For the equation $x^3 -y^3 + x-y=0$, how many real functions on $-infty <x < infty$ does this equation...
$begingroup$
I'm able to factor the equation $x^3-y^3+x-y=0$ to $(x-y)(x^2 + y^2 + xy + 1)=0$ to get that the only factor of the equation is $y=x$. But, I'm also told that there is one function, namely $f(x) = x$, that the equation defines. I'm kind of confused about what this is supposed to mean. Are we defining a function such that every $x in (-infty, infty)$ is mapped to $0$ by $g(x) = x^3-y^3+x-y$?
real-analysis
asked Dec 8 '18 at 1:37
K.MK.M
686412
$endgroup$
add a comment |
$begingroup$
I'm able to factor the equation $x^3-y^3+x-y=0$ to $(x-y)(x^2 + y^2 + xy + 1)=0$ to get that the only factor of the equation is $y=x$. But, I'm also told that there is one function, namely $f(x) = x$, that the equation defines. I'm kind of confused about what this is supposed to mean. Are we defining a function such that every $x in (-infty, infty)$ is mapped to $0$ by $g(x) = x^3-y^3+x-y$?
real-analysis
asked Dec 8 '18 at 1:37
K.MK.M
686412
$endgroup$
1
$begingroup$
Isn't $f(x)=x$ really just same as $y=x$ [i.e. $y-x=0$] in another notation?
$endgroup$
– coffeemath
Dec 8 '18 at 1:43
$begingroup$
The equation says $x^3 + x = y^3 + y$.
$endgroup$
– David G. Stork
Dec 8 '18 at 3:36
add a comment |
$begingroup$
I'm able to factor the equation $x^3-y^3+x-y=0$ to $(x-y)(x^2 + y^2 + xy + 1)=0$ to get that the only factor of the equation is $y=x$. But, I'm also told that there is one function, namely $f(x) = x$, that the equation defines. I'm kind of confused about what this is supposed to mean. Are we defining a function such that every $x in (-infty, infty)$ is mapped to $0$ by $g(x) = x^3-y^3+x-y$?
real-analysis
asked Dec 8 '18 at 1:37
K.MK.M
686412
$endgroup$
I'm able to factor the equation $x^3-y^3+x-y=0$ to $(x-y)(x^2 + y^2 + xy + 1)=0$ to get that the only factor of the equation is $y=x$. But, I'm also told that there is one function, namely $f(x) = x$, that the equation defines. I'm kind of confused about what this is supposed to mean. Are we defining a function such that every $x in (-infty, infty)$ is mapped to $0$ by $g(x) = x^3-y^3+x-y$?
real-analysis
real-analysis
asked Dec 8 '18 at 1:37
K.MK.M
686412
asked Dec 8 '18 at 1:37
K.MK.M
686412
asked Dec 8 '18 at 1:37
K.MK.M
686412
asked Dec 8 '18 at 1:37
K.MK.M
686412
asked Dec 8 '18 at 1:37
K.MK.M
686412
686412
1
$begingroup$
Isn't $f(x)=x$ really just same as $y=x$ [i.e. $y-x=0$] in another notation?
$endgroup$
– coffeemath
Dec 8 '18 at 1:43
$begingroup$
The equation says $x^3 + x = y^3 + y$.
$endgroup$
– David G. Stork
Dec 8 '18 at 3:36
add a comment |
1
$begingroup$
Isn't $f(x)=x$ really just same as $y=x$ [i.e. $y-x=0$] in another notation?
$endgroup$
– coffeemath
Dec 8 '18 at 1:43
$begingroup$
The equation says $x^3 + x = y^3 + y$.
$endgroup$
– David G. Stork
Dec 8 '18 at 3:36
1
1
$begingroup$
Isn't $f(x)=x$ really just same as $y=x$ [i.e. $y-x=0$] in another notation?
