Finding absolute and relative extrema of a function
up vote
0
down vote
favorite
I don't know if what I'm doing is correct.
Let $f:mathbb{R} to mathbb{R}$ be a function, $f(x)=e^{x^4-3x^2}$.
Choose the correct answer:
$f$ has an absolute maximum at $x=0$.
$f$ has a relative maximum at $x=0$ which isn't absolute.
$f$ has a absolute minimum at $x=1$ and doesn't have any relative minimum. (Does this even make sense?)
$f$ doesn't have relative maximum.
What I've been doing:
Since the question doesn't tell me the interval where I have to find the relative extrema, by the options the gave me I suppose that's $I=[0,1]$.
I know that when I have to find the relative extrema, I have to find the critical points for $f$ at the given interval, but this function doesn't have any. So how do I find it? (Algebraiclly).
When I look at the function it clearly does have a relative maximum at $x=0$ (Option 2), and it's clearly not absolute as $lim_{x to +infty} f(x) = +infty$ (is finding the limit as $x to +infty$ correct when trying to find the absolute extrema?).
So, how do I prove that option 2 is correct without looking at the graphic?
calculus real-analysis
asked Nov 22 at 16:46
parishilton
15910
add a comment |
up vote
0
down vote
favorite
I don't know if what I'm doing is correct.
Let $f:mathbb{R} to mathbb{R}$ be a function, $f(x)=e^{x^4-3x^2}$.
Choose the correct answer:
$f$ has an absolute maximum at $x=0$.
$f$ has a relative maximum at $x=0$ which isn't absolute.
$f$ has a absolute minimum at $x=1$ and doesn't have any relative minimum. (Does this even make sense?)
$f$ doesn't have relative maximum.
What I've been doing:
Since the question doesn't tell me the interval where I have to find the relative extrema, by the options the gave me I suppose that's $I=[0,1]$.
I know that when I have to find the relative extrema, I have to find the critical points for $f$ at the given interval, but this function doesn't have any. So how do I find it? (Algebraiclly).
When I look at the function it clearly does have a relative maximum at $x=0$ (Option 2), and it's clearly not absolute as $lim_{x to +infty} f(x) = +infty$ (is finding the limit as $x to +infty$ correct when trying to find the absolute extrema?).
So, how do I prove that option 2 is correct without looking at the graphic?
calculus real-analysis
asked Nov 22 at 16:46
parishilton
15910
I don't understand you approach. First, you tell us that the domain of $f$ is $mathbb R$. Then you write that “the question doesn't tell me the interval where I have to find the relative extrema”. Finally, you decree that that interval is $[0,1]$.
– José Carlos Santos
Nov 22 at 16:49
Yes it's kinda confussing, I thought that the interval was $[0,1]$ by the options they gave me. So I thought that I had to find the relative extrema at that interval.
– parishilton
Nov 22 at 16:52
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I don't know if what I'm doing is correct.
Let $f:mathbb{R} to mathbb{R}$ be a function, $f(x)=e^{x^4-3x^2}$.
Choose the correct answer:
$f$ has an absolute maximum at $x=0$.
$f$ has a relative maximum at $x=0$ which isn't absolute.
$f$ has a absolute minimum at $x=1$ and doesn't have any relative minimum. (Does this even make sense?)
$f$ doesn't have relative maximum.
What I've been doing:
Since the question doesn't tell me the interval where I have to find the relative extrema, by the options the gave me I suppose that's $I=[0,1]$.
I know that when I have to find the relative extrema, I have to find the critical points for $f$ at the given interval, but this function doesn't have any. So how do I find it? (Algebraiclly).
When I look at the function it clearly does have a relative maximum at $x=0$ (Option 2), and it's clearly not absolute as $lim_{x to +infty} f(x) = +infty$ (is finding the limit as $x to +infty$ correct when trying to find the absolute extrema?).
So, how do I prove that option 2 is correct without looking at the graphic?
calculus real-analysis
asked Nov 22 at 16:46
parishilton
15910
I don't know if what I'm doing is correct.
Let $f:mathbb{R} to mathbb{R}$ be a function, $f(x)=e^{x^4-3x^2}$.
Choose the correct answer:
$f$ has an absolute maximum at $x=0$.
$f$ has a relative maximum at $x=0$ which isn't absolute.
$f$ has a absolute minimum at $x=1$ and doesn't have any relative minimum. (Does this even make sense?)
$f$ doesn't have relative maximum.
What I've been doing:
Since the question doesn't tell me the interval where I have to find the relative extrema, by the options the gave me I suppose that's $I=[0,1]$.
I know that when I have to find the relative extrema, I have to find the critical points for $f$ at the given interval, but this function doesn't have any. So how do I find it? (Algebraiclly).
When I look at the function it clearly does have a relative maximum at $x=0$ (Option 2), and it's clearly not absolute as $lim_{x to +infty} f(x) = +infty$ (is finding the limit as $x to +infty$ correct when trying to find the absolute extrema?).
