State if the statement is True or False: The maximum value of $2x^3-9x^2-24x-20$ is $-7$.
State if the statement is True or False:
The maximum value of $2x^3-9x^2-24x-20$ is $-7$.
Let $f(x) = 2x^3-9x^2-24x-20$.
If we go by the derivative test:
$$f'(x) = 6x^2-18x-24 & f'(x) = 0 implies x=4,-1$$
At $x=-1$ we get $f(x)=-7$ and at $x=4$ we get $f(x) = -132$, so we have maximum value $-7$ by this method.
But this is a polynomial function, its value tends to infinity as $x to infty$.
So what can be said about the truth of the statement?
real-analysis analysis functions maxima-minima
asked Nov 22 '18 at 2:41
user8795
5,61961947
add a comment |
State if the statement is True or False:
The maximum value of $2x^3-9x^2-24x-20$ is $-7$.
Let $f(x) = 2x^3-9x^2-24x-20$.
If we go by the derivative test:
$$f'(x) = 6x^2-18x-24 & f'(x) = 0 implies x=4,-1$$
At $x=-1$ we get $f(x)=-7$ and at $x=4$ we get $f(x) = -132$, so we have maximum value $-7$ by this method.
But this is a polynomial function, its value tends to infinity as $x to infty$.
So what can be said about the truth of the statement?
real-analysis analysis functions maxima-minima
asked Nov 22 '18 at 2:41
user8795
5,61961947
Every local maximum $x$ of $f$ has $f'(x) = 0$, since $frac{f(x + h) - f(x)}{h}$ is negative for small $h > 0$ and positive for small $h < 0$; since $f$ is differentiable at $x$, the limit must equal $0$. That's necessary but not sufficient; as you note, $f$ is clearly unbounded.
– anomaly
Nov 22 '18 at 2:44
$f'(x) = 0 at local minimums also. Look at the dominating first term for the global max and mins.
– Phil H
Nov 22 '18 at 3:03
4
When they say "maximum" or "minimum" without qualification, they are generally referring to global phenomena, not local phenomena. So the statement is false.
– Deepak
Nov 22 '18 at 3:19
add a comment |
State if the statement is True or False:
The maximum value of $2x^3-9x^2-24x-20$ is $-7$.
Let $f(x) = 2x^3-9x^2-24x-20$.
If we go by the derivative test:
$$f'(x) = 6x^2-18x-24 & f'(x) = 0 implies x=4,-1$$
At $x=-1$ we get $f(x)=-7$ and at $x=4$ we get $f(x) = -132$, so we have maximum value $-7$ by this method.
But this is a polynomial function, its value tends to infinity as $x to infty$.
So what can be said about the truth of the statement?
real-analysis analysis functions maxima-minima
asked Nov 22 '18 at 2:41
user8795
5,61961947
State if the statement is True or False:
The maximum value of $2x^3-9x^2-24x-20$ is $-7$.
Let $f(x) = 2x^3-9x^2-24x-20$.
If we go by the derivative test:
$$f'(x) = 6x^2-18x-24 & f'(x) = 0 implies x=4,-1$$
At $x=-1$ we get $f(x)=-7$ and at $x=4$ we get $f(x) = -132$, so we have maximum value $-7$ by this method.
But this is a polynomial function, its value tends to infinity as $x to infty$.
So what can be said about the truth of the statement?
real-analysis analysis functions maxima-minima
real-analysis analysis functions maxima-minima
asked Nov 22 '18 at 2:41
user8795
5,61961947
asked Nov 22 '18 at 2:41
user8795
5,61961947
asked Nov 22 '18 at 2:41
user8795
5,61961947
asked Nov 22 '18 at 2:41
user8795
5,61961947
asked Nov 22 '18 at 2:41
user8795
5,61961947
5,61961947
Every local maximum $x$ of $f$ has $f'(x) = 0$, since $frac{f(x + h) - f(x)}{h}$ is negative for small $h > 0$ and positive for small $h < 0$; since $f$ is differentiable at $x$, the limit must equal $0$. That's necessary but not sufficient; as you note, $f$ is clearly unbounded.
– anomaly
Nov 22 '18 at 2:44
$f'(x) = 0 at local minimums also. Look at the dominating first term for the global max and mins.
– Phil H
Nov 22 '18 at 3:03
4
When they say "maximum" or "minimum" without qualification, they are generally referring to global phenomena, not local phenomena. So the statement is false.
