Why is union sign not allowed to denote intervals of function increasing?
On the calculus exam, we were given a function $$f(x) = frac{2x^2+x-1}{x^2-1}
$$ and asked to find intervals of function increasing and decreasing. In the post-exam solutions PDF, it was written that f(x) is increasing on $$(-infty; -1), (-1; 1), (1; +infty)$$
And specifically noted that writing the function increase interval as $$(-infty; -1) cup (-1; 1) cup (1; +infty)$$ is wrong and unacceptable, however, I fail to understand why so.
So, why is union-sign notation incorrect?
notation
add a comment |
On the calculus exam, we were given a function $$f(x) = frac{2x^2+x-1}{x^2-1}
$$ and asked to find intervals of function increasing and decreasing. In the post-exam solutions PDF, it was written that f(x) is increasing on $$(-infty; -1), (-1; 1), (1; +infty)$$
And specifically noted that writing the function increase interval as $$(-infty; -1) cup (-1; 1) cup (1; +infty)$$ is wrong and unacceptable, however, I fail to understand why so.
So, why is union-sign notation incorrect?
notation
9
Increasing on the union requires the function to be increasing across the discontinuities of the domain.
– Yves Daoust
Dec 22 at 17:14
2
@AndreasRejbrand: that should have been an answer :)
– Taladris
Dec 23 at 3:56
3
I'd say that the point is that the union of intervals fails to be an interval and the question is specifically asking for intervals.
– Dirk
Dec 23 at 16:12
1
There are several correct answers here. My comment is to suggest that the formulation you ask about causes unnecessary trouble by trying to use symbols rather than words to describe the answer. Reserve formal mathematical notation for the places where it is really called for.
– Ethan Bolker
Dec 23 at 16:23
add a comment |
On the calculus exam, we were given a function $$f(x) = frac{2x^2+x-1}{x^2-1}
$$ and asked to find intervals of function increasing and decreasing. In the post-exam solutions PDF, it was written that f(x) is increasing on $$(-infty; -1), (-1; 1), (1; +infty)$$
And specifically noted that writing the function increase interval as $$(-infty; -1) cup (-1; 1) cup (1; +infty)$$ is wrong and unacceptable, however, I fail to understand why so.
So, why is union-sign notation incorrect?
notation
On the calculus exam, we were given a function $$f(x) = frac{2x^2+x-1}{x^2-1}
$$ and asked to find intervals of function increasing and decreasing. In the post-exam solutions PDF, it was written that f(x) is increasing on $$(-infty; -1), (-1; 1), (1; +infty)$$
And specifically noted that writing the function increase interval as $$(-infty; -1) cup (-1; 1) cup (1; +infty)$$ is wrong and unacceptable, however, I fail to understand why so.
So, why is union-sign notation incorrect?
notation
notation
edited Dec 22 at 16:48
KM101
4,528418
4,528418
asked Dec 22 at 16:46
Rafael Sofi-Zadeh
1017
1017
9
Increasing on the union requires the function to be increasing across the discontinuities of the domain.
– Yves Daoust
Dec 22 at 17:14
2
@AndreasRejbrand: that should have been an answer :)
– Taladris
Dec 23 at 3:56
3
I'd say that the point is that the union of intervals fails to be an interval and the question is specifically asking for intervals.
– Dirk
Dec 23 at 16:12
1
There are several correct answers here. My comment is to suggest that the formulation you ask about causes unnecessary trouble by trying to use symbols rather than words to describe the answer. Reserve formal mathematical notation for the places where it is really called for.
– Ethan Bolker
Dec 23 at 16:23
add a comment |
9
Increasing on the union requires the function to be increasing across the discontinuities of the domain.
– Yves Daoust
Dec 22 at 17:14
2
@AndreasRejbrand: that should have been an answer :)
– Taladris
Dec 23 at 3:56
3
I'd say that the point is that the union of intervals fails to be an interval and the question is specifically asking for intervals.
– Dirk
Dec 23 at 16:12
1
There are several correct answers here. My comment is to suggest that the formulation you ask about causes unnecessary trouble by trying to use symbols rather than words to describe the answer. Reserve formal mathematical notation for the places where it is really called for.
– Ethan Bolker
Dec 23 at 16:23
9
9
Increasing on the union requires the function to be increasing across the discontinuities of the domain.
– Yves Daoust
Dec 22 at 17:14
Increasing on the union requires the function to be increasing across the discontinuities of the domain.
– Yves Daoust
Dec 22 at 17:14
2
2
@AndreasRejbrand: that should have been an answer :)
– Taladris
Dec 23 at 3:56
@AndreasRejbrand: that should have been an answer :)
– Taladris
Dec 23 at 3:56
3
3
I'd say that the point is that the union of intervals fails to be an interval and the question is specifically asking for intervals.
– Dirk
Dec 23 at 16:12
I'd say that the point is that the union of intervals fails to be an interval and the question is specifically asking for intervals.
– Dirk
Dec 23 at 16:12
1
1
There are several correct answers here. My comment is to suggest that the formulation you ask about causes unnecessary trouble by trying to use symbols rather than words to describe the answer. Reserve formal mathematical notation for the places where it is really called for.
– Ethan Bolker
Dec 23 at 16:23
There are several correct answers here. My comment is to suggest that the formulation you ask about causes unnecessary trouble by trying to use symbols rather than words to describe the answer. Reserve formal mathematical notation for the places where it is really called for.
– Ethan Bolker
Dec 23 at 16:23
add a comment |
5 Answers
5
active
oldest
votes
The function is decreasing (not increasing, as you said) on each of those three intervals separately. But it is not decreasing on their union because, for example, $f(5)>f(0)$.
Consider a similar case: my bank balance is decreasing from December 9 through December 14, and from December 16 through December 21. Is it therefore decreasing on the entire interval? No, because on December 15 I received my salary, so my bank balance increased sharply between December 14 and 16.
add a comment |
I have never seen one write an interval like $(a;b)$. The standard notation is $(a,b)$.
Consider a simpler example: $g(x)=frac{-1}{x}$. Note that $g$ is increasing on the interval $(-infty,0)$ and also increasing on the interval $(0,infty)$. However, $g$ is not increasing on the set
$$
S=(-infty,0)cup (0,infty).
$$
12
Note: where decimal separator is comma (not dot), it's good to write intervals as $(a;b)$, especially with numbers. $(1,2,3)$ is ambiguous while $(1,2;3)$ or $(1;2,3)$ is not.
