Mfcttrf

What does the copyright in a dissertation protect exactly?

What calendar would the Saturn nation use?

If an attacker targets a creature with the Sanctuary spell cast on them, but fails the Wisdom save, can they choose not to attack anyone else?

Learning how to read schematics, questions about fractional voltage in schematic

Convert Numbers To Emoji Math

What does “two-bit (jerk)” mean?

In a series of books, what happens after the coming of age?

Splitting polygons and dividing attribute value proportionally using ArcGIS Pro?

How does "politician" work as a job/career?

How to increase speed on my hybrid bike with flat handlebars and 700X35C tyres?

How to increase row height of a table and vertically "align middle"?

How do I minimise waste on a flight?

Drug Testing and Prescribed Medications

What's the difference between "ricochet" and "bounce"?

Convert a huge txt-file into a dataset

Why did Dr. Strange keep looking into the future after the snap?

Why is there a cap on 401k contributions?

What did Varys actually mean?

Concatenate all values of the same XML element using XPath/XQuery

Does restarting the SQL Services (on the machine) clear the server cache (for things like query plans and statistics)?

why it is 2>&1 and not 2>>&1 to append to a log file

How is it believable that Euron could so easily pull off this ambush?

Searching for a sentence that I only know part of it using Google's operators

Select list elements based on other list

Many one decreasing function?

many to one functionsDemonstrating that a function is monotonically increasing/decreasingHow to determine if this function is one-to-oneHow many times does the function $y=x+sin^2left(frac x3right)-3$ cross the x-axis?Having trouble finding the domain of this function?How to determine the new domain and range given the old domain and range?I got a question on whether the function is one-one or many-one and onto or into .Problem in finding whether function is one-one or many-oneShow that $left(dfrac1(xln x(ln(ln x))^2/3right)$ is decreasing for all $x > 3$principal root confusion

2

$begingroup$

The function $f(x) = frac1x-1 + frac2x-2 + frac3x-3$ is many-to-one, despite it having a strictly negative derivative (the domain being $mathbbR - 1,2,3$). Why is this so? Is there any way of knowing this without actually graphing $f(x)$, which seems rather difficult?

functions

share|cite|improve this question

edited 55 mins ago

David G. Stork

12.4k41836

asked 2 hours ago

HemaHema

6601313

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Umm... "my book"?
  $endgroup$
  – David G. Stork
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DavidG.Stork Resonance rank booster, it is a preparatory book for joint entrance advanced examination in India, I didn't think it would be well known so I didn't mention it.
  $endgroup$
  – Hema
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You DID mention it... that's the problem!
  $endgroup$
  – David G. Stork
  59 mins ago
  

  
  
  
  

add a comment | 

2

$begingroup$

The function $f(x) = frac1x-1 + frac2x-2 + frac3x-3$ is many-to-one, despite it having a strictly negative derivative (the domain being $mathbbR - 1,2,3$). Why is this so? Is there any way of knowing this without actually graphing $f(x)$, which seems rather difficult?

functions

share|cite|improve this question

edited 55 mins ago

David G. Stork

12.4k41836

asked 2 hours ago

HemaHema

6601313

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Umm... "my book"?
  $endgroup$
  – David G. Stork
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DavidG.Stork Resonance rank booster, it is a preparatory book for joint entrance advanced examination in India, I didn't think it would be well known so I didn't mention it.
  $endgroup$
  – Hema
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You DID mention it... that's the problem!
  $endgroup$
  – David G. Stork
  59 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

The function $f(x) = frac1x-1 + frac2x-2 + frac3x-3$ is many-to-one, despite it having a strictly negative derivative (the domain being $mathbbR - 1,2,3$). Why is this so? Is there any way of knowing this without actually graphing $f(x)$, which seems rather difficult?

functions

share|cite|improve this question

edited 55 mins ago

David G. Stork

12.4k41836

asked 2 hours ago

HemaHema

6601313

$endgroup$

share|cite|improve this question

edited 55 mins ago

David G. Stork

12.4k41836

asked 2 hours ago

HemaHema

6601313

share|cite|improve this question

edited 55 mins ago

David G. Stork

12.4k41836

asked 2 hours ago

HemaHema

6601313

edited 55 mins ago

David G. Stork

12.4k41836

edited 55 mins ago

David G. Stork

12.4k41836

asked 2 hours ago

HemaHema

6601313

asked 2 hours ago

HemaHema

6601313

2 Answers
2

active

oldest

votes

2

$begingroup$

When we approach $x=1$ very close from the negative direction, $x-1$ will be a very small negative value and so $frac1x-1$ will be a very large negative value. Approaching from the other side we have $frac1x-1$ becoming a very positive value. In fact, it can become as large or as small as you like, provided you get as close to $x=1$ as necessary.

By looking at the other terms, we see that the same thing happens at $x=2$ and $x=3$ and we see that, near each of these points, the function can attain any value we like. Therefore, there exist distinct $x$ values for which $f(x)$ is the same, and so the function is not one-to-one.

$$rule4cm0.4pt$$

By the way, the reason we can have a function with a strictly negative derivative that isn’t one-to-one is because it isn’t continuous. Specifically, this function isn’t continuous at the points $x=1, x=2, x=3$. Even if we remove these points from the domain as we have, we still find that our function can attain a particular value more than once without the derivative ever becoming zero or switching sign. This is because, at a certain point, the function ‘blows up’ to negative infinity and then magically returns again from positive infinity, so on the domain we have defined, the derivative is always negative.

share|cite|improve this answer

edited 2 hours ago

answered 2 hours ago

雨が好きな人雨が好きな人

1,426316

$endgroup$

add a comment | 

1

$begingroup$

It's because it goes from plus infinity to minus infinity on both $(1,2)$ and $(2,3)$ (so by Intermediate value theorem it assumes every real value at least twice) which you can see by computing the appropriate limits.

share|cite|improve this answer

answered 2 hours ago

BartekBartek

112

New contributor

Bartek is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Note that the function is continuous on the domain given and so the IVT applies.
  $endgroup$
  – 雨が好きな人
  2 hours ago
  

  
  
  
  

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%2f3216622%2fmany-one-decreasing-function%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

2

$begingroup$

When we approach $x=1$ very close from the negative direction, $x-1$ will be a very small negative value and so $frac1x-1$ will be a very large negative value. Approaching from the other side we have $frac1x-1$ becoming a very positive value. In fact, it can become as large or as small as you like, provided you get as close to $x=1$ as necessary.

