Using Rolle's theorem to show $e^x=1+x$ has only one real root
$begingroup$
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
edited Mar 31 at 7:56
YuiTo Cheng
2,40641037
asked Mar 31 at 5:58
pi-πpi-π
3,35021755
$endgroup$
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
edited Mar 31 at 7:56
YuiTo Cheng
2,40641037
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
edited Mar 31 at 7:56
YuiTo Cheng
2,40641037
asked Mar 31 at 5:58
pi-πpi-π
3,35021755
$endgroup$
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
edited Mar 31 at 7:56
YuiTo Cheng
2,40641037
asked Mar 31 at 5:58
pi-πpi-π
3,35021755
edited Mar 31 at 7:56
YuiTo Cheng
2,40641037
asked Mar 31 at 5:58
pi-πpi-π
3,35021755
edited Mar 31 at 7:56
YuiTo Cheng
2,40641037
edited Mar 31 at 7:56
YuiTo Cheng
2,40641037
edited Mar 31 at 7:56
YuiTo Cheng
2,40641037
2,40641037
asked Mar 31 at 5:58
pi-πpi-π
3,35021755
asked Mar 31 at 5:58
pi-πpi-π
3,35021755
asked Mar 31 at 5:58
pi-πpi-π
3,35021755
3,35021755
$begingroup$
It is $$exp(x)geq 1+x$$ for all real $x$
$endgroup$
– Dr. Sonnhard Graubner
Mar 31 at 6:31
add a comment |
$begingroup$
It is $$exp(x)geq 1+x$$ for all real $x$
$endgroup$
– Dr. Sonnhard Graubner
Mar 31 at 6:31
$begingroup$
It is $$exp(x)geq 1+x$$ for all real $x$
$endgroup$
– Dr. Sonnhard Graubner
Mar 31 at 6:31
$begingroup$
It is $$exp(x)geq 1+x$$ for all real $x$
$endgroup$
– Dr. Sonnhard Graubner
Mar 31 at 6:31
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
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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1 Answer
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1 Answer
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$begingroup$
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.
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.
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.
answered Mar 31 at 6:04
Eevee TrainerEevee Trainer
10.5k31842
$endgroup$
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.
answered Mar 31 at 6:04
Eevee TrainerEevee Trainer
10.5k31842
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
10.5k31842
$begingroup$
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
$begingroup$
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
$begingroup$
I don't understand the second para.
$endgroup$
– pi-π
Mar 31 at 6:07
$begingroup$
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
$begingroup$
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
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
$begingroup$
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
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
$begingroup$
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
$begingroup$
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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$begingroup$
It is $$exp(x)geq 1+x$$ for all real $x$
$endgroup$
– Dr. Sonnhard Graubner
Mar 31 at 6:31