– Kamil Maciorowski
Dec 22 at 19:28
1
@KamilMaciorowski: Indeed. Now I have seen it: en.wikipedia.org/wiki/… Thanks for pointing that out! And pardon my ignorance.
– user587192
Dec 23 at 2:29
@KamilMaciorowski Often another spacing is used as well, as in $(1{,}2;3)$ and $(1;2{,}3)$.
– Jeppe Stig Nielsen
Dec 24 at 10:27
add a comment |
Introduction
So, we are considering the function $f$ defined by
$$f(x) = frac{2x^2+x-1}{x^2-1}.$$
Clearly, this function is defined everywhere except at $x = pm 1$, so its domain is $D_f = mathbf{R} setminus {-1, +1}.$
Let us draw its graph, just so we can see very clearly what is going on:
Now, we are interested in knowing where the function is decreasing. Obviously, to answer this question, we need to know exactly what the word "decreasing" means in this context.
What does the word mean?
Likely your calculus textbook contains a definition very similar to this one:
A function $f : D_f to mathbf{R}$ is decreasing on the set $U subset D_f$ if, for all $x, y in U$, it holds that $x < y Rightarrow f(x) ge f(y)$.
So, fundamentally, the concept of "decreasing" applies not to a function alone -- or to a function at a particular point --, but to a function and a subset of its domain ($U$ in the definition above). For instance, it is meaningless to say that "$f$ is decreasing at $x = 5$" [you do not have two points to compare1!], but it makes sense to say that "$f$ is decreasing on $(-3, -2)$". (And that would be true in our case.)
Let's apply the definition
Now it is easy to see that our function $f$ is indeed decreasing on each of the subsets $(-infty, -1)$, $(-1, 1)$, and $(1, infty)$. For example, if you pick any two numbers $x$ and $y$ in $(-1, 1)$, it is obvious that if $x < y$, then $f(x) ge f(y)$.
Notice how beautifully the precise definition captures the essence of our intuitive idea of what it means to be "decreasing"!2
But with this precise knowledge, it also becomes very easy for us to see why $f$ isn't decreasing on $(-infty, -1) cup (-1, 1) cup (1, infty) = mathbf{R} setminus {-1, +1} = D_f$.
Think about it for a while.
For example, take $x = -2$ and $y = 2$. Clearly $x < y$, but $f(x) ge f(y)$ doesn't hold. [Intuitively: when we moved to the right, the value of the function increased!]
So it is very much not the case that $f$ is decreasing on $D_f$.
A small note on terminology
If the exam question asked for "the intervals where the function is decreasing", you should probably answer it by giving one or more intervals -- and not sets that aren't intervals. But the union clearly isn't an interval, so this also makes it obvious that that answer is incorrect.
[For example, there are no two real numbers $a$ and $b$ such that
$$mathbf{R} setminus {-1, +1} = (a, b)$$
so the union clearly isn't a bounded open interval.]
A subtle difference
A different exam question could have asked you to find the points in $D_f$ where the function's derivative is non-positive. This is a related but different question. The derivative of a function is also a function, and it has a value at each point where it is defined. So in contrast to the discussion above, it does make sense to say that the derivative of $f$ is non-positive at the particular point $x = 5$, say.
If you compute the derivative of $f$, you find a function $f'$ that is also defined on $D_f$. And its value is non-positive (even negative) everywhere. So it is true that the derivative is non-positive in each of the subsets $(-infty, -1)$, $(-1, 1)$, and $(1, infty)$. But this immediately implies that the derivative is non-positive in their union $D_f$.
Consequently, this hypothetical exam question could (indeed, should) be answered by giving the union in its most simplified form.
Footnotes
1 Exercise (very technical): What do you have to say about the statement "$f$ is decreasing on ${5}$"?
2 Yeah, this is one of the reasons why I love mathematics. And the $epsilondelta$ definitions in calculus ... Beautiful.
New contributor
(+1) Your effort to explain the answer in details and also explaining related material is incredible. However you should note that sometimes people would look for the most simple answer rather than the most detailed answer (even though the latter provides a better point of view on this mathematical area in general). To save your time I suggest you don't read what's not written in the question. The OP asks specifically about the difference in notations and it seems that he already understand the definition of increasing/decreasing function (or at least he don't ask about it).
– Yanko
Dec 23 at 15:10
@Yanko: It is hard to tell exactly what the OP understands and what (s)he doesn't fully understand. Still, I think it is good if a Q contains several answers with different approaches and levels of detail. Plus, I have all the time in the world! :)
– Andreas Rejbrand
Dec 23 at 15:14
Yes it's hard to tell what the OP understand. In fact sometimes detailed answers are well received and useful. This time I'm afraid that the OP is unlikely to read your answer but it might come in handy for other users. Anyway if you continue provide such answers and with "all the time in the world" you have I'm sure you'll become a great contributor for this site.
– Yanko
Dec 23 at 15:28
And the geometric interpretation of the formula $x<yRightarrow f(x)ge g(y)$ is the following. Whenever you pick two distinct points from the graph of $f$ within the domain in which we consider whether $f$ is decreasing, then the leftmost one of the two points (let us call it $(x,f(x))$) is located higher (in ordinate) than the rightmost one ($(y,f(y))$). In other words, if you pick the two distinct points on the graph, the secant passing through them has a negative (or zero) slope. If you pick points from different branches in your image, it is not hard to get a positive secant slope.
– Jeppe Stig Nielsen
Dec 24 at 12:22
add a comment |
Because $(-infty,-1)cup(-1,1)cup(1,+infty)$ is not an interval.
For the same equation, for concavity intervals, the answer was given using union-sign notation though. I just don't understand why writing like this would be considered incorrect.
– Rafael Sofi-Zadeh
Dec 22 at 16:54
I believe that I provide an answer to that question.
– José Carlos Santos
Dec 22 at 16:56
1
But it might be reasonable to say $dfrac{x^3-x}{x^2-1}$ is increasing on $(-infty,-1)cup(-1,1)cup(1,+infty)$
– Henry
Dec 22 at 21:05
4
But the question was about finding intervals in which the function is increasing.