By looking at the other terms, we see that the same thing happens at $x=2$ and $x=3$ and we see that, near each of these points, the function can attain any value we like. Therefore, there exist distinct $x$ values for which $f(x)$ is the same, and so the function is not one-to-one.

$$rule4cm0.4pt$$

By the way, the reason we can have a function with a strictly negative derivative that isn’t one-to-one is because it isn’t continuous. Specifically, this function isn’t continuous at the points $x=1, x=2, x=3$. Even if we remove these points from the domain as we have, we still find that our function can attain a particular value more than once without the derivative ever becoming zero or switching sign. This is because, at a certain point, the function ‘blows up’ to negative infinity and then magically returns again from positive infinity, so on the domain we have defined, the derivative is always negative.

share|cite|improve this answer

edited 2 hours ago

answered 2 hours ago

雨が好きな人雨が好きな人

1,426316

$endgroup$

add a comment | 

2

$begingroup$

When we approach $x=1$ very close from the negative direction, $x-1$ will be a very small negative value and so $frac1x-1$ will be a very large negative value. Approaching from the other side we have $frac1x-1$ becoming a very positive value. In fact, it can become as large or as small as you like, provided you get as close to $x=1$ as necessary.

By looking at the other terms, we see that the same thing happens at $x=2$ and $x=3$ and we see that, near each of these points, the function can attain any value we like. Therefore, there exist distinct $x$ values for which $f(x)$ is the same, and so the function is not one-to-one.

$$rule4cm0.4pt$$

By the way, the reason we can have a function with a strictly negative derivative that isn’t one-to-one is because it isn’t continuous. Specifically, this function isn’t continuous at the points $x=1, x=2, x=3$. Even if we remove these points from the domain as we have, we still find that our function can attain a particular value more than once without the derivative ever becoming zero or switching sign. This is because, at a certain point, the function ‘blows up’ to negative infinity and then magically returns again from positive infinity, so on the domain we have defined, the derivative is always negative.

share|cite|improve this answer

edited 2 hours ago

answered 2 hours ago

雨が好きな人雨が好きな人

1,426316

$endgroup$

add a comment | 

$begingroup$

When we approach $x=1$ very close from the negative direction, $x-1$ will be a very small negative value and so $frac1x-1$ will be a very large negative value. Approaching from the other side we have $frac1x-1$ becoming a very positive value. In fact, it can become as large or as small as you like, provided you get as close to $x=1$ as necessary.

By looking at the other terms, we see that the same thing happens at $x=2$ and $x=3$ and we see that, near each of these points, the function can attain any value we like. Therefore, there exist distinct $x$ values for which $f(x)$ is the same, and so the function is not one-to-one.

$$rule4cm0.4pt$$

By the way, the reason we can have a function with a strictly negative derivative that isn’t one-to-one is because it isn’t continuous. Specifically, this function isn’t continuous at the points $x=1, x=2, x=3$. Even if we remove these points from the domain as we have, we still find that our function can attain a particular value more than once without the derivative ever becoming zero or switching sign. This is because, at a certain point, the function ‘blows up’ to negative infinity and then magically returns again from positive infinity, so on the domain we have defined, the derivative is always negative.

share|cite|improve this answer

edited 2 hours ago

answered 2 hours ago

雨が好きな人雨が好きな人

1,426316

$endgroup$

share|cite|improve this answer

edited 2 hours ago

answered 2 hours ago

雨が好きな人雨が好きな人

1,426316

answered 2 hours ago

雨が好きな人雨が好きな人

1,426316

answered 2 hours ago

雨が好きな人雨が好きな人

1,426316

1

$begingroup$

It's because it goes from plus infinity to minus infinity on both $(1,2)$ and $(2,3)$ (so by Intermediate value theorem it assumes every real value at least twice) which you can see by computing the appropriate limits.

share|cite|improve this answer

answered 2 hours ago

BartekBartek

112

New contributor

Bartek is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Note that the function is continuous on the domain given and so the IVT applies.
  $endgroup$
  – 雨が好きな人
  2 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

It's because it goes from plus infinity to minus infinity on both $(1,2)$ and $(2,3)$ (so by Intermediate value theorem it assumes every real value at least twice) which you can see by computing the appropriate limits.

share|cite|improve this answer

answered 2 hours ago

BartekBartek

112

New contributor

Bartek is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Note that the function is continuous on the domain given and so the IVT applies.
  $endgroup$
  – 雨が好きな人
  2 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

It's because it goes from plus infinity to minus infinity on both $(1,2)$ and $(2,3)$ (so by Intermediate value theorem it assumes every real value at least twice) which you can see by computing the appropriate limits.

share|cite|improve this answer

answered 2 hours ago

BartekBartek

112

New contributor

Bartek is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this answer

answered 2 hours ago

BartekBartek

112

New contributor

Bartek is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered 2 hours ago

BartekBartek

112

New contributor

Bartek is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered 2 hours ago

BartekBartek

112