Root Function Inequalities
$begingroup$
I have tried many times to identify my error, but I can't recognize it. I think it may have to do with my properties. This is the inequality that I tried to solve by having the same denominator and using a sign diagram. I didn't have x=2 as part of the sign diagram.
$$
2x^{-1/3}(x-3)^{1/3}+x^{2/3}(x-3)^{-2/3}ge 0
$$
algebra-precalculus inequality
edited Dec 3 '18 at 4:39
gt6989b
33.9k22455
asked Dec 3 '18 at 4:35
anonymousanonymous
185
$endgroup$
add a comment |
$begingroup$
I have tried many times to identify my error, but I can't recognize it. I think it may have to do with my properties. This is the inequality that I tried to solve by having the same denominator and using a sign diagram. I didn't have x=2 as part of the sign diagram.
$$
2x^{-1/3}(x-3)^{1/3}+x^{2/3}(x-3)^{-2/3}ge 0
$$
algebra-precalculus inequality
edited Dec 3 '18 at 4:39
gt6989b
33.9k22455
asked Dec 3 '18 at 4:35
anonymousanonymous
185
$endgroup$
add a comment |
$begingroup$
I have tried many times to identify my error, but I can't recognize it. I think it may have to do with my properties. This is the inequality that I tried to solve by having the same denominator and using a sign diagram. I didn't have x=2 as part of the sign diagram.
$$
2x^{-1/3}(x-3)^{1/3}+x^{2/3}(x-3)^{-2/3}ge 0
$$
algebra-precalculus inequality
edited Dec 3 '18 at 4:39
gt6989b
33.9k22455
asked Dec 3 '18 at 4:35
anonymousanonymous
185
$endgroup$
I have tried many times to identify my error, but I can't recognize it. I think it may have to do with my properties. This is the inequality that I tried to solve by having the same denominator and using a sign diagram. I didn't have x=2 as part of the sign diagram.
$$
2x^{-1/3}(x-3)^{1/3}+x^{2/3}(x-3)^{-2/3}ge 0
$$
algebra-precalculus inequality
algebra-precalculus inequality
edited Dec 3 '18 at 4:39
gt6989b
33.9k22455
asked Dec 3 '18 at 4:35
anonymousanonymous
185
edited Dec 3 '18 at 4:39
gt6989b
33.9k22455
asked Dec 3 '18 at 4:35
anonymousanonymous
185
edited Dec 3 '18 at 4:39
gt6989b
33.9k22455
edited Dec 3 '18 at 4:39
gt6989b
33.9k22455
edited Dec 3 '18 at 4:39
gt6989b
33.9k22455
33.9k22455
asked Dec 3 '18 at 4:35
anonymousanonymous
185
asked Dec 3 '18 at 4:35
anonymousanonymous
185
asked Dec 3 '18 at 4:35
anonymousanonymous
185
185
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The easiest way to see this is to let $z = x^{-1/3}(x-3)^{1/3}$, then your inequality says $$2z+z^2 ge 0\z(2+z) ge 0$$ which implies that either $z ge 0$ and $z+2ge 0$ or $z le 0$ and $z+2 le 0$.
Combine 2 inequalities in each case into the stronger one, and solve both cases for $x$, plugging in the original definition of $z$.
answered Dec 3 '18 at 4:41
gt6989bgt6989b
33.9k22455
$endgroup$
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%2f3023635%2froot-function-inequalities%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
$begingroup$
The easiest way to see this is to let $z = x^{-1/3}(x-3)^{1/3}$, then your inequality says $$2z+z^2 ge 0\z(2+z) ge 0$$ which implies that either $z ge 0$ and $z+2ge 0$ or $z le 0$ and $z+2 le 0$.
Combine 2 inequalities in each case into the stronger one, and solve both cases for $x$, plugging in the original definition of $z$.
answered Dec 3 '18 at 4:41
gt6989bgt6989b
33.9k22455
$endgroup$
add a comment |
$begingroup$
The easiest way to see this is to let $z = x^{-1/3}(x-3)^{1/3}$, then your inequality says $$2z+z^2 ge 0\z(2+z) ge 0$$ which implies that either $z ge 0$ and $z+2ge 0$ or $z le 0$ and $z+2 le 0$.
Combine 2 inequalities in each case into the stronger one, and solve both cases for $x$, plugging in the original definition of $z$.
answered Dec 3 '18 at 4:41
gt6989bgt6989b
33.9k22455
$endgroup$
add a comment |
$begingroup$
The easiest way to see this is to let $z = x^{-1/3}(x-3)^{1/3}$, then your inequality says $$2z+z^2 ge 0\z(2+z) ge 0$$ which implies that either $z ge 0$ and $z+2ge 0$ or $z le 0$ and $z+2 le 0$.
Combine 2 inequalities in each case into the stronger one, and solve both cases for $x$, plugging in the original definition of $z$.
answered Dec 3 '18 at 4:41
gt6989bgt6989b
33.9k22455
$endgroup$
The easiest way to see this is to let $z = x^{-1/3}(x-3)^{1/3}$, then your inequality says $$2z+z^2 ge 0\z(2+z) ge 0$$ which implies that either $z ge 0$ and $z+2ge 0$ or $z le 0$ and $z+2 le 0$.
Combine 2 inequalities in each case into the stronger one, and solve both cases for $x$, plugging in the original definition of $z$.
answered Dec 3 '18 at 4:41
gt6989bgt6989b
33.9k22455
answered Dec 3 '18 at 4:41
gt6989bgt6989b
33.9k22455
answered Dec 3 '18 at 4:41
gt6989bgt6989b
33.9k22455
answered Dec 3 '18 at 4:41
gt6989bgt6989b
33.9k22455
33.9k22455
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%2f3023635%2froot-function-inequalities%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