– José Carlos Santos
Dec 22 at 21:37
add a comment |
The existing answers have focused on the fact that $f$ is not increasing from one interval to the next. An even better example would be $g(x)$ defined as $-1/x$ when $x<0$ and $sqrt{x}$ when $xge 0$. Then $g$ is increasing on $(-infty,0);[0,infty)$. But we definitely can't say that $g$ is increasing on $(-infty,0)cup[0,infty)=(-infty,infty)$.
add a comment |
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
});
}
});
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%2f3049628%2fwhy-is-union-sign-not-allowed-to-denote-intervals-of-function-increasing%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
The function is decreasing (not increasing, as you said) on each of those three intervals separately. But it is not decreasing on their union because, for example, $f(5)>f(0)$.
Consider a similar case: my bank balance is decreasing from December 9 through December 14, and from December 16 through December 21. Is it therefore decreasing on the entire interval? No, because on December 15 I received my salary, so my bank balance increased sharply between December 14 and 16.
add a comment |
The function is decreasing (not increasing, as you said) on each of those three intervals separately. But it is not decreasing on their union because, for example, $f(5)>f(0)$.
Consider a similar case: my bank balance is decreasing from December 9 through December 14, and from December 16 through December 21. Is it therefore decreasing on the entire interval? No, because on December 15 I received my salary, so my bank balance increased sharply between December 14 and 16.
add a comment |
The function is decreasing (not increasing, as you said) on each of those three intervals separately. But it is not decreasing on their union because, for example, $f(5)>f(0)$.
Consider a similar case: my bank balance is decreasing from December 9 through December 14, and from December 16 through December 21. Is it therefore decreasing on the entire interval? No, because on December 15 I received my salary, so my bank balance increased sharply between December 14 and 16.
The function is decreasing (not increasing, as you said) on each of those three intervals separately. But it is not decreasing on their union because, for example, $f(5)>f(0)$.
Consider a similar case: my bank balance is decreasing from December 9 through December 14, and from December 16 through December 21. Is it therefore decreasing on the entire interval? No, because on December 15 I received my salary, so my bank balance increased sharply between December 14 and 16.
edited Dec 22 at 17:18
community wiki
2 revs
MJD
add a comment |
add a comment |
I have never seen one write an interval like $(a;b)$. The standard notation is $(a,b)$.
Consider a simpler example: $g(x)=frac{-1}{x}$. Note that $g$ is increasing on the interval $(-infty,0)$ and also increasing on the interval $(0,infty)$. However, $g$ is not increasing on the set
$$
S=(-infty,0)cup (0,infty).
$$
12
Note: where decimal separator is comma (not dot), it's good to write intervals as $(a;b)$, especially with numbers. $(1,2,3)$ is ambiguous while $(1,2;3)$ or $(1;2,3)$ is not.
– Kamil Maciorowski
Dec 22 at 19:28
1
@KamilMaciorowski: Indeed. Now I have seen it: en.wikipedia.org/wiki/… Thanks for pointing that out! And pardon my ignorance.
– user587192
Dec 23 at 2:29
@KamilMaciorowski Often another spacing is used as well, as in $(1{,}2;3)$ and $(1;2{,}3)$.
– Jeppe Stig Nielsen
Dec 24 at 10:27
add a comment |
I have never seen one write an interval like $(a;b)$. The standard notation is $(a,b)$.
Consider a simpler example: $g(x)=frac{-1}{x}$. Note that $g$ is increasing on the interval $(-infty,0)$ and also increasing on the interval $(0,infty)$. However, $g$ is not increasing on the set
$$
S=(-infty,0)cup (0,infty).
$$
12
Note: where decimal separator is comma (not dot), it's good to write intervals as $(a;b)$, especially with numbers. $(1,2,3)$ is ambiguous while $(1,2;3)$ or $(1;2,3)$ is not.
– Kamil Maciorowski
Dec 22 at 19:28
1
@KamilMaciorowski: Indeed. Now I have seen it: en.wikipedia.org/wiki/… Thanks for pointing that out! And pardon my ignorance.
– user587192
Dec 23 at 2:29
@KamilMaciorowski Often another spacing is used as well, as in $(1{,}2;3)$ and $(1;2{,}3)$.
– Jeppe Stig Nielsen
Dec 24 at 10:27
add a comment |
I have never seen one write an interval like $(a;b)$. The standard notation is $(a,b)$.
Consider a simpler example: $g(x)=frac{-1}{x}$. Note that $g$ is increasing on the interval $(-infty,0)$ and also increasing on the interval $(0,infty)$. However, $g$ is not increasing on the set
$$
S=(-infty,0)cup (0,infty).
$$
I have never seen one write an interval like $(a;b)$. The standard notation is $(a,b)$.
Consider a simpler example: $g(x)=frac{-1}{x}$. Note that $g$ is increasing on the interval $(-infty,0)$ and also increasing on the interval $(0,infty)$. However, $g$ is not increasing on the set
$$
S=(-infty,0)cup (0,infty).
$$
edited Dec 23 at 2:28
answered Dec 22 at 17:10
user587192
1,723213
1,723213
12
Note: where decimal separator is comma (not dot), it's good to write intervals as $(a;b)$, especially with numbers. $(1,2,3)$ is ambiguous while $(1,2;3)$ or $(1;2,3)$ is not.
– Kamil Maciorowski
Dec 22 at 19:28
1
@KamilMaciorowski: Indeed. Now I have seen it: en.wikipedia.org/wiki/… Thanks for pointing that out! And pardon my ignorance.
– user587192
Dec 23 at 2:29
@KamilMaciorowski Often another spacing is used as well, as in $(1{,}2;3)$ and $(1;2{,}3)$.
– Jeppe Stig Nielsen
Dec 24 at 10:27
add a comment |
12
Note: where decimal separator is comma (not dot), it's good to write intervals as $(a;b)$, especially with numbers. $(1,2,3)$ is ambiguous while $(1,2;3)$ or $(1;2,3)$ is not.
– Kamil Maciorowski
Dec 22 at 19:28
1
@KamilMaciorowski: Indeed. Now I have seen it: en.wikipedia.org/wiki/… Thanks for pointing that out! And pardon my ignorance.
