Example of function which is twice differentiable with $f,f''$ strictly increasing but $lim_{xto...
up vote
2
down vote
favorite
I wanted to find Example of function which is twice differentiable with $f,f''$ strictly increasing but $lim_{xto infty}f(x)neq infty$.
My usual notion fails for above statement .
As I thought if $f$ is strictly increasing and $f''$ strictly incresing means $f'$ should also in increase.
Where is my mistake in my thinking ?
Any help will be appreciated
real-analysis derivatives
edited Nov 27 at 9:03
Tianlalu
3,0101938
asked Nov 27 at 9:01
Shubham
1,5921519
add a comment |
up vote
2
down vote
favorite
I wanted to find Example of function which is twice differentiable with $f,f''$ strictly increasing but $lim_{xto infty}f(x)neq infty$.
My usual notion fails for above statement .
As I thought if $f$ is strictly increasing and $f''$ strictly incresing means $f'$ should also in increase.
Where is my mistake in my thinking ?
Any help will be appreciated
real-analysis derivatives
edited Nov 27 at 9:03
Tianlalu
3,0101938
asked Nov 27 at 9:01
Shubham
1,5921519
1
If Jerry Mouse's grandfather is a funny character and Jerry Mouse is also a funny character, it isn't necessary that Jerry Mouse's father is a funny character. (Sorry if this is childish :P)
– L. F.
Nov 27 at 9:21
1
The correct intuition should be convex ($f''>0$) to get $f'$ increasing.
– Calvin Khor
Nov 27 at 9:25
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I wanted to find Example of function which is twice differentiable with $f,f''$ strictly increasing but $lim_{xto infty}f(x)neq infty$.
My usual notion fails for above statement .
As I thought if $f$ is strictly increasing and $f''$ strictly incresing means $f'$ should also in increase.
Where is my mistake in my thinking ?
Any help will be appreciated
real-analysis derivatives
edited Nov 27 at 9:03
Tianlalu
3,0101938
asked Nov 27 at 9:01
Shubham
1,5921519
I wanted to find Example of function which is twice differentiable with $f,f''$ strictly increasing but $lim_{xto infty}f(x)neq infty$.
My usual notion fails for above statement .
As I thought if $f$ is strictly increasing and $f''$ strictly incresing means $f'$ should also in increase.
Where is my mistake in my thinking ?
Any help will be appreciated
real-analysis derivatives
real-analysis derivatives
edited Nov 27 at 9:03
Tianlalu
3,0101938
asked Nov 27 at 9:01
Shubham
1,5921519
edited Nov 27 at 9:03
Tianlalu
3,0101938
asked Nov 27 at 9:01
Shubham
1,5921519
edited Nov 27 at 9:03
Tianlalu
3,0101938
edited Nov 27 at 9:03
Tianlalu
3,0101938
edited Nov 27 at 9:03
Tianlalu
3,0101938
3,0101938
asked Nov 27 at 9:01
Shubham
1,5921519
asked Nov 27 at 9:01
Shubham
1,5921519
asked Nov 27 at 9:01
Shubham
1,5921519
1,5921519
1
If Jerry Mouse's grandfather is a funny character and Jerry Mouse is also a funny character, it isn't necessary that Jerry Mouse's father is a funny character. (Sorry if this is childish :P)
– L. F.
Nov 27 at 9:21
1
The correct intuition should be convex ($f''>0$) to get $f'$ increasing.
– Calvin Khor
Nov 27 at 9:25
add a comment |
1
If Jerry Mouse's grandfather is a funny character and Jerry Mouse is also a funny character, it isn't necessary that Jerry Mouse's father is a funny character. (Sorry if this is childish :P)
– L. F.
Nov 27 at 9:21
1
The correct intuition should be convex ($f''>0$) to get $f'$ increasing.
– Calvin Khor
Nov 27 at 9:25
1
1
If Jerry Mouse's grandfather is a funny character and Jerry Mouse is also a funny character, it isn't necessary that Jerry Mouse's father is a funny character. (Sorry if this is childish :P)
– L. F.
Nov 27 at 9:21
If Jerry Mouse's grandfather is a funny character and Jerry Mouse is also a funny character, it isn't necessary that Jerry Mouse's father is a funny character. (Sorry if this is childish :P)
– L. F.
Nov 27 at 9:21
1
1
The correct intuition should be convex ($f''>0$) to get $f'$ increasing.
– Calvin Khor
Nov 27 at 9:25
The correct intuition should be convex ($f''>0$) to get $f'$ increasing.
– Calvin Khor
Nov 27 at 9:25
add a comment |
3 Answers
3
active
oldest
votes
up vote
10
down vote
accepted
Isn't $-e^{-x}$ such an example?
answered Nov 27 at 9:04
Kavi Rama Murthy
47.4k31854
Excuse me, if I replace 'strictly increasing' by 'strictly positive', then what would be your answer ??
– Anik Bhowmick
Dec 3 at 16:08
add a comment |
up vote
4
down vote
What about the simpler
$$f(x)=-frac1x$$
answered Nov 27 at 9:21
gimusi
93k94495
add a comment |
up vote
4
down vote
Your mistake in your thinking is to believe that "strictly increasing" means "positive".
