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How is the many-to-one function $f(x) = frac1x-1 + frac2x-2 + frac3x-3$ be decreasing?

Can we talk about the continuity/discontinuity of a function at a point which is not in its domain?many to one functionsHow to show that functions of this type are strictly 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?How to find the range of a function $y=frac(x-1)(x+4)(x-2)(x-3)$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-oneprincipal root confusion

3

1

$begingroup$

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 May 7 at 8:39

YuiTo Cheng

3,25371245

asked May 7 at 1:59

HemaHema

6731314

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Umm... "my book"?
  $endgroup$
  – David G. Stork
  May 7 at 2:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  May 7 at 3:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You DID mention it... that's the problem!
  $endgroup$
  – David G. Stork
  May 7 at 3:34
  

  
  
  
  

add a comment | 

3

1

$begingroup$

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 May 7 at 8:39

YuiTo Cheng

3,25371245

asked May 7 at 1:59

HemaHema

6731314

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Umm... "my book"?
  $endgroup$
  – David G. Stork
  May 7 at 2:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  May 7 at 3:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You DID mention it... that's the problem!
  $endgroup$
  – David G. Stork
  May 7 at 3:34
  

  
  
  
  

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 May 7 at 8:39

YuiTo Cheng

3,25371245

asked May 7 at 1:59

HemaHema

6731314

$endgroup$

share|cite|improve this question

edited May 7 at 8:39

YuiTo Cheng

3,25371245

asked May 7 at 1:59

HemaHema

6731314

share|cite|improve this question

edited May 7 at 8:39

YuiTo Cheng

3,25371245

asked May 7 at 1:59

HemaHema

6731314

edited May 7 at 8:39

YuiTo Cheng

3,25371245

edited May 7 at 8:39

YuiTo Cheng

3,25371245

asked May 7 at 1:59

HemaHema

6731314

asked May 7 at 1:59

HemaHema

6731314

3 Answers
3

active

oldest

votes

5

$begingroup$

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.

---

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 May 8 at 0:18

answered May 7 at 2:17

雨が好きな人雨が好きな人

2,065417

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The function is continuous where it's defined, see math.stackexchange.com/questions/1482787/…
  $endgroup$
  – Michael Hoppe
  May 7 at 10:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are right—of course. But I don’t think raising that issue here is going to do anyone any favours. My answer serves as a short, intuitive picture.
  $endgroup$
  – 雨が好きな人
  May 7 at 12:07
  

  
  
  
  

add a comment | 

2

$begingroup$

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 May 7 at 2:09

BartekBartek

3439

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Note that the function is continuous on the domain given and so the IVT applies.
  $endgroup$
  – 雨が好きな人
  May 7 at 2:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The IVM demands also that the domain is an interval.
  $endgroup$
  – Michael Hoppe
  May 7 at 10:43
  

  
  
  
  

add a comment | 

1

$begingroup$

If a $C^1$-function's derivative is negative then it's decreasing on any interval where it's defined.

share|cite|improve this answer

answered May 7 at 10:45

Michael HoppeMichael Hoppe

11.4k31837

$endgroup$

add a comment | 

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5

$begingroup$

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.

---

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 May 8 at 0:18

answered May 7 at 2:17

雨が好きな人雨が好きな人

2,065417

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The function is continuous where it's defined, see math.stackexchange.com/questions/1482787/…
  $endgroup$
  – Michael Hoppe
  May 7 at 10:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are right—of course. But I don’t think raising that issue here is going to do anyone any favours. My answer serves as a short, intuitive picture.
  $endgroup$
  – 雨が好きな人
  May 7 at 12:07
  

  
  
  
  

add a comment | 

5

$begingroup$

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.

---

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 May 8 at 0:18

answered May 7 at 2:17

雨が好きな人雨が好きな人

2,065417

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The function is continuous where it's defined, see math.stackexchange.com/questions/1482787/…
  $endgroup$
  – Michael Hoppe
  May 7 at 10:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are right—of course. But I don’t think raising that issue here is going to do anyone any favours. My answer serves as a short, intuitive picture.
  $endgroup$
  – 雨が好きな人
  May 7 at 12:07
  

  
  
  
  

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.

---

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 May 8 at 0:18

answered May 7 at 2:17

雨が好きな人雨が好きな人

2,065417

$endgroup$

share|cite|improve this answer

edited May 8 at 0:18

answered May 7 at 2:17

雨が好きな人雨が好きな人

2,065417

answered May 7 at 2:17

雨が好きな人雨が好きな人

2,065417

answered May 7 at 2:17

雨が好きな人雨が好きな人

2,065417

2

$begingroup$

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 May 7 at 2:09

BartekBartek

3439

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Note that the function is continuous on the domain given and so the IVT applies.
  $endgroup$
  – 雨が好きな人
  May 7 at 2:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The IVM demands also that the domain is an interval.
  $endgroup$
  – Michael Hoppe
  May 7 at 10:43
  

  
  
  
  

add a comment | 

2

$begingroup$

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 May 7 at 2:09

BartekBartek

3439

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Note that the function is continuous on the domain given and so the IVT applies.
  $endgroup$
  – 雨が好きな人
  May 7 at 2:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The IVM demands also that the domain is an interval.
  $endgroup$
  – Michael Hoppe
  May 7 at 10:43
  

  
  
  
  

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 May 7 at 2:09

BartekBartek

3439

$endgroup$

share|cite|improve this answer

answered May 7 at 2:09

BartekBartek

3439

answered May 7 at 2:09

BartekBartek

3439

answered May 7 at 2:09

BartekBartek

3439

1

$begingroup$

If a $C^1$-function's derivative is negative then it's decreasing on any interval where it's defined.

share|cite|improve this answer

answered May 7 at 10:45

Michael HoppeMichael Hoppe

11.4k31837

$endgroup$

add a comment | 

1

$begingroup$

If a $C^1$-function's derivative is negative then it's decreasing on any interval where it's defined.

share|cite|improve this answer

answered May 7 at 10:45

Michael HoppeMichael Hoppe

11.4k31837

$endgroup$

add a comment | 

$begingroup$

If a $C^1$-function's derivative is negative then it's decreasing on any interval where it's defined.

share|cite|improve this answer

answered May 7 at 10:45

Michael HoppeMichael Hoppe

11.4k31837

$endgroup$

share|cite|improve this answer

answered May 7 at 10:45

Michael HoppeMichael Hoppe

11.4k31837

answered May 7 at 10:45

Michael HoppeMichael Hoppe

11.4k31837

answered May 7 at 10:45

Michael HoppeMichael Hoppe

11.4k31837