Mixing
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
user8795user8795
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
user8795user8795
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
user8795user8795
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
user8795user8795
5,61961947
asked Nov 22 '18 at 2:41
user8795user8795
5,61961947
asked Nov 22 '18 at 2:41
user8795user8795
5,61961947
asked Nov 22 '18 at 2:41
user8795user8795
5,61961947
asked Nov 22 '18 at 2:41
user8795user8795
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
ShaunShaun
8,820113681
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
AnupamAnupam
2,3741824
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 HughesRhys Hughes
5,0541427
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 AyazMostafa Ayaz
14.2k3937
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
AmbretteOrriseyAmbretteOrrisey
57410
add a comment |
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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
ShaunShaun
8,820113681
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
ShaunShaun
8,820113681
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
ShaunShaun
8,820113681
Hint: Let $x=10^{20}$. That should do it.
answered Nov 22 '18 at 2:45
ShaunShaun
8,820113681
answered Nov 22 '18 at 2:45
ShaunShaun
8,820113681
answered Nov 22 '18 at 2:45
ShaunShaun
8,820113681
answered Nov 22 '18 at 2:45
ShaunShaun
8,820113681
8,820113681
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
AnupamAnupam
2,3741824
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
AnupamAnupam
2,3741824
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
AnupamAnupam
2,3741824
$f^{primeprime}(-1)=12(-1)-18=-30<0$. 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
AnupamAnupam
2,3741824
answered Nov 22 '18 at 3:14
AnupamAnupam
2,3741824
answered Nov 22 '18 at 3:14
AnupamAnupam
2,3741824
answered Nov 22 '18 at 3:14
AnupamAnupam
2,3741824
2,3741824
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 HughesRhys Hughes
5,0541427
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 HughesRhys Hughes
5,0541427
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 HughesRhys Hughes
5,0541427
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 HughesRhys Hughes
5,0541427
answered Nov 22 '18 at 5:59
Rhys HughesRhys Hughes
5,0541427
answered Nov 22 '18 at 5:59
Rhys HughesRhys Hughes
5,0541427
answered Nov 22 '18 at 5:59
Rhys HughesRhys Hughes
5,0541427
5,0541427
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 AyazMostafa Ayaz
14.2k3937
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 AyazMostafa Ayaz
14.2k3937
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 AyazMostafa Ayaz
14.2k3937
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 AyazMostafa Ayaz
14.2k3937
answered Nov 22 '18 at 6:31
Mostafa AyazMostafa Ayaz
14.2k3937
answered Nov 22 '18 at 6:31
Mostafa AyazMostafa Ayaz
14.2k3937
answered Nov 22 '18 at 6:31
Mostafa AyazMostafa Ayaz
14.2k3937
14.2k3937
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
AmbretteOrriseyAmbretteOrrisey
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
AmbretteOrriseyAmbretteOrrisey
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
AmbretteOrriseyAmbretteOrrisey
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
AmbretteOrriseyAmbretteOrrisey
57410
edited Nov 22 '18 at 8:04
answered Nov 22 '18 at 5:39
AmbretteOrriseyAmbretteOrrisey
57410
answered Nov 22 '18 at 5:39
AmbretteOrriseyAmbretteOrrisey
57410
answered Nov 22 '18 at 5:39
AmbretteOrriseyAmbretteOrrisey
57410
57410
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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