Using Rolle's theorem to show $e^x=1+x$ 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 theoremProof 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 theoremApplication of Derivatives (Rolle's Theorem?)Prove, 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 $e^x=1+x$ 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 theoremProof 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 theoremApplication of Derivatives (Rolle's Theorem?)Prove, without using Rolle's theorem, that a polynomial $f$ with $f'(a) = 0 = f'(b)$ for some $a < b$, has at most one 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 yesterday
YuiTo Cheng
2,1863937
asked yesterday
blue_eyed_...blue_eyed_...
3,32121755
$endgroup$
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 yesterday
YuiTo Cheng
2,1863937
asked yesterday
blue_eyed_...blue_eyed_...
3,32121755
$endgroup$
$begingroup$
It is $$exp(x)geq 1+x$$ for all real $x$
$endgroup$
– Dr. Sonnhard Graubner
yesterday
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 yesterday
YuiTo Cheng
2,1863937
asked yesterday
blue_eyed_...blue_eyed_...
3,32121755
$endgroup$
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 yesterday
YuiTo Cheng
2,1863937
asked yesterday
blue_eyed_...blue_eyed_...
3,32121755
edited yesterday
YuiTo Cheng
2,1863937
asked yesterday
blue_eyed_...blue_eyed_...
3,32121755
edited yesterday
YuiTo Cheng
2,1863937
edited yesterday
YuiTo Cheng
2,1863937
edited yesterday
YuiTo Cheng
2,1863937
2,1863937
asked yesterday
blue_eyed_...blue_eyed_...
3,32121755
asked yesterday
blue_eyed_...blue_eyed_...
3,32121755
asked yesterday
blue_eyed_...blue_eyed_...
3,32121755
3,32121755
$begingroup$
It is $$exp(x)geq 1+x$$ for all real $x$
$endgroup$
– Dr. Sonnhard Graubner
yesterday
add a comment |
$begingroup$
It is $$exp(x)geq 1+x$$ for all real $x$
$endgroup$
– Dr. Sonnhard Graubner
yesterday
$begingroup$
It is $$exp(x)geq 1+x$$ for all real $x$
$endgroup$
– Dr. Sonnhard Graubner
yesterday
$begingroup$
It is $$exp(x)geq 1+x$$ for all real $x$
$endgroup$
– Dr. Sonnhard Graubner
yesterday
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 yesterday
Eevee TrainerEevee Trainer
9,23331640
$endgroup$
$begingroup$
I don't understand the second para.
$endgroup$
– blue_eyed_...
yesterday
$begingroup$
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
yesterday
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_...
yesterday
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
yesterday
$begingroup$
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_...
yesterday
|
show 2 more comments
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
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 yesterday
Eevee TrainerEevee Trainer
9,23331640
$endgroup$
$begingroup$
I don't understand the second para.
$endgroup$
– blue_eyed_...
yesterday
$begingroup$
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
yesterday
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_...
yesterday
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
yesterday
$begingroup$
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_...
yesterday
|
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 yesterday
Eevee TrainerEevee Trainer
9,23331640
$endgroup$
$begingroup$
I don't understand the second para.
$endgroup$
– blue_eyed_...
yesterday
$begingroup$
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
yesterday
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_...
yesterday
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
yesterday
$begingroup$
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_...
yesterday
|
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 yesterday
Eevee TrainerEevee Trainer
9,23331640
$endgroup$
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 yesterday
Eevee TrainerEevee Trainer
9,23331640
edited yesterday
answered yesterday
Eevee TrainerEevee Trainer
9,23331640
answered yesterday
Eevee TrainerEevee Trainer
9,23331640
answered yesterday
Eevee TrainerEevee Trainer
9,23331640
9,23331640
$begingroup$
I don't understand the second para.
$endgroup$
– blue_eyed_...
yesterday
$begingroup$
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
yesterday
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_...
yesterday
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
yesterday
$begingroup$
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_...
yesterday
|
show 2 more comments
$begingroup$
I don't understand the second para.
$endgroup$
– blue_eyed_...
yesterday
$begingroup$
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
yesterday
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_...
yesterday
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
yesterday
$begingroup$
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_...
yesterday
$begingroup$
I don't understand the second para.
$endgroup$
– blue_eyed_...
yesterday
$begingroup$
I don't understand the second para.
$endgroup$
– blue_eyed_...
yesterday
$begingroup$
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
yesterday
$begingroup$
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
yesterday
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_...
yesterday
$begingroup$
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_...
yesterday
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
yesterday
$begingroup$
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
yesterday
$begingroup$
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_...
yesterday
$begingroup$
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_...
yesterday
|
show 2 more comments
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$begingroup$
It is $$exp(x)geq 1+x$$ for all real $x$
$endgroup$
– Dr. Sonnhard Graubner
yesterday