– Deepak
Nov 22 '18 at 3:19
add a comment |
Every local maximum $x$ of $f$ has $f'(x) = 0$, since $frac{f(x + h) - f(x)}{h}$ is negative for small $h > 0$ and positive for small $h < 0$; since $f$ is differentiable at $x$, the limit must equal $0$. That's necessary but not sufficient; as you note, $f$ is clearly unbounded.
– anomaly
Nov 22 '18 at 2:44
$f'(x) = 0 at local minimums also. Look at the dominating first term for the global max and mins.
– Phil H
Nov 22 '18 at 3:03
4
When they say "maximum" or "minimum" without qualification, they are generally referring to global phenomena, not local phenomena. So the statement is false.
– Deepak
Nov 22 '18 at 3:19
Every local maximum $x$ of $f$ has $f'(x) = 0$, since $frac{f(x + h) - f(x)}{h}$ is negative for small $h > 0$ and positive for small $h < 0$; since $f$ is differentiable at $x$, the limit must equal $0$. That's necessary but not sufficient; as you note, $f$ is clearly unbounded.
– anomaly
Nov 22 '18 at 2:44
Every local maximum $x$ of $f$ has $f'(x) = 0$, since $frac{f(x + h) - f(x)}{h}$ is negative for small $h > 0$ and positive for small $h < 0$; since $f$ is differentiable at $x$, the limit must equal $0$. That's necessary but not sufficient; as you note, $f$ is clearly unbounded.
– anomaly
Nov 22 '18 at 2:44
$f'(x) = 0 at local minimums also. Look at the dominating first term for the global max and mins.
– Phil H
Nov 22 '18 at 3:03
$f'(x) = 0 at local minimums also. Look at the dominating first term for the global max and mins.
– Phil H
Nov 22 '18 at 3:03
4
4
When they say "maximum" or "minimum" without qualification, they are generally referring to global phenomena, not local phenomena. So the statement is false.
– Deepak
Nov 22 '18 at 3:19
When they say "maximum" or "minimum" without qualification, they are generally referring to global phenomena, not local phenomena. So the statement is false.
– Deepak
Nov 22 '18 at 3:19
add a comment |
5 Answers
5
active
oldest
votes
Hint: Let $x=10^{20}$. That should do it.
answered Nov 22 '18 at 2:45
Shaun
8,810113680
1
x=10 will also give positive value!
– user8795
Nov 22 '18 at 2:49
Well, there's your answer, @user8795.
– Shaun
Nov 22 '18 at 2:51
I think @anomaly's comment should be an answer and my answer should have been a comment. If you're looking for a quick, simple answer, though, mine will do the job if you recall the definition of what a maximum is.
– Shaun
Nov 22 '18 at 2:55
add a comment |
$f^{primeprime}(-1)=12(-1)-18=-30<0$. Thus $f$ has a local maximum at $-1$ and the local maximum value is $f(-1)=-7$. But it does not mean $f$ can not assume any value greater than $-7$. In fact the function $f$ is clearly unbounded. Here $f^{prime}(x)>0$ for all $xin (-infty,-1)$ and for all $xin (4,infty)$ and $f^{prime}(x)<0$ for all $xin (-1,4)$. Thus $f$ is strictly increasing in $(-infty,-1)cup (4,infty)$ and is strictly decreasing in $(-1,4)$. At $-1$ it has a local maximum and at $4$ it has a local minimum.
answered Nov 22 '18 at 3:14
Anupam
2,3741823
so the conclusion is false?
– user8795
Nov 22 '18 at 3:15
2
Unless the word "local" is mentioned, it is certainly false.
– Anupam
Nov 22 '18 at 3:17
add a comment |
Assume the opposite:
$$2x^3-9x^2-24x-20>-7$$
$$to 2x^3-9x^2-24x-13>0$$
$$to (2x-13)(x+1)^2>0$$
Pretty blatant from this.
answered Nov 22 '18 at 5:59
Rhys Hughes
5,0091427
Your use of implication (" $to$ ") is not enough. One must be able to go from your last inequality to your first. Indeed, this is possible, but you haven't made that clear.
– Shaun
Nov 23 '18 at 3:47
add a comment |
For large enough $x$, the polynomial grows unboundedly, therefore no finite maximum exists for the function. The points then you found, are just local maximum or minimum not global
answered Nov 22 '18 at 6:31
Mostafa Ayaz
14k3936
add a comment |
The differentiation & setting of the derivative to zero yields stationary points. A range-of-x ought to have been specified for this question, as without it it is ill-posed.