– user587192
Dec 23 at 2:29
@KamilMaciorowski Often another spacing is used as well, as in $(1{,}2;3)$ and $(1;2{,}3)$.
– Jeppe Stig Nielsen
Dec 24 at 10:27
12
12
Note: where decimal separator is comma (not dot), it's good to write intervals as $(a;b)$, especially with numbers. $(1,2,3)$ is ambiguous while $(1,2;3)$ or $(1;2,3)$ is not.
– Kamil Maciorowski
Dec 22 at 19:28
Note: where decimal separator is comma (not dot), it's good to write intervals as $(a;b)$, especially with numbers. $(1,2,3)$ is ambiguous while $(1,2;3)$ or $(1;2,3)$ is not.
– Kamil Maciorowski
Dec 22 at 19:28
1
1
@KamilMaciorowski: Indeed. Now I have seen it: en.wikipedia.org/wiki/… Thanks for pointing that out! And pardon my ignorance.
– user587192
Dec 23 at 2:29
@KamilMaciorowski: Indeed. Now I have seen it: en.wikipedia.org/wiki/… Thanks for pointing that out! And pardon my ignorance.
– user587192
Dec 23 at 2:29
@KamilMaciorowski Often another spacing is used as well, as in $(1{,}2;3)$ and $(1;2{,}3)$.
– Jeppe Stig Nielsen
Dec 24 at 10:27
@KamilMaciorowski Often another spacing is used as well, as in $(1{,}2;3)$ and $(1;2{,}3)$.
– Jeppe Stig Nielsen
Dec 24 at 10:27
add a comment |
Introduction
So, we are considering the function $f$ defined by
$$f(x) = frac{2x^2+x-1}{x^2-1}.$$
Clearly, this function is defined everywhere except at $x = pm 1$, so its domain is $D_f = mathbf{R} setminus {-1, +1}.$
Let us draw its graph, just so we can see very clearly what is going on:
Now, we are interested in knowing where the function is decreasing. Obviously, to answer this question, we need to know exactly what the word "decreasing" means in this context.
What does the word mean?
Likely your calculus textbook contains a definition very similar to this one:
A function $f : D_f to mathbf{R}$ is decreasing on the set $U subset D_f$ if, for all $x, y in U$, it holds that $x < y Rightarrow f(x) ge f(y)$.
So, fundamentally, the concept of "decreasing" applies not to a function alone -- or to a function at a particular point --, but to a function and a subset of its domain ($U$ in the definition above). For instance, it is meaningless to say that "$f$ is decreasing at $x = 5$" [you do not have two points to compare1!], but it makes sense to say that "$f$ is decreasing on $(-3, -2)$". (And that would be true in our case.)
Let's apply the definition
Now it is easy to see that our function $f$ is indeed decreasing on each of the subsets $(-infty, -1)$, $(-1, 1)$, and $(1, infty)$. For example, if you pick any two numbers $x$ and $y$ in $(-1, 1)$, it is obvious that if $x < y$, then $f(x) ge f(y)$.
Notice how beautifully the precise definition captures the essence of our intuitive idea of what it means to be "decreasing"!2
But with this precise knowledge, it also becomes very easy for us to see why $f$ isn't decreasing on $(-infty, -1) cup (-1, 1) cup (1, infty) = mathbf{R} setminus {-1, +1} = D_f$.
Think about it for a while.
For example, take $x = -2$ and $y = 2$. Clearly $x < y$, but $f(x) ge f(y)$ doesn't hold. [Intuitively: when we moved to the right, the value of the function increased!]
So it is very much not the case that $f$ is decreasing on $D_f$.
A small note on terminology
If the exam question asked for "the intervals where the function is decreasing", you should probably answer it by giving one or more intervals -- and not sets that aren't intervals. But the union clearly isn't an interval, so this also makes it obvious that that answer is incorrect.
[For example, there are no two real numbers $a$ and $b$ such that
$$mathbf{R} setminus {-1, +1} = (a, b)$$
so the union clearly isn't a bounded open interval.]
A subtle difference
A different exam question could have asked you to find the points in $D_f$ where the function's derivative is non-positive. This is a related but different question. The derivative of a function is also a function, and it has a value at each point where it is defined. So in contrast to the discussion above, it does make sense to say that the derivative of $f$ is non-positive at the particular point $x = 5$, say.
If you compute the derivative of $f$, you find a function $f'$ that is also defined on $D_f$. And its value is non-positive (even negative) everywhere. So it is true that the derivative is non-positive in each of the subsets $(-infty, -1)$, $(-1, 1)$, and $(1, infty)$. But this immediately implies that the derivative is non-positive in their union $D_f$.
Consequently, this hypothetical exam question could (indeed, should) be answered by giving the union in its most simplified form.
Footnotes
1 Exercise (very technical): What do you have to say about the statement "$f$ is decreasing on ${5}$"?
2 Yeah, this is one of the reasons why I love mathematics. And the $epsilondelta$ definitions in calculus ... Beautiful.
New contributor
(+1) Your effort to explain the answer in details and also explaining related material is incredible. However you should note that sometimes people would look for the most simple answer rather than the most detailed answer (even though the latter provides a better point of view on this mathematical area in general). To save your time I suggest you don't read what's not written in the question. The OP asks specifically about the difference in notations and it seems that he already understand the definition of increasing/decreasing function (or at least he don't ask about it).
– Yanko
Dec 23 at 15:10
@Yanko: It is hard to tell exactly what the OP understands and what (s)he doesn't fully understand. Still, I think it is good if a Q contains several answers with different approaches and levels of detail. Plus, I have all the time in the world! :)
– Andreas Rejbrand
Dec 23 at 15:14
Yes it's hard to tell what the OP understand. In fact sometimes detailed answers are well received and useful. This time I'm afraid that the OP is unlikely to read your answer but it might come in handy for other users. Anyway if you continue provide such answers and with "all the time in the world" you have I'm sure you'll become a great contributor for this site.
– Yanko
Dec 23 at 15:28
And the geometric interpretation of the formula $x<yRightarrow f(x)ge g(y)$ is the following. Whenever you pick two distinct points from the graph of $f$ within the domain in which we consider whether $f$ is decreasing, then the leftmost one of the two points (let us call it $(x,f(x))$) is located higher (in ordinate) than the rightmost one ($(y,f(y))$). In other words, if you pick the two distinct points on the graph, the secant passing through them has a negative (or zero) slope. If you pick points from different branches in your image, it is not hard to get a positive secant slope.
