755

asked Mar 31 at 5:58

pi-πpi-π

3,35021755

1 Answer
1

active

oldest

votes

7

$begingroup$

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 Mar 31 at 6:26

answered Mar 31 at 6:04

Eevee TrainerEevee Trainer

10.5k31842

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't understand the second para.
  $endgroup$
  – pi-π
  Mar 31 at 6:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Mar 31 at 6:09
  
  

  
  
  
  


• 
  
  
  

  
  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?
  $endgroup$
  – pi-π
  Mar 31 at 6:14
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Mar 31 at 6:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  With regard to my course, we need to prove those conditions of Rolle's Theorem everytime we are willing to use it.
  $endgroup$
  – pi-π
  Mar 31 at 6:28
  

  
  
  
  

 | 
show 2 more comments

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7

$begingroup$

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 Mar 31 at 6:26

answered Mar 31 at 6:04

Eevee TrainerEevee Trainer

10.5k31842

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't understand the second para.
  $endgroup$
  – pi-π
  Mar 31 at 6:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Mar 31 at 6:09
  
  

  
  
  
  


• 
  
  
  

  
  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?
  $endgroup$
  – pi-π
  Mar 31 at 6:14
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Mar 31 at 6:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  With regard to my course, we need to prove those conditions of Rolle's Theorem everytime we are willing to use it.
  $endgroup$
  – pi-π
  Mar 31 at 6:28
  

  
  
  
  

 | 
show 2 more comments

7

$begingroup$

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 Mar 31 at 6:26

answered Mar 31 at 6:04

Eevee TrainerEevee Trainer

10.5k31842

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't understand the second para.
  $endgroup$
  – pi-π
  Mar 31 at 6:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Mar 31 at 6:09
  
  

  
  
  
  


• 
  
  
  

  
  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?
  $endgroup$
  – pi-π
  Mar 31 at 6:14
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Mar 31 at 6:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  With regard to my course, we need to prove those conditions of Rolle's Theorem everytime we are willing to use it.
  $endgroup$
  – pi-π
  Mar 31 at 6:28
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

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 Mar 31 at 6:26

answered Mar 31 at 6:04

Eevee TrainerEevee Trainer

10.5k31842

$endgroup$

share|cite|improve this answer

edited Mar 31 at 6:26

answered Mar 31 at 6:04

Eevee TrainerEevee Trainer

10.5k31842

answered Mar 31 at 6:04

Eevee TrainerEevee Trainer

10.5k31842

answered Mar 31 at 6:04

Eevee TrainerEevee Trainer

10.5k31842