So, how do I prove that option 2 is correct without looking at the graphic?
calculus real-analysis
calculus real-analysis
asked Nov 22 at 16:46
parishilton
15910
asked Nov 22 at 16:46
parishilton
15910
asked Nov 22 at 16:46
parishilton
15910
asked Nov 22 at 16:46
parishilton
15910
asked Nov 22 at 16:46
parishilton
15910
15910
I don't understand you approach. First, you tell us that the domain of $f$ is $mathbb R$. Then you write that “the question doesn't tell me the interval where I have to find the relative extrema”. Finally, you decree that that interval is $[0,1]$.
– José Carlos Santos
Nov 22 at 16:49
Yes it's kinda confussing, I thought that the interval was $[0,1]$ by the options they gave me. So I thought that I had to find the relative extrema at that interval.
– parishilton
Nov 22 at 16:52
add a comment |
I don't understand you approach. First, you tell us that the domain of $f$ is $mathbb R$. Then you write that “the question doesn't tell me the interval where I have to find the relative extrema”. Finally, you decree that that interval is $[0,1]$.
– José Carlos Santos
Nov 22 at 16:49
Yes it's kinda confussing, I thought that the interval was $[0,1]$ by the options they gave me. So I thought that I had to find the relative extrema at that interval.
– parishilton
Nov 22 at 16:52
I don't understand you approach. First, you tell us that the domain of $f$ is $mathbb R$. Then you write that “the question doesn't tell me the interval where I have to find the relative extrema”. Finally, you decree that that interval is $[0,1]$.
– José Carlos Santos
Nov 22 at 16:49
I don't understand you approach. First, you tell us that the domain of $f$ is $mathbb R$. Then you write that “the question doesn't tell me the interval where I have to find the relative extrema”. Finally, you decree that that interval is $[0,1]$.
– José Carlos Santos
Nov 22 at 16:49
Yes it's kinda confussing, I thought that the interval was $[0,1]$ by the options they gave me. So I thought that I had to find the relative extrema at that interval.
– parishilton
Nov 22 at 16:52
Yes it's kinda confussing, I thought that the interval was $[0,1]$ by the options they gave me. So I thought that I had to find the relative extrema at that interval.
– parishilton
Nov 22 at 16:52
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
HINT
We have that
$$f(x)=e^{x^4-3x^2}>0 implies f'(x)=(4x^3-6x)e^{x^4-3x^2}=0 implies x=0 ,lor , x=pm sqrt{frac23}$$
then consider the sign of $f'(x)$.
answered Nov 22 at 16:49
gimusi
92.7k94495
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%2f3009339%2ffinding-absolute-and-relative-extrema-of-a-function%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
HINT
We have that
$$f(x)=e^{x^4-3x^2}>0 implies f'(x)=(4x^3-6x)e^{x^4-3x^2}=0 implies x=0 ,lor , x=pm sqrt{frac23}$$
then consider the sign of $f'(x)$.
answered Nov 22 at 16:49
gimusi
92.7k94495
add a comment |
up vote
1
down vote
accepted
HINT
We have that
$$f(x)=e^{x^4-3x^2}>0 implies f'(x)=(4x^3-6x)e^{x^4-3x^2}=0 implies x=0 ,lor , x=pm sqrt{frac23}$$
then consider the sign of $f'(x)$.
answered Nov 22 at 16:49
gimusi
92.7k94495
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
HINT
We have that
$$f(x)=e^{x^4-3x^2}>0 implies f'(x)=(4x^3-6x)e^{x^4-3x^2}=0 implies x=0 ,lor , x=pm sqrt{frac23}$$
then consider the sign of $f'(x)$.
answered Nov 22 at 16:49
gimusi
92.7k94495
HINT
We have that
$$f(x)=e^{x^4-3x^2}>0 implies f'(x)=(4x^3-6x)e^{x^4-3x^2}=0 implies x=0 ,lor , x=pm sqrt{frac23}$$
then consider the sign of $f'(x)$.
answered Nov 22 at 16:49
gimusi
92.7k94495
answered Nov 22 at 16:49
gimusi
92.7k94495
answered Nov 22 at 16:49
gimusi
92.7k94495
answered Nov 22 at 16:49
gimusi
92.7k94495
92.7k94495
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%2f3009339%2ffinding-absolute-and-relative-extrema-of-a-function%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
I don't understand you approach. First, you tell us that the domain of $f$ is $mathbb R$. Then you write that “the question doesn't tell me the interval where I have to find the relative extrema”. Finally, you decree that that interval is $[0,1]$.
– José Carlos Santos
Nov 22 at 16:49
Yes it's kinda confussing, I thought that the interval was $[0,1]$ by the options they gave me. So I thought that I had to find the relative extrema at that interval.
– parishilton
Nov 22 at 16:52