Fdtxjr

What does this Jacques Hadamard quote mean?

How to convince students of the implication truth values?

How do pianists reach extremely loud dynamics?

How could we fake a moon landing now?

Why do we bend a book to keep it straight?

Circuit to "zoom in" on mV fluctuations of a DC signal?

Fundamental Solution of the Pell Equation

What are the out-of-universe reasons for the references to Toby Maguire-era Spider-Man in ITSV

When the Haste spell ends on a creature, do attackers have advantage against that creature?

Generate an RGB colour grid

Delete nth line from bottom

Is it cost-effective to upgrade an old-ish Giant Escape R3 commuter bike with entry-level branded parts (wheels, drivetrain)?

Can an alien society believe that their star system is the universe?

Can a party unilaterally change candidates in preparation for a General election?

Significance of Cersei's obsession with elephants?

When a candle burns, why does the top of wick glow if bottom of flame is hottest?

Do I really need to have a message in a novel to appeal to readers?

Using et al. for a last / senior author rather than for a first author

Does classifying an integer as a discrete log require it be part of a multiplicative group?

Is there a kind of relay only consumes power when switching?

Trademark violation for app?

Denied boarding although I have proper visa and documentation. To whom should I make a complaint?

How do I find out the mythology and history of my Fortress?

Can you use the Shield Master feat to shove someone before you make an attack by using a Readied action?

Determining the domain of $f(t) = 2t + sqrt25 - t^2 $

0

$begingroup$

I'm not sure what to do here. I want to sqrt the $$25-t^2$$ but t is negative. I know I could factor it, but that wouldn't do any good either.

$$f(t) = 2t + sqrt25 - t^2 $$

I need to determine why this function is continuous and also state its domain. What should I do? How do I solve this?

algebra-precalculus

share|cite|improve this question

edited Feb 4 '14 at 21:04

Nicky Hekster

29.1k63556

asked Feb 4 '14 at 14:19

chopper draw lion4chopper draw lion4

4411723

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  the graph is a tilted semi circle of radius $5$ centered at the origin.
  $endgroup$
  – abel
  May 3 '15 at 17:20
  

  
  
  
  

add a comment | 

0

$begingroup$

I'm not sure what to do here. I want to sqrt the $$25-t^2$$ but t is negative. I know I could factor it, but that wouldn't do any good either.

$$f(t) = 2t + sqrt25 - t^2 $$

I need to determine why this function is continuous and also state its domain. What should I do? How do I solve this?

algebra-precalculus

share|cite|improve this question

edited Feb 4 '14 at 21:04

Nicky Hekster

29.1k63556

asked Feb 4 '14 at 14:19

chopper draw lion4chopper draw lion4

4411723

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  the graph is a tilted semi circle of radius $5$ centered at the origin.
  $endgroup$
  – abel
  May 3 '15 at 17:20
  

  
  
  
  

add a comment | 

$begingroup$

I'm not sure what to do here. I want to sqrt the $$25-t^2$$ but t is negative. I know I could factor it, but that wouldn't do any good either.

$$f(t) = 2t + sqrt25 - t^2 $$

I need to determine why this function is continuous and also state its domain. What should I do? How do I solve this?