– Jeppe Stig Nielsen
Dec 24 at 12:22
add a comment |
Introduction
So, we are considering the function $f$ defined by
$$f(x) = frac{2x^2+x-1}{x^2-1}.$$
Clearly, this function is defined everywhere except at $x = pm 1$, so its domain is $D_f = mathbf{R} setminus {-1, +1}.$
Let us draw its graph, just so we can see very clearly what is going on:
Now, we are interested in knowing where the function is decreasing. Obviously, to answer this question, we need to know exactly what the word "decreasing" means in this context.
What does the word mean?
Likely your calculus textbook contains a definition very similar to this one:
A function $f : D_f to mathbf{R}$ is decreasing on the set $U subset D_f$ if, for all $x, y in U$, it holds that $x < y Rightarrow f(x) ge f(y)$.
So, fundamentally, the concept of "decreasing" applies not to a function alone -- or to a function at a particular point --, but to a function and a subset of its domain ($U$ in the definition above). For instance, it is meaningless to say that "$f$ is decreasing at $x = 5$" [you do not have two points to compare1!], but it makes sense to say that "$f$ is decreasing on $(-3, -2)$". (And that would be true in our case.)
Let's apply the definition
Now it is easy to see that our function $f$ is indeed decreasing on each of the subsets $(-infty, -1)$, $(-1, 1)$, and $(1, infty)$. For example, if you pick any two numbers $x$ and $y$ in $(-1, 1)$, it is obvious that if $x < y$, then $f(x) ge f(y)$.
Notice how beautifully the precise definition captures the essence of our intuitive idea of what it means to be "decreasing"!2
But with this precise knowledge, it also becomes very easy for us to see why $f$ isn't decreasing on $(-infty, -1) cup (-1, 1) cup (1, infty) = mathbf{R} setminus {-1, +1} = D_f$.
Think about it for a while.
For example, take $x = -2$ and $y = 2$. Clearly $x < y$, but $f(x) ge f(y)$ doesn't hold. [Intuitively: when we moved to the right, the value of the function increased!]
So it is very much not the case that $f$ is decreasing on $D_f$.
A small note on terminology
If the exam question asked for "the intervals where the function is decreasing", you should probably answer it by giving one or more intervals -- and not sets that aren't intervals. But the union clearly isn't an interval, so this also makes it obvious that that answer is incorrect.
[For example, there are no two real numbers $a$ and $b$ such that
$$mathbf{R} setminus {-1, +1} = (a, b)$$
so the union clearly isn't a bounded open interval.]
A subtle difference
A different exam question could have asked you to find the points in $D_f$ where the function's derivative is non-positive. This is a related but different question. The derivative of a function is also a function, and it has a value at each point where it is defined. So in contrast to the discussion above, it does make sense to say that the derivative of $f$ is non-positive at the particular point $x = 5$, say.
If you compute the derivative of $f$, you find a function $f'$ that is also defined on $D_f$. And its value is non-positive (even negative) everywhere. So it is true that the derivative is non-positive in each of the subsets $(-infty, -1)$, $(-1, 1)$, and $(1, infty)$. But this immediately implies that the derivative is non-positive in their union $D_f$.
Consequently, this hypothetical exam question could (indeed, should) be answered by giving the union in its most simplified form.
Footnotes
1 Exercise (very technical): What do you have to say about the statement "$f$ is decreasing on ${5}$"?
2 Yeah, this is one of the reasons why I love mathematics. And the $epsilondelta$ definitions in calculus ... Beautiful.
New contributor
(+1) Your effort to explain the answer in details and also explaining related material is incredible. However you should note that sometimes people would look for the most simple answer rather than the most detailed answer (even though the latter provides a better point of view on this mathematical area in general). To save your time I suggest you don't read what's not written in the question. The OP asks specifically about the difference in notations and it seems that he already understand the definition of increasing/decreasing function (or at least he don't ask about it).
– Yanko
Dec 23 at 15:10
@Yanko: It is hard to tell exactly what the OP understands and what (s)he doesn't fully understand. Still, I think it is good if a Q contains several answers with different approaches and levels of detail. Plus, I have all the time in the world! :)
– Andreas Rejbrand
Dec 23 at 15:14
Yes it's hard to tell what the OP understand. In fact sometimes detailed answers are well received and useful. This time I'm afraid that the OP is unlikely to read your answer but it might come in handy for other users. Anyway if you continue provide such answers and with "all the time in the world" you have I'm sure you'll become a great contributor for this site.
– Yanko
Dec 23 at 15:28
And the geometric interpretation of the formula $x<yRightarrow f(x)ge g(y)$ is the following. Whenever you pick two distinct points from the graph of $f$ within the domain in which we consider whether $f$ is decreasing, then the leftmost one of the two points (let us call it $(x,f(x))$) is located higher (in ordinate) than the rightmost one ($(y,f(y))$). In other words, if you pick the two distinct points on the graph, the secant passing through them has a negative (or zero) slope. If you pick points from different branches in your image, it is not hard to get a positive secant slope.
– Jeppe Stig Nielsen
Dec 24 at 12:22
add a comment |
Introduction
So, we are considering the function $f$ defined by
$$f(x) = frac{2x^2+x-1}{x^2-1}.$$
Clearly, this function is defined everywhere except at $x = pm 1$, so its domain is $D_f = mathbf{R} setminus {-1, +1}.$
Let us draw its graph, just so we can see very clearly what is going on:
Now, we are interested in knowing where the function is decreasing. Obviously, to answer this question, we need to know exactly what the word "decreasing" means in this context.
What does the word mean?
Likely your calculus textbook contains a definition very similar to this one:
A function $f : D_f to mathbf{R}$ is decreasing on the set $U subset D_f$ if, for all $x, y in U$, it holds that $x < y Rightarrow f(x) ge f(y)$.
So, fundamentally, the concept of "decreasing" applies not to a function alone -- or to a function at a particular point --, but to a function and a subset of its domain ($U$ in the definition above). For instance, it is meaningless to say that "$f$ is decreasing at $x = 5$" [you do not have two points to compare1!], but it makes sense to say that "$f$ is decreasing on $(-3, -2)$". (And that would be true in our case.)