To say that the function has a maximum value of ∞ at x=∞ is a scrambling of the meaning of what's going-on, which is that the function increases without limit as x increases without limit, & therefore does not have a maximum value. If a range of x had been given, you would compare the -7 gotten as a local maximum to the value taken by the function at the upper end of the range of x (you need not consider the lower end of the range of x as the function is decreasing without limit in that direction) ... and whichever were the greater would be the answer to the question.
A fussy little point though: but it's important in many kinds of problem: the range would have to be a closed range - ie of the form $$aleq xleq b ,$$ often denoted $$xin[a,b], $$ rather than what is called an open range $$a< x< b ,$$ often denoted $$xin(a,b) ;$$ as if it were the latter that were specified, you would technically still have the problem of the non-existence of a maximum value of the function if the upper limit of the range were such that it could exceed -7 for x still less than it.
answered Nov 22 '18 at 5:39
AmbretteOrrisey
57410
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%2f3008677%2fstate-if-the-statement-is-true-or-false-the-maximum-value-of-2x3-9x2-24x-20%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
Hint: Let $x=10^{20}$. That should do it.
answered Nov 22 '18 at 2:45
Shaun
8,810113680
1
x=10 will also give positive value!
– user8795
Nov 22 '18 at 2:49
Well, there's your answer, @user8795.
– Shaun
Nov 22 '18 at 2:51
I think @anomaly's comment should be an answer and my answer should have been a comment. If you're looking for a quick, simple answer, though, mine will do the job if you recall the definition of what a maximum is.
– Shaun
Nov 22 '18 at 2:55
add a comment |
Hint: Let $x=10^{20}$. That should do it.
answered Nov 22 '18 at 2:45
Shaun
8,810113680
1
x=10 will also give positive value!
– user8795
Nov 22 '18 at 2:49
Well, there's your answer, @user8795.
– Shaun
Nov 22 '18 at 2:51
I think @anomaly's comment should be an answer and my answer should have been a comment. If you're looking for a quick, simple answer, though, mine will do the job if you recall the definition of what a maximum is.
– Shaun
Nov 22 '18 at 2:55
add a comment |
Hint: Let $x=10^{20}$. That should do it.
answered Nov 22 '18 at 2:45
Shaun
8,810113680
Hint: Let $x=10^{20}$. That should do it.
answered Nov 22 '18 at 2:45
Shaun
8,810113680
answered Nov 22 '18 at 2:45
Shaun
8,810113680
answered Nov 22 '18 at 2:45
Shaun
8,810113680
answered Nov 22 '18 at 2:45
Shaun
8,810113680
8,810113680
1
x=10 will also give positive value!
– user8795
Nov 22 '18 at 2:49
Well, there's your answer, @user8795.
– Shaun
Nov 22 '18 at 2:51
I think @anomaly's comment should be an answer and my answer should have been a comment. If you're looking for a quick, simple answer, though, mine will do the job if you recall the definition of what a maximum is.
– Shaun
Nov 22 '18 at 2:55
add a comment |
1
x=10 will also give positive value!
– user8795
Nov 22 '18 at 2:49
Well, there's your answer, @user8795.
– Shaun
Nov 22 '18 at 2:51
I think @anomaly's comment should be an answer and my answer should have been a comment. If you're looking for a quick, simple answer, though, mine will do the job if you recall the definition of what a maximum is.
– Shaun
Nov 22 '18 at 2:55
1
1
x=10 will also give positive value!
– user8795
Nov 22 '18 at 2:49
x=10 will also give positive value!
– user8795
Nov 22 '18 at 2:49
Well, there's your answer, @user8795.
– Shaun
Nov 22 '18 at 2:51
Well, there's your answer, @user8795.
– Shaun
Nov 22 '18 at 2:51
I think @anomaly's comment should be an answer and my answer should have been a comment. If you're looking for a quick, simple answer, though, mine will do the job if you recall the definition of what a maximum is.
– Shaun
Nov 22 '18 at 2:55
I think @anomaly's comment should be an answer and my answer should have been a comment. If you're looking for a quick, simple answer, though, mine will do the job if you recall the definition of what a maximum is.
