Equation of secant line in mean value theorem proof
up vote
1
down vote
favorite
I'm going through a proof for the mean value theorem.
We have a function $f(x)$ continuous on $[a, b]$ and differentiable on $(a, b)$.
Then we define a function $g(x)$ to be the secant line passing through $(a, f(a))$ and $(b, f(b))$.
The slope of said secant is:
$$m=frac{f(b)-f(a)}{b-a}$$
That is clear. Now the proof I'm following defines $g(x)$ like so:
$$g(x) = left[ frac{f(b)-f(a)}{b-a} right](x-a)+f(a)$$
What confuses me: why is the coefficient defined to be $(x-a)$ and not simply $(x)$.
calculus derivatives proof-explanation
add a comment |
up vote
1
down vote
favorite
I'm going through a proof for the mean value theorem.
We have a function $f(x)$ continuous on $[a, b]$ and differentiable on $(a, b)$.
Then we define a function $g(x)$ to be the secant line passing through $(a, f(a))$ and $(b, f(b))$.
The slope of said secant is:
$$m=frac{f(b)-f(a)}{b-a}$$
That is clear. Now the proof I'm following defines $g(x)$ like so:
$$g(x) = left[ frac{f(b)-f(a)}{b-a} right](x-a)+f(a)$$
What confuses me: why is the coefficient defined to be $(x-a)$ and not simply $(x)$.
calculus derivatives proof-explanation
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm going through a proof for the mean value theorem.
We have a function $f(x)$ continuous on $[a, b]$ and differentiable on $(a, b)$.
Then we define a function $g(x)$ to be the secant line passing through $(a, f(a))$ and $(b, f(b))$.
The slope of said secant is:
$$m=frac{f(b)-f(a)}{b-a}$$
That is clear. Now the proof I'm following defines $g(x)$ like so:
$$g(x) = left[ frac{f(b)-f(a)}{b-a} right](x-a)+f(a)$$
What confuses me: why is the coefficient defined to be $(x-a)$ and not simply $(x)$.
calculus derivatives proof-explanation
I'm going through a proof for the mean value theorem.
We have a function $f(x)$ continuous on $[a, b]$ and differentiable on $(a, b)$.
Then we define a function $g(x)$ to be the secant line passing through $(a, f(a))$ and $(b, f(b))$.
The slope of said secant is:
$$m=frac{f(b)-f(a)}{b-a}$$
That is clear. Now the proof I'm following defines $g(x)$ like so:
$$g(x) = left[ frac{f(b)-f(a)}{b-a} right](x-a)+f(a)$$
What confuses me: why is the coefficient defined to be $(x-a)$ and not simply $(x)$.
calculus derivatives proof-explanation
calculus derivatives proof-explanation
asked 3 hours ago
Max
585519
585519
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
The equation of the straight line passing through $(a,f(a)),(b,f(b))$ is given as:
$displaystylefrac{y-f(a)}{x-a}=frac{f(b)-f(a)}{b-a}$
$displaystyleimplies g(x)=y=Big[frac{f(b)-f(a)}{b-a}Big](x-a)+f(a)$
add a comment |
up vote
2
down vote
First of all, it is not correct to assert that $g(x)$ is the secant line passing through $bigl(a,f(a)bigr)$ and $bigl(b,f(b)bigr)$. It is the function whose graph is the line segment uniting those two points.
Now, concerning your question: it is so that $g(a)=a$ and $g(b)=b$.
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',
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%2f3042405%2fequation-of-secant-line-in-mean-value-theorem-proof%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The equation of the straight line passing through $(a,f(a)),(b,f(b))$ is given as:
$displaystylefrac{y-f(a)}{x-a}=frac{f(b)-f(a)}{b-a}$
$displaystyleimplies g(x)=y=Big[frac{f(b)-f(a)}{b-a}Big](x-a)+f(a)$
add a comment |
up vote
2
down vote
accepted
The equation of the straight line passing through $(a,f(a)),(b,f(b))$ is given as:
$displaystylefrac{y-f(a)}{x-a}=frac{f(b)-f(a)}{b-a}$
$displaystyleimplies g(x)=y=Big[frac{f(b)-f(a)}{b-a}Big](x-a)+f(a)$
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The equation of the straight line passing through $(a,f(a)),(b,f(b))$ is given as:
$displaystylefrac{y-f(a)}{x-a}=frac{f(b)-f(a)}{b-a}$
$displaystyleimplies g(x)=y=Big[frac{f(b)-f(a)}{b-a}Big](x-a)+f(a)$
The equation of the straight line passing through $(a,f(a)),(b,f(b))$ is given as:
$displaystylefrac{y-f(a)}{x-a}=frac{f(b)-f(a)}{b-a}$
$displaystyleimplies g(x)=y=Big[frac{f(b)-f(a)}{b-a}Big](x-a)+f(a)$
answered 3 hours ago
Shubham Johri
2,547413
2,547413
add a comment |
add a comment |
up vote
2
down vote
First of all, it is not correct to assert that $g(x)$ is the secant line passing through $bigl(a,f(a)bigr)$ and $bigl(b,f(b)bigr)$. It is the function whose graph is the line segment uniting those two points.
Now, concerning your question: it is so that $g(a)=a$ and $g(b)=b$.
add a comment |
up vote
2
down vote
First of all, it is not correct to assert that $g(x)$ is the secant line passing through $bigl(a,f(a)bigr)$ and $bigl(b,f(b)bigr)$. It is the function whose graph is the line segment uniting those two points.
Now, concerning your question: it is so that $g(a)=a$ and $g(b)=b$.
add a comment |
up vote
2
down vote
up vote
2
down vote
First of all, it is not correct to assert that $g(x)$ is the secant line passing through $bigl(a,f(a)bigr)$ and $bigl(b,f(b)bigr)$. It is the function whose graph is the line segment uniting those two points.
Now, concerning your question: it is so that $g(a)=a$ and $g(b)=b$.
First of all, it is not correct to assert that $g(x)$ is the secant line passing through $bigl(a,f(a)bigr)$ and $bigl(b,f(b)bigr)$. It is the function whose graph is the line segment uniting those two points.
Now, concerning your question: it is so that $g(a)=a$ and $g(b)=b$.
answered 3 hours ago
José Carlos Santos
146k22117217
146k22117217
add a comment |
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3042405%2fequation-of-secant-line-in-mean-value-theorem-proof%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