algebra-precalculus

share|cite|improve this question

edited Feb 4 '14 at 21:04

Nicky Hekster

29.1k63556

asked Feb 4 '14 at 14:19

chopper draw lion4chopper draw lion4

4411723

$endgroup$

share|cite|improve this question

edited Feb 4 '14 at 21:04

Nicky Hekster

29.1k63556

asked Feb 4 '14 at 14:19

chopper draw lion4chopper draw lion4

4411723

share|cite|improve this question

edited Feb 4 '14 at 21:04

Nicky Hekster

29.1k63556

asked Feb 4 '14 at 14:19

chopper draw lion4chopper draw lion4

4411723

edited Feb 4 '14 at 21:04

Nicky Hekster

29.1k63556

edited Feb 4 '14 at 21:04

Nicky Hekster

29.1k63556

asked Feb 4 '14 at 14:19

chopper draw lion4chopper draw lion4

4411723

asked Feb 4 '14 at 14:19

chopper draw lion4chopper draw lion4

4411723

2 Answers
2

active

oldest

votes

4

$begingroup$

Note that $sqrtg(t)$ is defined where $g(t)ge 0.$ In particular, note that $$25-t^2ge0\5^2-t^2ge0\(5+t)(5-t)ge0.$$ The product of two real numbers $a$ and $b$ is nonnegative exactly when one of them is zero or they have the same sign--that is, $abge0$ if and only if $a,bge0$ or $a,b< 0.$ In order for both $5+t,5-tge0,$ we need $-5le tle 5;$ we cannot have $5+t,5-t<0.$ (Why?) Hence, the domain of $f$ is $-5le tle 5.$

As for continuity, you should be able to show it readily using results about sums and compositions of continuous functions.

share|cite|improve this answer

edited Mar 27 at 1:01

answered Feb 4 '14 at 14:38

Cameron BuieCameron Buie

87k773161

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I understand what you mean about the domain, but what do you mean when you say 'using results about sums and compositions of continuous functions.'
  $endgroup$
  – chopper draw lion4
  Feb 4 '14 at 14:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If $f$ and $g$ are functions that are continuous at a point $x_0,$ what can you say about the function $f+g$ at $x_0$? If $f$ is continuous at $x_0$ and $g$ is continuous at $f(x_0),$ what can you say about the function $gcirc f$ at $x_0$?
  $endgroup$
  – Cameron Buie
  Feb 4 '14 at 14:42
  

  
  
  
  

add a comment | 

1

$begingroup$

To find the domain ask yourself, where this function is defined. Are you familiar with square roots? For a function $sqrtf(x)$ the domain is $f(x) geq 0$. Can you handle from here?

share|cite|improve this answer

answered Feb 4 '14 at 14:23

AlexAlex

14.2k42134

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah, so t <= 5, otherwise it would be negative under square root and therefore undefined. I see now. So domain is [-5, 5]
  $endgroup$
  – chopper draw lion4
  Feb 4 '14 at 14:25
  
  

  
  
  
  

add a comment | 

Your Answer

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

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f663339%2fdetermining-the-domain-of-ft-2t-sqrt25-t2%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

4

$begingroup$

Note that $sqrtg(t)$ is defined where $g(t)ge 0.$ In particular, note that $$25-t^2ge0\5^2-t^2ge0\(5+t)(5-t)ge0.$$ The product of two real numbers $a$ and $b$ is nonnegative exactly when one of them is zero or they have the same sign--that is, $abge0$ if and only if $a,bge0$ or $a,b< 0.$ In order for both $5+t,5-tge0,$ we need $-5le tle 5;$ we cannot have $5+t,5-t<0.$ (Why?) Hence, the domain of $f$ is $-5le tle 5.$

As for continuity, you should be able to show it readily using results about sums and compositions of continuous functions.

share|cite|improve this answer

edited Mar 27 at 1:01

answered Feb 4 '14 at 14:38

Cameron BuieCameron Buie

87k773161

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I understand what you mean about the domain, but what do you mean when you say 'using results about sums and compositions of continuous functions.'
  $endgroup$
  – chopper draw lion4
  Feb 4 '14 at 14:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If $f$ and $g$ are functions that are continuous at a point $x_0,$ what can you say about the function $f+g$ at $x_0$? If $f$ is continuous at $x_0$ and $g$ is continuous at $f(x_0),$ what can you say about the function $gcirc f$ at $x_0$?
  $endgroup$
  – Cameron Buie
  Feb 4 '14 at 14:42
  

  
  