– Shaun
Nov 22 '18 at 2:55
add a comment |
$f^{primeprime}(-1)=12(-1)-18=-30<0$. Thus $f$ has a local maximum at $-1$ and the local maximum value is $f(-1)=-7$. But it does not mean $f$ can not assume any value greater than $-7$. In fact the function $f$ is clearly unbounded. Here $f^{prime}(x)>0$ for all $xin (-infty,-1)$ and for all $xin (4,infty)$ and $f^{prime}(x)<0$ for all $xin (-1,4)$. Thus $f$ is strictly increasing in $(-infty,-1)cup (4,infty)$ and is strictly decreasing in $(-1,4)$. At $-1$ it has a local maximum and at $4$ it has a local minimum.
answered Nov 22 '18 at 3:14
Anupam
2,3741823
so the conclusion is false?
– user8795
Nov 22 '18 at 3:15
2
Unless the word "local" is mentioned, it is certainly false.
– Anupam
Nov 22 '18 at 3:17
add a comment |
$f^{primeprime}(-1)=12(-1)-18=-30<0$. Thus $f$ has a local maximum at $-1$ and the local maximum value is $f(-1)=-7$. But it does not mean $f$ can not assume any value greater than $-7$. In fact the function $f$ is clearly unbounded. Here $f^{prime}(x)>0$ for all $xin (-infty,-1)$ and for all $xin (4,infty)$ and $f^{prime}(x)<0$ for all $xin (-1,4)$. Thus $f$ is strictly increasing in $(-infty,-1)cup (4,infty)$ and is strictly decreasing in $(-1,4)$. At $-1$ it has a local maximum and at $4$ it has a local minimum.
answered Nov 22 '18 at 3:14
Anupam
2,3741823
so the conclusion is false?
– user8795
Nov 22 '18 at 3:15
2
Unless the word "local" is mentioned, it is certainly false.
– Anupam
Nov 22 '18 at 3:17
add a comment |
$f^{primeprime}(-1)=12(-1)-18=-30<0$. Thus $f$ has a local maximum at $-1$ and the local maximum value is $f(-1)=-7$. But it does not mean $f$ can not assume any value greater than $-7$. In fact the function $f$ is clearly unbounded. Here $f^{prime}(x)>0$ for all $xin (-infty,-1)$ and for all $xin (4,infty)$ and $f^{prime}(x)<0$ for all $xin (-1,4)$. Thus $f$ is strictly increasing in $(-infty,-1)cup (4,infty)$ and is strictly decreasing in $(-1,4)$. At $-1$ it has a local maximum and at $4$ it has a local minimum.
answered Nov 22 '18 at 3:14
Anupam
2,3741823
$f^{primeprime}(-1)=12(-1)-18=-30<0$. Thus $f$ has a local maximum at $-1$ and the local maximum value is $f(-1)=-7$. But it does not mean $f$ can not assume any value greater than $-7$. In fact the function $f$ is clearly unbounded. Here $f^{prime}(x)>0$ for all $xin (-infty,-1)$ and for all $xin (4,infty)$ and $f^{prime}(x)<0$ for all $xin (-1,4)$. Thus $f$ is strictly increasing in $(-infty,-1)cup (4,infty)$ and is strictly decreasing in $(-1,4)$. At $-1$ it has a local maximum and at $4$ it has a local minimum.
answered Nov 22 '18 at 3:14
Anupam
2,3741823
answered Nov 22 '18 at 3:14
Anupam
2,3741823
answered Nov 22 '18 at 3:14
Anupam
2,3741823
answered Nov 22 '18 at 3:14
Anupam
2,3741823
2,3741823
so the conclusion is false?
– user8795
Nov 22 '18 at 3:15
2
Unless the word "local" is mentioned, it is certainly false.
– Anupam
Nov 22 '18 at 3:17
add a comment |
so the conclusion is false?
– user8795
Nov 22 '18 at 3:15
2
Unless the word "local" is mentioned, it is certainly false.
– Anupam
Nov 22 '18 at 3:17
so the conclusion is false?
– user8795
Nov 22 '18 at 3:15
so the conclusion is false?
– user8795
Nov 22 '18 at 3:15
2
2
Unless the word "local" is mentioned, it is certainly false.
– Anupam
Nov 22 '18 at 3:17
Unless the word "local" is mentioned, it is certainly false.
– Anupam
Nov 22 '18 at 3:17
add a comment |
Assume the opposite:
$$2x^3-9x^2-24x-20>-7$$
$$to 2x^3-9x^2-24x-13>0$$
$$to (2x-13)(x+1)^2>0$$
Pretty blatant from this.
answered Nov 22 '18 at 5:59
Rhys Hughes
5,0091427
Your use of implication (" $to$ ") is not enough. One must be able to go from your last inequality to your first. Indeed, this is possible, but you haven't made that clear.
