How do I use the Composite of Continuous Functions theorem to show that a function is continuous?
0
$begingroup$
$$f(x)= sqrt frac{x}{x+1}$$
I know that I have to split it into two separate functions:
$$ g(x)= sqrt x$$
$$h(x)=frac{x}{x+1}$$
I'm just not sure what to do next, and I can't seem to find anything that explains it well enough (or simply enough, I guess). Am I just an idiot? Any help would really be appreciated!
calculus functions continuity function-and-relation-composition
asked Nov 28 '18 at 0:02
dumpster firedumpster fire
1
$endgroup$
add a comment |
0
$begingroup$
$$f(x)= sqrt frac{x}{x+1}$$
I know that I have to split it into two separate functions:
$$ g(x)= sqrt x$$
$$h(x)=frac{x}{x+1}$$
I'm just not sure what to do next, and I can't seem to find anything that explains it well enough (or simply enough, I guess). Am I just an idiot? Any help would really be appreciated!
calculus functions continuity function-and-relation-composition
asked Nov 28 '18 at 0:02
dumpster firedumpster fire
1
$endgroup$
$begingroup$
I would suggest you use the definition of continuity. In case you were just told "you can't pick up your pencil", try the definition that $lim_{x rightarrow x_0} f(x) = f(x_0)$.
$endgroup$
– tekay-squared
Nov 28 '18 at 0:05
1
$begingroup$
continuous on what domain?
$endgroup$
– qbert
Nov 28 '18 at 0:06
$begingroup$
$g$ and $h$ are continuous and $f(x) = g(h(x)$ (which you figured out). Then the theorem you quote says $f$ is continuous where it's defined- nothing more is required, unless you are supposed to find the domain of $f$ too. Can you figure out what values of $x$ will cause trouble?
$endgroup$
– Ethan Bolker
Nov 28 '18 at 0:09
2
$begingroup$
If you've already shown that $sqrt{x}$ is continuous then you're done the first step. Otherwise, prove it, and then also prove that $h$ is continuous. This can be done by recalling the theorem that says that the quotient of two continuous functions is continuous whenever the denominator is nonzero (notice that $g$ is just the quotient of $x$ and $x+1$, polynomials).
$endgroup$
– AlkaKadri
Nov 28 '18 at 0:11
$begingroup$
@EthanBolker Oh, that makes sense. Thank you!
$endgroup$
– dumpster fire
Nov 28 '18 at 0:27
add a comment |
0
0
0
0
$begingroup$
$$f(x)= sqrt frac{x}{x+1}$$
I know that I have to split it into two separate functions:
$$ g(x)= sqrt x$$
$$h(x)=frac{x}{x+1}$$
I'm just not sure what to do next, and I can't seem to find anything that explains it well enough (or simply enough, I guess). Am I just an idiot? Any help would really be appreciated!
calculus functions continuity function-and-relation-composition
asked Nov 28 '18 at 0:02
dumpster firedumpster fire
1
$endgroup$
$$f(x)= sqrt frac{x}{x+1}$$
I know that I have to split it into two separate functions:
$$ g(x)= sqrt x$$
$$h(x)=frac{x}{x+1}$$
I'm just not sure what to do next, and I can't seem to find anything that explains it well enough (or simply enough, I guess). Am I just an idiot? Any help would really be appreciated!
calculus functions continuity function-and-relation-composition
calculus functions continuity function-and-relation-composition
asked Nov 28 '18 at 0:02
dumpster firedumpster fire
1
asked Nov 28 '18 at 0:02
dumpster firedumpster fire
1
asked Nov 28 '18 at 0:02
dumpster firedumpster fire
1
asked Nov 28 '18 at 0:02
dumpster firedumpster fire
1
asked Nov 28 '18 at 0:02
dumpster firedumpster fire
1
1
$begingroup$
I would suggest you use the definition of continuity. In case you were just told "you can't pick up your pencil", try the definition that $lim_{x rightarrow x_0} f(x) = f(x_0)$.
$endgroup$
– tekay-squared
Nov 28 '18 at 0:05
1
$begingroup$
continuous on what domain?
$endgroup$
– qbert
Nov 28 '18 at 0:06
$begingroup$
$g$ and $h$ are continuous and $f(x) = g(h(x)$ (which you figured out). Then the theorem you quote says $f$ is continuous where it's defined- nothing more is required, unless you are supposed to find the domain of $f$ too. Can you figure out what values of $x$ will cause trouble?
$endgroup$
– Ethan Bolker
Nov 28 '18 at 0:09
2
$begingroup$
If you've already shown that $sqrt{x}$ is continuous then you're done the first step. Otherwise, prove it, and then also prove that $h$ is continuous. This can be done by recalling the theorem that says that the quotient of two continuous functions is continuous whenever the denominator is nonzero (notice that $g$ is just the quotient of $x$ and $x+1$, polynomials).
$endgroup$
– AlkaKadri
Nov 28 '18 at 0:11
$begingroup$
@EthanBolker Oh, that makes sense. Thank you!
$endgroup$
– dumpster fire
Nov 28 '18 at 0:27
add a comment |
$begingroup$
I would suggest you use the definition of continuity. In case you were just told "you can't pick up your pencil", try the definition that $lim_{x rightarrow x_0} f(x) = f(x_0)$.
$endgroup$
– tekay-squared
Nov 28 '18 at 0:05
1
$begingroup$
continuous on what domain?