$endgroup$
– coffeemath
Dec 8 '18 at 1:43
$begingroup$
Isn't $f(x)=x$ really just same as $y=x$ [i.e. $y-x=0$] in another notation?
$endgroup$
– coffeemath
Dec 8 '18 at 1:43
$begingroup$
The equation says $x^3 + x = y^3 + y$.
$endgroup$
– David G. Stork
Dec 8 '18 at 3:36
$begingroup$
The equation says $x^3 + x = y^3 + y$.
$endgroup$
– David G. Stork
Dec 8 '18 at 3:36
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Because for any real numbers $x,y$,
$$x^2+y^2+xy+1ge frac{1}{2}[(x+y)^2+x^2+y^2]+1ge 1,$$
we know that
$$x^3-y^3+x-y=0iff x-y=0.$$
Hence, the subset of $mathbb{R}^2$ defined by the equation is exactly the line $y=x$. It is the graph of $f(x)=x$.
answered Dec 8 '18 at 1:44
Eclipse SunEclipse Sun
6,9841437
$endgroup$
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%2f3030601%2ffor-the-equation-x3-y3-x-y-0-how-many-real-functions-on-infty-x-i%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
$begingroup$
Because for any real numbers $x,y$,
$$x^2+y^2+xy+1ge frac{1}{2}[(x+y)^2+x^2+y^2]+1ge 1,$$
we know that
$$x^3-y^3+x-y=0iff x-y=0.$$
Hence, the subset of $mathbb{R}^2$ defined by the equation is exactly the line $y=x$. It is the graph of $f(x)=x$.
answered Dec 8 '18 at 1:44
Eclipse SunEclipse Sun
6,9841437
$endgroup$
add a comment |
$begingroup$
Because for any real numbers $x,y$,
$$x^2+y^2+xy+1ge frac{1}{2}[(x+y)^2+x^2+y^2]+1ge 1,$$
we know that
$$x^3-y^3+x-y=0iff x-y=0.$$
Hence, the subset of $mathbb{R}^2$ defined by the equation is exactly the line $y=x$. It is the graph of $f(x)=x$.
answered Dec 8 '18 at 1:44
Eclipse SunEclipse Sun
6,9841437
$endgroup$
add a comment |
$begingroup$
Because for any real numbers $x,y$,
$$x^2+y^2+xy+1ge frac{1}{2}[(x+y)^2+x^2+y^2]+1ge 1,$$
we know that
$$x^3-y^3+x-y=0iff x-y=0.$$
Hence, the subset of $mathbb{R}^2$ defined by the equation is exactly the line $y=x$. It is the graph of $f(x)=x$.
answered Dec 8 '18 at 1:44
Eclipse SunEclipse Sun
6,9841437
$endgroup$
Because for any real numbers $x,y$,
$$x^2+y^2+xy+1ge frac{1}{2}[(x+y)^2+x^2+y^2]+1ge 1,$$
we know that
$$x^3-y^3+x-y=0iff x-y=0.$$
Hence, the subset of $mathbb{R}^2$ defined by the equation is exactly the line $y=x$. It is the graph of $f(x)=x$.
answered Dec 8 '18 at 1:44
Eclipse SunEclipse Sun
6,9841437
answered Dec 8 '18 at 1:44
Eclipse SunEclipse Sun
6,9841437
answered Dec 8 '18 at 1:44
Eclipse SunEclipse Sun
6,9841437
answered Dec 8 '18 at 1:44
Eclipse SunEclipse Sun
6,9841437
6,9841437
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.
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%2f3030601%2ffor-the-equation-x3-y3-x-y-0-how-many-real-functions-on-infty-x-i%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
1
$begingroup$
Isn't $f(x)=x$ really just same as $y=x$ [i.e. $y-x=0$] in another notation?
$endgroup$
– coffeemath
Dec 8 '18 at 1:43
$begingroup$
The equation says $x^3 + x = y^3 + y$.
$endgroup$
– David G. Stork
Dec 8 '18 at 3:36