– Shaun
Nov 23 '18 at 3:47
add a comment |
Assume the opposite:
$$2x^3-9x^2-24x-20>-7$$
$$to 2x^3-9x^2-24x-13>0$$
$$to (2x-13)(x+1)^2>0$$
Pretty blatant from this.
answered Nov 22 '18 at 5:59
Rhys Hughes
5,0091427
Your use of implication (" $to$ ") is not enough. One must be able to go from your last inequality to your first. Indeed, this is possible, but you haven't made that clear.
– Shaun
Nov 23 '18 at 3:47
add a comment |
Assume the opposite:
$$2x^3-9x^2-24x-20>-7$$
$$to 2x^3-9x^2-24x-13>0$$
$$to (2x-13)(x+1)^2>0$$
Pretty blatant from this.
answered Nov 22 '18 at 5:59
Rhys Hughes
5,0091427
Assume the opposite:
$$2x^3-9x^2-24x-20>-7$$
$$to 2x^3-9x^2-24x-13>0$$
$$to (2x-13)(x+1)^2>0$$
Pretty blatant from this.
answered Nov 22 '18 at 5:59
Rhys Hughes
5,0091427
answered Nov 22 '18 at 5:59
Rhys Hughes
5,0091427
answered Nov 22 '18 at 5:59
Rhys Hughes
5,0091427
answered Nov 22 '18 at 5:59
Rhys Hughes
5,0091427
5,0091427
Your use of implication (" $to$ ") is not enough. One must be able to go from your last inequality to your first. Indeed, this is possible, but you haven't made that clear.
– Shaun
Nov 23 '18 at 3:47
add a comment |
Your use of implication (" $to$ ") is not enough. One must be able to go from your last inequality to your first. Indeed, this is possible, but you haven't made that clear.
– Shaun
Nov 23 '18 at 3:47
Your use of implication (" $to$ ") is not enough. One must be able to go from your last inequality to your first. Indeed, this is possible, but you haven't made that clear.
– Shaun
Nov 23 '18 at 3:47
Your use of implication (" $to$ ") is not enough. One must be able to go from your last inequality to your first. Indeed, this is possible, but you haven't made that clear.
– Shaun
Nov 23 '18 at 3:47
add a comment |
For large enough $x$, the polynomial grows unboundedly, therefore no finite maximum exists for the function. The points then you found, are just local maximum or minimum not global
answered Nov 22 '18 at 6:31
Mostafa Ayaz
14k3936
add a comment |
For large enough $x$, the polynomial grows unboundedly, therefore no finite maximum exists for the function. The points then you found, are just local maximum or minimum not global
answered Nov 22 '18 at 6:31
Mostafa Ayaz
14k3936
add a comment |
For large enough $x$, the polynomial grows unboundedly, therefore no finite maximum exists for the function. The points then you found, are just local maximum or minimum not global
answered Nov 22 '18 at 6:31
Mostafa Ayaz
14k3936
For large enough $x$, the polynomial grows unboundedly, therefore no finite maximum exists for the function. The points then you found, are just local maximum or minimum not global
answered Nov 22 '18 at 6:31
Mostafa Ayaz
14k3936
answered Nov 22 '18 at 6:31
Mostafa Ayaz
14k3936
answered Nov 22 '18 at 6:31
Mostafa Ayaz
14k3936
answered Nov 22 '18 at 6:31
Mostafa Ayaz
14k3936
14k3936
add a comment |
add a comment |
The differentiation & setting of the derivative to zero yields stationary points. A range-of-x ought to have been specified for this question, as without it it is ill-posed.
To say that the function has a maximum value of ∞ at x=∞ is a scrambling of the meaning of what's going-on, which is that the function increases without limit as x increases without limit, & therefore does not have a maximum value. If a range of x had been given, you would compare the -7 gotten as a local maximum to the value taken by the function at the upper end of the range of x (you need not consider the lower end of the range of x as the function is decreasing without limit in that direction) ... and whichever were the greater would be the answer to the question.
A fussy little point though: but it's important in many kinds of problem: the range would have to be a closed range - ie of the form $$aleq xleq b ,$$ often denoted $$xin[a,b], $$ rather than what is called an open range $$a< x< b ,$$ often denoted $$xin(a,b) ;$$ as if it were the latter that were specified, you would technically still have the problem of the non-existence of a maximum value of the function if the upper limit of the range were such that it could exceed -7 for x still less than it.
answered Nov 22 '18 at 5:39
AmbretteOrrisey
57410
add a comment |
The differentiation & setting of the derivative to zero yields stationary points. A range-of-x ought to have been specified for this question, as without it it is ill-posed.
