Using Rolle's theorem to show an equation 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 theoremProve that the equation $x + cos(x) + e^x = 0$ has *exactly* one rootProof 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 theoremProve, without using Rolle's theorem, that a polynomial $f$ with $f'(a) = 0 = f'(b)$ for some $a < b$, has at most one root
Prepend last line of stdin to entire stdin
Proper way to express "He disappeared them"
Domestic-to-international connection at Orlando (MCO)
What connection does MS Office have to Netscape Navigator?
WOW air has ceased operation, can I get my tickets refunded?
Running a General Election and the European Elections together
Why the difference in type-inference over the as-pattern in two similar function definitions?
Necessary condition on homology group for a set to be contractible
Grabbing quick drinks
What steps are necessary to read a Modern SSD in Medieval Europe?
Is French Guiana a (hard) EU border?
Is a distribution that is normal, but highly skewed considered Gaussian?
Is it professional to write unrelated content in an almost-empty email?
What does "Its cash flow is deeply negative" mean?
Why isn't the Mueller report being released completely and unredacted?
Why does the flight controls check come before arming the autobrake on the A320?
Is it convenient to ask the journal's editor for two additional days to complete a review?
Find non-case sensitive string in a mixed list of elements?
RigExpert AA-35 - Interpreting The Information
I believe this to be a fraud - hired, then asked to cash check and send cash as Bitcoin
Dominated convergence theorem - what sequence?
Do I need to write [sic] when a number is less than 10 but isn't written out?
Why didn't Khan get resurrected in the Genesis Explosion?
TikZ: How to reverse arrow direction without switching start/end point?
Using Rolle's theorem to show an equation 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 theoremProve that the equation $x + cos(x) + e^x = 0$ has *exactly* one rootProof 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 theoremProve, 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 36 mins ago
Eevee Trainer
9,06731640
asked 45 mins ago
blue_eyed_...blue_eyed_...
3,30221755
$endgroup$
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 36 mins ago
Eevee Trainer
9,06731640
asked 45 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
11 mins 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
edited 36 mins ago
Eevee Trainer
9,06731640
asked 45 mins ago
blue_eyed_...blue_eyed_...
3,30221755
$endgroup$
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 36 mins ago
Eevee Trainer
9,06731640
asked 45 mins ago
blue_eyed_...blue_eyed_...
3,30221755
edited 36 mins ago
Eevee Trainer
9,06731640
asked 45 mins ago
blue_eyed_...blue_eyed_...
3,30221755
edited 36 mins ago
Eevee Trainer
9,06731640
edited 36 mins ago
Eevee Trainer
9,06731640
edited 36 mins ago
Eevee Trainer
9,06731640
9,06731640
asked 45 mins ago
blue_eyed_...blue_eyed_...
3,30221755
asked 45 mins ago
blue_eyed_...blue_eyed_...
3,30221755
asked 45 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
11 mins ago
add a comment |
$begingroup$
It is $$exp(x)geq 1+x$$ for all real $x$
$endgroup$
– Dr. Sonnhard Graubner
11 mins ago
$begingroup$
It is $$exp(x)geq 1+x$$ for all real $x$
$endgroup$
– Dr. Sonnhard Graubner
11 mins ago
$begingroup$
It is $$exp(x)geq 1+x$$ for all real $x$
$endgroup$
– Dr. Sonnhard Graubner
11 mins ago
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 39 mins ago
Eevee TrainerEevee Trainer
9,06731640
$endgroup$
$begingroup$
I don't understand the second para.
$endgroup$
– blue_eyed_...
35 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
33 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_...
29 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
18 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_...
15 mins ago
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3169097%2fusing-rolles-theorem-to-show-an-equation-has-only-one-real-root%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
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 39 mins ago
Eevee TrainerEevee Trainer
9,06731640
$endgroup$
$begingroup$
I don't understand the second para.
$endgroup$
– blue_eyed_...
35 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
33 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_...
29 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
18 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_...
15 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.
answered 39 mins ago
Eevee TrainerEevee Trainer
9,06731640
$endgroup$
$begingroup$
I don't understand the second para.
$endgroup$
– blue_eyed_...
35 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
33 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_...
29 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
18 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_...
15 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.
answered 39 mins ago
Eevee TrainerEevee Trainer
9,06731640
$endgroup$
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 39 mins ago
Eevee TrainerEevee Trainer
9,06731640
edited 16 mins ago
answered 39 mins ago
Eevee TrainerEevee Trainer
9,06731640
answered 39 mins ago
Eevee TrainerEevee Trainer
9,06731640
answered 39 mins ago
Eevee TrainerEevee Trainer
9,06731640
9,06731640
$begingroup$
I don't understand the second para.
$endgroup$
– blue_eyed_...
35 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
33 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_...
29 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
18 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_...
15 mins ago
add a comment |
$begingroup$
I don't understand the second para.
$endgroup$
– blue_eyed_...
35 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
33 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_...
29 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
18 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_...
15 mins ago
$begingroup$
I don't understand the second para.
$endgroup$
– blue_eyed_...
35 mins ago
$begingroup$
I don't understand the second para.
$endgroup$
– blue_eyed_...
35 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
33 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
33 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_...
29 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_...
29 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
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
18 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_...
15 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_...
15 mins ago
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3169097%2fusing-rolles-theorem-to-show-an-equation-has-only-one-real-root%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
It is $$exp(x)geq 1+x$$ for all real $x$
$endgroup$
– Dr. Sonnhard Graubner
11 mins ago