Let's apply the definition
Now it is easy to see that our function $f$ is indeed decreasing on each of the subsets $(-infty, -1)$, $(-1, 1)$, and $(1, infty)$. For example, if you pick any two numbers $x$ and $y$ in $(-1, 1)$, it is obvious that if $x < y$, then $f(x) ge f(y)$.
Notice how beautifully the precise definition captures the essence of our intuitive idea of what it means to be "decreasing"!2
But with this precise knowledge, it also becomes very easy for us to see why $f$ isn't decreasing on $(-infty, -1) cup (-1, 1) cup (1, infty) = mathbf{R} setminus {-1, +1} = D_f$.
Think about it for a while.
For example, take $x = -2$ and $y = 2$. Clearly $x < y$, but $f(x) ge f(y)$ doesn't hold. [Intuitively: when we moved to the right, the value of the function increased!]
So it is very much not the case that $f$ is decreasing on $D_f$.
A small note on terminology
If the exam question asked for "the intervals where the function is decreasing", you should probably answer it by giving one or more intervals -- and not sets that aren't intervals. But the union clearly isn't an interval, so this also makes it obvious that that answer is incorrect.
[For example, there are no two real numbers $a$ and $b$ such that
$$mathbf{R} setminus {-1, +1} = (a, b)$$
so the union clearly isn't a bounded open interval.]
A subtle difference
A different exam question could have asked you to find the points in $D_f$ where the function's derivative is non-positive. This is a related but different question. The derivative of a function is also a function, and it has a value at each point where it is defined. So in contrast to the discussion above, it does make sense to say that the derivative of $f$ is non-positive at the particular point $x = 5$, say.
If you compute the derivative of $f$, you find a function $f'$ that is also defined on $D_f$. And its value is non-positive (even negative) everywhere. So it is true that the derivative is non-positive in each of the subsets $(-infty, -1)$, $(-1, 1)$, and $(1, infty)$. But this immediately implies that the derivative is non-positive in their union $D_f$.
Consequently, this hypothetical exam question could (indeed, should) be answered by giving the union in its most simplified form.
Footnotes
1 Exercise (very technical): What do you have to say about the statement "$f$ is decreasing on ${5}$"?
2 Yeah, this is one of the reasons why I love mathematics. And the $epsilondelta$ definitions in calculus ... Beautiful.
New contributor
Introduction
So, we are considering the function $f$ defined by
$$f(x) = frac{2x^2+x-1}{x^2-1}.$$
Clearly, this function is defined everywhere except at $x = pm 1$, so its domain is $D_f = mathbf{R} setminus {-1, +1}.$
Let us draw its graph, just so we can see very clearly what is going on:
Now, we are interested in knowing where the function is decreasing. Obviously, to answer this question, we need to know exactly what the word "decreasing" means in this context.
What does the word mean?
Likely your calculus textbook contains a definition very similar to this one:
A function $f : D_f to mathbf{R}$ is decreasing on the set $U subset D_f$ if, for all $x, y in U$, it holds that $x < y Rightarrow f(x) ge f(y)$.
So, fundamentally, the concept of "decreasing" applies not to a function alone -- or to a function at a particular point --, but to a function and a subset of its domain ($U$ in the definition above). For instance, it is meaningless to say that "$f$ is decreasing at $x = 5$" [you do not have two points to compare1!], but it makes sense to say that "$f$ is decreasing on $(-3, -2)$". (And that would be true in our case.)
Let's apply the definition
Now it is easy to see that our function $f$ is indeed decreasing on each of the subsets $(-infty, -1)$, $(-1, 1)$, and $(1, infty)$. For example, if you pick any two numbers $x$ and $y$ in $(-1, 1)$, it is obvious that if $x < y$, then $f(x) ge f(y)$.
Notice how beautifully the precise definition captures the essence of our intuitive idea of what it means to be "decreasing"!2
But with this precise knowledge, it also becomes very easy for us to see why $f$ isn't decreasing on $(-infty, -1) cup (-1, 1) cup (1, infty) = mathbf{R} setminus {-1, +1} = D_f$.
Think about it for a while.
For example, take $x = -2$ and $y = 2$. Clearly $x < y$, but $f(x) ge f(y)$ doesn't hold. [Intuitively: when we moved to the right, the value of the function increased!]
So it is very much not the case that $f$ is decreasing on $D_f$.
A small note on terminology
If the exam question asked for "the intervals where the function is decreasing", you should probably answer it by giving one or more intervals -- and not sets that aren't intervals. But the union clearly isn't an interval, so this also makes it obvious that that answer is incorrect.
[For example, there are no two real numbers $a$ and $b$ such that
$$mathbf{R} setminus {-1, +1} = (a, b)$$
so the union clearly isn't a bounded open interval.]
A subtle difference
A different exam question could have asked you to find the points in $D_f$ where the function's derivative is non-positive. This is a related but different question. The derivative of a function is also a function, and it has a value at each point where it is defined. So in contrast to the discussion above, it does make sense to say that the derivative of $f$ is non-positive at the particular point $x = 5$, say.
If you compute the derivative of $f$, you find a function $f'$ that is also defined on $D_f$. And its value is non-positive (even negative) everywhere. So it is true that the derivative is non-positive in each of the subsets $(-infty, -1)$, $(-1, 1)$, and $(1, infty)$. But this immediately implies that the derivative is non-positive in their union $D_f$.
Consequently, this hypothetical exam question could (indeed, should) be answered by giving the union in its most simplified form.
Footnotes
1 Exercise (very technical): What do you have to say about the statement "$f$ is decreasing on ${5}$"?
2 Yeah, this is one of the reasons why I love mathematics. And the $epsilondelta$ definitions in calculus ... Beautiful.
New contributor
edited Dec 24 at 13:34
New contributor
answered Dec 23 at 13:33
Andreas Rejbrand
1915
1915
New contributor
New contributor
(+1) Your effort to explain the answer in details and also explaining related material is incredible. However you should note that sometimes people would look for the most simple answer rather than the most detailed answer (even though the latter provides a better point of view on this mathematical area in general). To save your time I suggest you don't read what's not written in the question. The OP asks specifically about the difference in notations and it seems that he already understand the definition of increasing/decreasing function (or at least he don't ask about it).
