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
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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
$begingroup$
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
edited 55 mins ago
David G. Stork
12.4k41836
asked 2 hours ago
HemaHema
6601313
$endgroup$
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
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
edited 55 mins ago
David G. Stork
12.4k41836
asked 2 hours ago
HemaHema
6601313
$endgroup$
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
functions
edited 55 mins ago
David G. Stork
12.4k41836
asked 2 hours ago
HemaHema
6601313
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
edited 55 mins ago
David G. Stork
12.4k41836
12.4k41836
asked 2 hours ago
HemaHema
6601313
asked 2 hours ago
HemaHema
6601313
asked 2 hours ago
HemaHema
6601313
6601313
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 |
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
1
1
$begingroup$
Umm... "my book"?
$endgroup$
– David G. Stork
2 hours ago
$begingroup$
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
$begingroup$
@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
$begingroup$
You DID mention it... that's the problem!
$endgroup$
– David G. Stork
59 mins ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
answered 2 hours ago
雨が好きな人雨が好きな人
1,426316
$endgroup$
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.
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 |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
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.
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.
answered 2 hours ago
雨が好きな人雨が好きな人
1,426316
$endgroup$
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.
answered 2 hours ago
雨が好きな人雨が好きな人
1,426316
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
1,426316
add a comment |
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.
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.
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.
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.
$endgroup$
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.
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
answered 2 hours ago
BartekBartek
112
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.
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.
$begingroup$
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 |
$begingroup$
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
$begingroup$
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
$begingroup$
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 |
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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