To say that the function has a maximum value of ∞ at x=∞ is a scrambling of the meaning of what's going-on, which is that the function increases without limit as x increases without limit, & therefore does not have a maximum value. If a range of x had been given, you would compare the -7 gotten as a local maximum to the value taken by the function at the upper end of the range of x (you need not consider the lower end of the range of x as the function is decreasing without limit in that direction) ... and whichever were the greater would be the answer to the question.
A fussy little point though: but it's important in many kinds of problem: the range would have to be a closed range - ie of the form $$aleq xleq b ,$$ often denoted $$xin[a,b], $$ rather than what is called an open range $$a< x< b ,$$ often denoted $$xin(a,b) ;$$ as if it were the latter that were specified, you would technically still have the problem of the non-existence of a maximum value of the function if the upper limit of the range were such that it could exceed -7 for x still less than it.
answered Nov 22 '18 at 5:39
AmbretteOrrisey
57410
add a comment |
The differentiation & setting of the derivative to zero yields stationary points. A range-of-x ought to have been specified for this question, as without it it is ill-posed.
To say that the function has a maximum value of ∞ at x=∞ is a scrambling of the meaning of what's going-on, which is that the function increases without limit as x increases without limit, & therefore does not have a maximum value. If a range of x had been given, you would compare the -7 gotten as a local maximum to the value taken by the function at the upper end of the range of x (you need not consider the lower end of the range of x as the function is decreasing without limit in that direction) ... and whichever were the greater would be the answer to the question.
A fussy little point though: but it's important in many kinds of problem: the range would have to be a closed range - ie of the form $$aleq xleq b ,$$ often denoted $$xin[a,b], $$ rather than what is called an open range $$a< x< b ,$$ often denoted $$xin(a,b) ;$$ as if it were the latter that were specified, you would technically still have the problem of the non-existence of a maximum value of the function if the upper limit of the range were such that it could exceed -7 for x still less than it.
answered Nov 22 '18 at 5:39
AmbretteOrrisey
57410
The differentiation & setting of the derivative to zero yields stationary points. A range-of-x ought to have been specified for this question, as without it it is ill-posed.
To say that the function has a maximum value of ∞ at x=∞ is a scrambling of the meaning of what's going-on, which is that the function increases without limit as x increases without limit, & therefore does not have a maximum value. If a range of x had been given, you would compare the -7 gotten as a local maximum to the value taken by the function at the upper end of the range of x (you need not consider the lower end of the range of x as the function is decreasing without limit in that direction) ... and whichever were the greater would be the answer to the question.
A fussy little point though: but it's important in many kinds of problem: the range would have to be a closed range - ie of the form $$aleq xleq b ,$$ often denoted $$xin[a,b], $$ rather than what is called an open range $$a< x< b ,$$ often denoted $$xin(a,b) ;$$ as if it were the latter that were specified, you would technically still have the problem of the non-existence of a maximum value of the function if the upper limit of the range were such that it could exceed -7 for x still less than it.
answered Nov 22 '18 at 5:39
AmbretteOrrisey
57410
edited Nov 22 '18 at 8:04
answered Nov 22 '18 at 5:39
AmbretteOrrisey
57410
answered Nov 22 '18 at 5:39
AmbretteOrrisey
57410
answered Nov 22 '18 at 5:39
AmbretteOrrisey
57410
57410
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%2f3008677%2fstate-if-the-statement-is-true-or-false-the-maximum-value-of-2x3-9x2-24x-20%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
Every local maximum $x$ of $f$ has $f'(x) = 0$, since $frac{f(x + h) - f(x)}{h}$ is negative for small $h > 0$ and positive for small $h < 0$; since $f$ is differentiable at $x$, the limit must equal $0$. That's necessary but not sufficient; as you note, $f$ is clearly unbounded.
– anomaly
Nov 22 '18 at 2:44
$f'(x) = 0 at local minimums also. Look at the dominating first term for the global max and mins.
– Phil H
Nov 22 '18 at 3:03
4
When they say "maximum" or "minimum" without qualification, they are generally referring to global phenomena, not local phenomena. So the statement is false.
– Deepak
Nov 22 '18 at 3:19