  
  

add a comment | 

4

$begingroup$

Note that $sqrtg(t)$ is defined where $g(t)ge 0.$ In particular, note that $$25-t^2ge0\5^2-t^2ge0\(5+t)(5-t)ge0.$$ The product of two real numbers $a$ and $b$ is nonnegative exactly when one of them is zero or they have the same sign--that is, $abge0$ if and only if $a,bge0$ or $a,b< 0.$ In order for both $5+t,5-tge0,$ we need $-5le tle 5;$ we cannot have $5+t,5-t<0.$ (Why?) Hence, the domain of $f$ is $-5le tle 5.$

As for continuity, you should be able to show it readily using results about sums and compositions of continuous functions.

share|cite|improve this answer

edited Mar 27 at 1:01

answered Feb 4 '14 at 14:38

Cameron BuieCameron Buie

87k773161

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I understand what you mean about the domain, but what do you mean when you say 'using results about sums and compositions of continuous functions.'
  $endgroup$
  – chopper draw lion4
  Feb 4 '14 at 14:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If $f$ and $g$ are functions that are continuous at a point $x_0,$ what can you say about the function $f+g$ at $x_0$? If $f$ is continuous at $x_0$ and $g$ is continuous at $f(x_0),$ what can you say about the function $gcirc f$ at $x_0$?
  $endgroup$
  – Cameron Buie
  Feb 4 '14 at 14:42
  

  
  
  
  

add a comment | 

$begingroup$

Note that $sqrtg(t)$ is defined where $g(t)ge 0.$ In particular, note that $$25-t^2ge0\5^2-t^2ge0\(5+t)(5-t)ge0.$$ The product of two real numbers $a$ and $b$ is nonnegative exactly when one of them is zero or they have the same sign--that is, $abge0$ if and only if $a,bge0$ or $a,b< 0.$ In order for both $5+t,5-tge0,$ we need $-5le tle 5;$ we cannot have $5+t,5-t<0.$ (Why?) Hence, the domain of $f$ is $-5le tle 5.$

As for continuity, you should be able to show it readily using results about sums and compositions of continuous functions.

share|cite|improve this answer

edited Mar 27 at 1:01

answered Feb 4 '14 at 14:38

Cameron BuieCameron Buie

87k773161

$endgroup$

share|cite|improve this answer

edited Mar 27 at 1:01

answered Feb 4 '14 at 14:38

Cameron BuieCameron Buie

87k773161

answered Feb 4 '14 at 14:38

Cameron BuieCameron Buie

87k773161

answered Feb 4 '14 at 14:38

Cameron BuieCameron Buie

87k773161

1

$begingroup$

To find the domain ask yourself, where this function is defined. Are you familiar with square roots? For a function $sqrtf(x)$ the domain is $f(x) geq 0$. Can you handle from here?

share|cite|improve this answer

answered Feb 4 '14 at 14:23

AlexAlex

14.2k42134

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah, so t <= 5, otherwise it would be negative under square root and therefore undefined. I see now. So domain is [-5, 5]
  $endgroup$
  – chopper draw lion4
  Feb 4 '14 at 14:25
  
  

  
  
  
  

add a comment | 

1

$begingroup$

To find the domain ask yourself, where this function is defined. Are you familiar with square roots? For a function $sqrtf(x)$ the domain is $f(x) geq 0$. Can you handle from here?

share|cite|improve this answer

answered Feb 4 '14 at 14:23

AlexAlex

14.2k42134

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah, so t <= 5, otherwise it would be negative under square root and therefore undefined. I see now. So domain is [-5, 5]
  $endgroup$
  – chopper draw lion4
  Feb 4 '14 at 14:25
  
  

  
  
  
  

add a comment | 

$begingroup$

To find the domain ask yourself, where this function is defined. Are you familiar with square roots? For a function $sqrtf(x)$ the domain is $f(x) geq 0$. Can you handle from here?

share|cite|improve this answer

answered Feb 4 '14 at 14:23

AlexAlex

14.2k42134

$endgroup$

share|cite|improve this answer

answered Feb 4 '14 at 14:23

AlexAlex

14.2k42134

answered Feb 4 '14 at 14:23

AlexAlex

14.2k42134

answered Feb 4 '14 at 14:23

AlexAlex

14.2k42134