$endgroup$
– qbert
Nov 28 '18 at 0:06
$begingroup$
$g$ and $h$ are continuous and $f(x) = g(h(x)$ (which you figured out). Then the theorem you quote says $f$ is continuous where it's defined- nothing more is required, unless you are supposed to find the domain of $f$ too. Can you figure out what values of $x$ will cause trouble?
$endgroup$
– Ethan Bolker
Nov 28 '18 at 0:09
2
$begingroup$
If you've already shown that $sqrt{x}$ is continuous then you're done the first step. Otherwise, prove it, and then also prove that $h$ is continuous. This can be done by recalling the theorem that says that the quotient of two continuous functions is continuous whenever the denominator is nonzero (notice that $g$ is just the quotient of $x$ and $x+1$, polynomials).
$endgroup$
– AlkaKadri
Nov 28 '18 at 0:11
$begingroup$
@EthanBolker Oh, that makes sense. Thank you!
$endgroup$
– dumpster fire
Nov 28 '18 at 0:27
$begingroup$
I would suggest you use the definition of continuity. In case you were just told "you can't pick up your pencil", try the definition that $lim_{x rightarrow x_0} f(x) = f(x_0)$.
$endgroup$
– tekay-squared
Nov 28 '18 at 0:05
$begingroup$
I would suggest you use the definition of continuity. In case you were just told "you can't pick up your pencil", try the definition that $lim_{x rightarrow x_0} f(x) = f(x_0)$.
$endgroup$
– tekay-squared
Nov 28 '18 at 0:05
1
1
$begingroup$
continuous on what domain?
$endgroup$
– qbert
Nov 28 '18 at 0:06
$begingroup$
continuous on what domain?
$endgroup$
– qbert
Nov 28 '18 at 0:06
$begingroup$
$g$ and $h$ are continuous and $f(x) = g(h(x)$ (which you figured out). Then the theorem you quote says $f$ is continuous where it's defined- nothing more is required, unless you are supposed to find the domain of $f$ too. Can you figure out what values of $x$ will cause trouble?
$endgroup$
– Ethan Bolker
Nov 28 '18 at 0:09
$begingroup$
$g$ and $h$ are continuous and $f(x) = g(h(x)$ (which you figured out). Then the theorem you quote says $f$ is continuous where it's defined- nothing more is required, unless you are supposed to find the domain of $f$ too. Can you figure out what values of $x$ will cause trouble?
$endgroup$
– Ethan Bolker
Nov 28 '18 at 0:09
2
2
$begingroup$
If you've already shown that $sqrt{x}$ is continuous then you're done the first step. Otherwise, prove it, and then also prove that $h$ is continuous. This can be done by recalling the theorem that says that the quotient of two continuous functions is continuous whenever the denominator is nonzero (notice that $g$ is just the quotient of $x$ and $x+1$, polynomials).
$endgroup$
– AlkaKadri
Nov 28 '18 at 0:11
$begingroup$
If you've already shown that $sqrt{x}$ is continuous then you're done the first step. Otherwise, prove it, and then also prove that $h$ is continuous. This can be done by recalling the theorem that says that the quotient of two continuous functions is continuous whenever the denominator is nonzero (notice that $g$ is just the quotient of $x$ and $x+1$, polynomials).
$endgroup$
– AlkaKadri
Nov 28 '18 at 0:11
$begingroup$
@EthanBolker Oh, that makes sense. Thank you!
$endgroup$
– dumpster fire
Nov 28 '18 at 0:27
$begingroup$
@EthanBolker Oh, that makes sense. Thank you!
$endgroup$
– dumpster fire
Nov 28 '18 at 0:27
add a comment |
0
active
oldest
votes
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
});
}
});
draft saved
draft discarded
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%2f3016508%2fhow-do-i-use-the-composite-of-continuous-functions-theorem-to-show-that-a-functi%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
draft saved
draft discarded
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.
draft saved
draft discarded
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%2f3016508%2fhow-do-i-use-the-composite-of-continuous-functions-theorem-to-show-that-a-functi%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
$begingroup$
I would suggest you use the definition of continuity. In case you were just told "you can't pick up your pencil", try the definition that $lim_{x rightarrow x_0} f(x) = f(x_0)$.
$endgroup$
– tekay-squared
Nov 28 '18 at 0:05
1
$begingroup$
continuous on what domain?
$endgroup$
– qbert
Nov 28 '18 at 0:06
$begingroup$
$g$ and $h$ are continuous and $f(x) = g(h(x)$ (which you figured out). Then the theorem you quote says $f$ is continuous where it's defined- nothing more is required, unless you are supposed to find the domain of $f$ too. Can you figure out what values of $x$ will cause trouble?
$endgroup$
– Ethan Bolker
Nov 28 '18 at 0:09
2
$begingroup$
If you've already shown that $sqrt{x}$ is continuous then you're done the first step. Otherwise, prove it, and then also prove that $h$ is continuous. This can be done by recalling the theorem that says that the quotient of two continuous functions is continuous whenever the denominator is nonzero (notice that $g$ is just the quotient of $x$ and $x+1$, polynomials).
$endgroup$
– AlkaKadri
Nov 28 '18 at 0:11
$begingroup$
@EthanBolker Oh, that makes sense. Thank you!
$endgroup$
– dumpster fire
Nov 28 '18 at 0:27