– Yanko
Dec 23 at 15:10
@Yanko: It is hard to tell exactly what the OP understands and what (s)he doesn't fully understand. Still, I think it is good if a Q contains several answers with different approaches and levels of detail. Plus, I have all the time in the world! :)
– Andreas Rejbrand
Dec 23 at 15:14
Yes it's hard to tell what the OP understand. In fact sometimes detailed answers are well received and useful. This time I'm afraid that the OP is unlikely to read your answer but it might come in handy for other users. Anyway if you continue provide such answers and with "all the time in the world" you have I'm sure you'll become a great contributor for this site.
– Yanko
Dec 23 at 15:28
And the geometric interpretation of the formula $x<yRightarrow f(x)ge g(y)$ is the following. Whenever you pick two distinct points from the graph of $f$ within the domain in which we consider whether $f$ is decreasing, then the leftmost one of the two points (let us call it $(x,f(x))$) is located higher (in ordinate) than the rightmost one ($(y,f(y))$). In other words, if you pick the two distinct points on the graph, the secant passing through them has a negative (or zero) slope. If you pick points from different branches in your image, it is not hard to get a positive secant slope.
– Jeppe Stig Nielsen
Dec 24 at 12:22
add a comment |
(+1) Your effort to explain the answer in details and also explaining related material is incredible. However you should note that sometimes people would look for the most simple answer rather than the most detailed answer (even though the latter provides a better point of view on this mathematical area in general). To save your time I suggest you don't read what's not written in the question. The OP asks specifically about the difference in notations and it seems that he already understand the definition of increasing/decreasing function (or at least he don't ask about it).
– Yanko
Dec 23 at 15:10
@Yanko: It is hard to tell exactly what the OP understands and what (s)he doesn't fully understand. Still, I think it is good if a Q contains several answers with different approaches and levels of detail. Plus, I have all the time in the world! :)
– Andreas Rejbrand
Dec 23 at 15:14
Yes it's hard to tell what the OP understand. In fact sometimes detailed answers are well received and useful. This time I'm afraid that the OP is unlikely to read your answer but it might come in handy for other users. Anyway if you continue provide such answers and with "all the time in the world" you have I'm sure you'll become a great contributor for this site.
– Yanko
Dec 23 at 15:28
And the geometric interpretation of the formula $x<yRightarrow f(x)ge g(y)$ is the following. Whenever you pick two distinct points from the graph of $f$ within the domain in which we consider whether $f$ is decreasing, then the leftmost one of the two points (let us call it $(x,f(x))$) is located higher (in ordinate) than the rightmost one ($(y,f(y))$). In other words, if you pick the two distinct points on the graph, the secant passing through them has a negative (or zero) slope. If you pick points from different branches in your image, it is not hard to get a positive secant slope.
– Jeppe Stig Nielsen
Dec 24 at 12:22
(+1) Your effort to explain the answer in details and also explaining related material is incredible. However you should note that sometimes people would look for the most simple answer rather than the most detailed answer (even though the latter provides a better point of view on this mathematical area in general). To save your time I suggest you don't read what's not written in the question. The OP asks specifically about the difference in notations and it seems that he already understand the definition of increasing/decreasing function (or at least he don't ask about it).
– Yanko
Dec 23 at 15:10
(+1) Your effort to explain the answer in details and also explaining related material is incredible. However you should note that sometimes people would look for the most simple answer rather than the most detailed answer (even though the latter provides a better point of view on this mathematical area in general). To save your time I suggest you don't read what's not written in the question. The OP asks specifically about the difference in notations and it seems that he already understand the definition of increasing/decreasing function (or at least he don't ask about it).
– Yanko
Dec 23 at 15:10
@Yanko: It is hard to tell exactly what the OP understands and what (s)he doesn't fully understand. Still, I think it is good if a Q contains several answers with different approaches and levels of detail. Plus, I have all the time in the world! :)
– Andreas Rejbrand
Dec 23 at 15:14
@Yanko: It is hard to tell exactly what the OP understands and what (s)he doesn't fully understand. Still, I think it is good if a Q contains several answers with different approaches and levels of detail. Plus, I have all the time in the world! :)
– Andreas Rejbrand
Dec 23 at 15:14
Yes it's hard to tell what the OP understand. In fact sometimes detailed answers are well received and useful. This time I'm afraid that the OP is unlikely to read your answer but it might come in handy for other users. Anyway if you continue provide such answers and with "all the time in the world" you have I'm sure you'll become a great contributor for this site.
– Yanko
Dec 23 at 15:28
Yes it's hard to tell what the OP understand. In fact sometimes detailed answers are well received and useful. This time I'm afraid that the OP is unlikely to read your answer but it might come in handy for other users. Anyway if you continue provide such answers and with "all the time in the world" you have I'm sure you'll become a great contributor for this site.
– Yanko
Dec 23 at 15:28
And the geometric interpretation of the formula $x<yRightarrow f(x)ge g(y)$ is the following. Whenever you pick two distinct points from the graph of $f$ within the domain in which we consider whether $f$ is decreasing, then the leftmost one of the two points (let us call it $(x,f(x))$) is located higher (in ordinate) than the rightmost one ($(y,f(y))$). In other words, if you pick the two distinct points on the graph, the secant passing through them has a negative (or zero) slope. If you pick points from different branches in your image, it is not hard to get a positive secant slope.
– Jeppe Stig Nielsen
Dec 24 at 12:22
And the geometric interpretation of the formula $x<yRightarrow f(x)ge g(y)$ is the following. Whenever you pick two distinct points from the graph of $f$ within the domain in which we consider whether $f$ is decreasing, then the leftmost one of the two points (let us call it $(x,f(x))$) is located higher (in ordinate) than the rightmost one ($(y,f(y))$). In other words, if you pick the two distinct points on the graph, the secant passing through them has a negative (or zero) slope. If you pick points from different branches in your image, it is not hard to get a positive secant slope.
– Jeppe Stig Nielsen
Dec 24 at 12:22
add a comment |
Because $(-infty,-1)cup(-1,1)cup(1,+infty)$ is not an interval.
For the same equation, for concavity intervals, the answer was given using union-sign notation though. I just don't understand why writing like this would be considered incorrect.
– Rafael Sofi-Zadeh
Dec 22 at 16:54
I believe that I provide an answer to that question.