Going from negative to less negative is increasing. So you can have $f''$ strictly increasing but negative, with a decreasing $f'$ (but positive), hence an increasing $f$ as well.
You need to choose a function that does all that (see Kavi example).
answered Nov 27 at 9:21
Martigan
5,205917
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',
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%2f3015527%2fexample-of-function-which-is-twice-differentiable-with-f-f-strictly-increasi%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
10
down vote
accepted
Isn't $-e^{-x}$ such an example?
answered Nov 27 at 9:04
Kavi Rama Murthy
47.4k31854
Excuse me, if I replace 'strictly increasing' by 'strictly positive', then what would be your answer ??
– Anik Bhowmick
Dec 3 at 16:08
add a comment |
up vote
10
down vote
accepted
Isn't $-e^{-x}$ such an example?
answered Nov 27 at 9:04
Kavi Rama Murthy
47.4k31854
Excuse me, if I replace 'strictly increasing' by 'strictly positive', then what would be your answer ??
– Anik Bhowmick
Dec 3 at 16:08
add a comment |
up vote
10
down vote
accepted
up vote
10
down vote
accepted
Isn't $-e^{-x}$ such an example?
answered Nov 27 at 9:04
Kavi Rama Murthy
47.4k31854
Isn't $-e^{-x}$ such an example?
answered Nov 27 at 9:04
Kavi Rama Murthy
47.4k31854
answered Nov 27 at 9:04
Kavi Rama Murthy
47.4k31854
answered Nov 27 at 9:04
Kavi Rama Murthy
47.4k31854
answered Nov 27 at 9:04
Kavi Rama Murthy
47.4k31854
47.4k31854
Excuse me, if I replace 'strictly increasing' by 'strictly positive', then what would be your answer ??
– Anik Bhowmick
Dec 3 at 16:08
add a comment |
Excuse me, if I replace 'strictly increasing' by 'strictly positive', then what would be your answer ??
– Anik Bhowmick
Dec 3 at 16:08
Excuse me, if I replace 'strictly increasing' by 'strictly positive', then what would be your answer ??
– Anik Bhowmick
Dec 3 at 16:08
Excuse me, if I replace 'strictly increasing' by 'strictly positive', then what would be your answer ??
– Anik Bhowmick
Dec 3 at 16:08
add a comment |
up vote
4
down vote
What about the simpler
$$f(x)=-frac1x$$
answered Nov 27 at 9:21
gimusi
93k94495
add a comment |
up vote
4
down vote
What about the simpler
$$f(x)=-frac1x$$
answered Nov 27 at 9:21
gimusi
93k94495
add a comment |
up vote
4
down vote
up vote
4
down vote
What about the simpler
$$f(x)=-frac1x$$
answered Nov 27 at 9:21
gimusi
93k94495
What about the simpler
$$f(x)=-frac1x$$
answered Nov 27 at 9:21
gimusi
93k94495
answered Nov 27 at 9:21
gimusi
93k94495
answered Nov 27 at 9:21
gimusi
93k94495
answered Nov 27 at 9:21
gimusi
93k94495
93k94495
add a comment |
add a comment |
up vote
4
down vote
Your mistake in your thinking is to believe that "strictly increasing" means "positive".
Going from negative to less negative is increasing. So you can have $f''$ strictly increasing but negative, with a decreasing $f'$ (but positive), hence an increasing $f$ as well.
You need to choose a function that does all that (see Kavi example).
answered Nov 27 at 9:21
Martigan
5,205917
add a comment |
up vote
4
down vote
Your mistake in your thinking is to believe that "strictly increasing" means "positive".
Going from negative to less negative is increasing. So you can have $f''$ strictly increasing but negative, with a decreasing $f'$ (but positive), hence an increasing $f$ as well.
You need to choose a function that does all that (see Kavi example).
answered Nov 27 at 9:21
Martigan
5,205917
add a comment |
up vote
4
down vote
up vote
4
down vote
Your mistake in your thinking is to believe that "strictly increasing" means "positive".
Going from negative to less negative is increasing. So you can have $f''$ strictly increasing but negative, with a decreasing $f'$ (but positive), hence an increasing $f$ as well.
You need to choose a function that does all that (see Kavi example).
answered Nov 27 at 9:21
Martigan
5,205917
Your mistake in your thinking is to believe that "strictly increasing" means "positive".
Going from negative to less negative is increasing. So you can have $f''$ strictly increasing but negative, with a decreasing $f'$ (but positive), hence an increasing $f$ as well.
You need to choose a function that does all that (see Kavi example).
answered Nov 27 at 9:21
Martigan
5,205917
answered Nov 27 at 9:21
Martigan
5,205917
answered Nov 27 at 9:21
Martigan
5,205917
answered Nov 27 at 9:21
Martigan
5,205917
5,205917
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%2f3015527%2fexample-of-function-which-is-twice-differentiable-with-f-f-strictly-increasi%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
1
If Jerry Mouse's grandfather is a funny character and Jerry Mouse is also a funny character, it isn't necessary that Jerry Mouse's father is a funny character. (Sorry if this is childish :P)
– L. F.
Nov 27 at 9:21
1
The correct intuition should be convex ($f''>0$) to get $f'$ increasing.
– Calvin Khor
Nov 27 at 9:25