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







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$endgroup$












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






calculus applications rolles-theorem







share|cite|improve this question











$endgroup$












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






calculus applications rolles-theorem







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




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

9,06731640




9,06731640










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
    1 min ago




















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


















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






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
















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_...
    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














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 6 mins ago

























answered 28 mins ago









Eevee TrainerEevee Trainer

9,06731640




9,06731640












  • $begingroup$
    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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