– José Carlos Santos
Dec 22 at 16:56
1
But it might be reasonable to say $dfrac{x^3-x}{x^2-1}$ is increasing on $(-infty,-1)cup(-1,1)cup(1,+infty)$
– Henry
Dec 22 at 21:05
4
But the question was about finding intervals in which the function is increasing.
– José Carlos Santos
Dec 22 at 21:37
add a comment |
Because $(-infty,-1)cup(-1,1)cup(1,+infty)$ is not an interval.
For the same equation, for concavity intervals, the answer was given using union-sign notation though. I just don't understand why writing like this would be considered incorrect.
– Rafael Sofi-Zadeh
Dec 22 at 16:54
I believe that I provide an answer to that question.
– José Carlos Santos
Dec 22 at 16:56
1
But it might be reasonable to say $dfrac{x^3-x}{x^2-1}$ is increasing on $(-infty,-1)cup(-1,1)cup(1,+infty)$
– Henry
Dec 22 at 21:05
4
But the question was about finding intervals in which the function is increasing.
– José Carlos Santos
Dec 22 at 21:37
add a comment |
Because $(-infty,-1)cup(-1,1)cup(1,+infty)$ is not an interval.
Because $(-infty,-1)cup(-1,1)cup(1,+infty)$ is not an interval.
answered Dec 22 at 16:47
José Carlos Santos
149k22117219
149k22117219
For the same equation, for concavity intervals, the answer was given using union-sign notation though. I just don't understand why writing like this would be considered incorrect.
– Rafael Sofi-Zadeh
Dec 22 at 16:54
I believe that I provide an answer to that question.
– José Carlos Santos
Dec 22 at 16:56
1
But it might be reasonable to say $dfrac{x^3-x}{x^2-1}$ is increasing on $(-infty,-1)cup(-1,1)cup(1,+infty)$
– Henry
Dec 22 at 21:05
4
But the question was about finding intervals in which the function is increasing.
– José Carlos Santos
Dec 22 at 21:37
add a comment |
For the same equation, for concavity intervals, the answer was given using union-sign notation though. I just don't understand why writing like this would be considered incorrect.
– Rafael Sofi-Zadeh
Dec 22 at 16:54
I believe that I provide an answer to that question.
– José Carlos Santos
Dec 22 at 16:56
1
But it might be reasonable to say $dfrac{x^3-x}{x^2-1}$ is increasing on $(-infty,-1)cup(-1,1)cup(1,+infty)$
– Henry
Dec 22 at 21:05
4
But the question was about finding intervals in which the function is increasing.
– José Carlos Santos
Dec 22 at 21:37
For the same equation, for concavity intervals, the answer was given using union-sign notation though. I just don't understand why writing like this would be considered incorrect.
– Rafael Sofi-Zadeh
Dec 22 at 16:54
For the same equation, for concavity intervals, the answer was given using union-sign notation though. I just don't understand why writing like this would be considered incorrect.
– Rafael Sofi-Zadeh
Dec 22 at 16:54
I believe that I provide an answer to that question.
– José Carlos Santos
Dec 22 at 16:56
I believe that I provide an answer to that question.
– José Carlos Santos
Dec 22 at 16:56
1
1
But it might be reasonable to say $dfrac{x^3-x}{x^2-1}$ is increasing on $(-infty,-1)cup(-1,1)cup(1,+infty)$
– Henry
Dec 22 at 21:05
But it might be reasonable to say $dfrac{x^3-x}{x^2-1}$ is increasing on $(-infty,-1)cup(-1,1)cup(1,+infty)$
– Henry
Dec 22 at 21:05
4
4
But the question was about finding intervals in which the function is increasing.
– José Carlos Santos
Dec 22 at 21:37
But the question was about finding intervals in which the function is increasing.
– José Carlos Santos
Dec 22 at 21:37
add a comment |
The existing answers have focused on the fact that $f$ is not increasing from one interval to the next. An even better example would be $g(x)$ defined as $-1/x$ when $x<0$ and $sqrt{x}$ when $xge 0$. Then $g$ is increasing on $(-infty,0);[0,infty)$. But we definitely can't say that $g$ is increasing on $(-infty,0)cup[0,infty)=(-infty,infty)$.
add a comment |
The existing answers have focused on the fact that $f$ is not increasing from one interval to the next. An even better example would be $g(x)$ defined as $-1/x$ when $x<0$ and $sqrt{x}$ when $xge 0$. Then $g$ is increasing on $(-infty,0);[0,infty)$. But we definitely can't say that $g$ is increasing on $(-infty,0)cup[0,infty)=(-infty,infty)$.
add a comment |
The existing answers have focused on the fact that $f$ is not increasing from one interval to the next. An even better example would be $g(x)$ defined as $-1/x$ when $x<0$ and $sqrt{x}$ when $xge 0$. Then $g$ is increasing on $(-infty,0);[0,infty)$. But we definitely can't say that $g$ is increasing on $(-infty,0)cup[0,infty)=(-infty,infty)$.
The existing answers have focused on the fact that $f$ is not increasing from one interval to the next. An even better example would be $g(x)$ defined as $-1/x$ when $x<0$ and $sqrt{x}$ when $xge 0$. Then $g$ is increasing on $(-infty,0);[0,infty)$. But we definitely can't say that $g$ is increasing on $(-infty,0)cup[0,infty)=(-infty,infty)$.
answered Dec 23 at 21:14
Teepeemm
69259
69259
add a comment |
add a comment |
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
To learn more, see our tips on writing great answers.
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%2f3049628%2fwhy-is-union-sign-not-allowed-to-denote-intervals-of-function-increasing%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
9
Increasing on the union requires the function to be increasing across the discontinuities of the domain.
– Yves Daoust
Dec 22 at 17:14
2
@AndreasRejbrand: that should have been an answer :)
– Taladris
Dec 23 at 3:56
3
I'd say that the point is that the union of intervals fails to be an interval and the question is specifically asking for intervals.
– Dirk
Dec 23 at 16:12
1
There are several correct answers here. My comment is to suggest that the formulation you ask about causes unnecessary trouble by trying to use symbols rather than words to describe the answer. Reserve formal mathematical notation for the places where it is really called for.
– Ethan Bolker
Dec 23 at 16:23