Maximum Value of $g(x)=(8+x)^3(8-x)^4$
I think there can not be a maximum value of this, as if I plug $x=$1000, it will increase the value of the function in leaps and bounds. The answer says that the maximum value will occur at $x=-8/7$. What am I missing here?
functions
edited Nov 28 '18 at 5:39
Moo
5,53131020
asked Nov 28 '18 at 5:38
Sherlock WatsonSherlock Watson
3422413
|
show 2 more comments
I think there can not be a maximum value of this, as if I plug $x=$1000, it will increase the value of the function in leaps and bounds. The answer says that the maximum value will occur at $x=-8/7$. What am I missing here?
functions
edited Nov 28 '18 at 5:39
Moo
5,53131020
asked Nov 28 '18 at 5:38
Sherlock WatsonSherlock Watson
3422413
Yes Moo. Thank you very much.
– Sherlock Watson
Nov 28 '18 at 5:40
There is a local maximum and no global maximum - do a plot to see that.
– Moo
Nov 28 '18 at 5:41
I plotted it, the max value is going towards the infinite. X has to be a real value as well. How do we know whether question asks about local or global maximum?
– Sherlock Watson
Nov 28 '18 at 5:44
That has to be given, generally, to avoid such ambiguities. I was actually writing a comment that hinted at how the question is poorly phrased for this reason.
– Eevee Trainer
Nov 28 '18 at 5:44
1
Are you familiar with differentiation / the notion of maxima,minima of a differentiable function?
– астон вілла олоф мэллбэрг
Nov 28 '18 at 5:49
|
show 2 more comments
I think there can not be a maximum value of this, as if I plug $x=$1000, it will increase the value of the function in leaps and bounds. The answer says that the maximum value will occur at $x=-8/7$. What am I missing here?
functions
edited Nov 28 '18 at 5:39
Moo
5,53131020
asked Nov 28 '18 at 5:38
Sherlock WatsonSherlock Watson
3422413
I think there can not be a maximum value of this, as if I plug $x=$1000, it will increase the value of the function in leaps and bounds. The answer says that the maximum value will occur at $x=-8/7$. What am I missing here?
functions
functions
edited Nov 28 '18 at 5:39
Moo
5,53131020
asked Nov 28 '18 at 5:38
Sherlock WatsonSherlock Watson
3422413
edited Nov 28 '18 at 5:39
Moo
5,53131020
asked Nov 28 '18 at 5:38
Sherlock WatsonSherlock Watson
3422413
edited Nov 28 '18 at 5:39
Moo
5,53131020
edited Nov 28 '18 at 5:39
Moo
5,53131020
edited Nov 28 '18 at 5:39
Moo
5,53131020
5,53131020
asked Nov 28 '18 at 5:38
Sherlock WatsonSherlock Watson
3422413
asked Nov 28 '18 at 5:38
Sherlock WatsonSherlock Watson
3422413
asked Nov 28 '18 at 5:38
Sherlock WatsonSherlock Watson
3422413
3422413
Yes Moo. Thank you very much.
– Sherlock Watson
Nov 28 '18 at 5:40
There is a local maximum and no global maximum - do a plot to see that.
– Moo
Nov 28 '18 at 5:41
I plotted it, the max value is going towards the infinite. X has to be a real value as well. How do we know whether question asks about local or global maximum?
– Sherlock Watson
Nov 28 '18 at 5:44
That has to be given, generally, to avoid such ambiguities. I was actually writing a comment that hinted at how the question is poorly phrased for this reason.
– Eevee Trainer
Nov 28 '18 at 5:44
1
Are you familiar with differentiation / the notion of maxima,minima of a differentiable function?
– астон вілла олоф мэллбэрг
Nov 28 '18 at 5:49
|
show 2 more comments
Yes Moo. Thank you very much.
– Sherlock Watson
Nov 28 '18 at 5:40
There is a local maximum and no global maximum - do a plot to see that.
– Moo
Nov 28 '18 at 5:41
I plotted it, the max value is going towards the infinite. X has to be a real value as well. How do we know whether question asks about local or global maximum?
– Sherlock Watson
Nov 28 '18 at 5:44
That has to be given, generally, to avoid such ambiguities. I was actually writing a comment that hinted at how the question is poorly phrased for this reason.
– Eevee Trainer
Nov 28 '18 at 5:44
1
Are you familiar with differentiation / the notion of maxima,minima of a differentiable function?
– астон вілла олоф мэллбэрг
Nov 28 '18 at 5:49
Yes Moo. Thank you very much.
– Sherlock Watson
Nov 28 '18 at 5:40
Yes Moo. Thank you very much.
– Sherlock Watson
Nov 28 '18 at 5:40
There is a local maximum and no global maximum - do a plot to see that.
– Moo
Nov 28 '18 at 5:41
There is a local maximum and no global maximum - do a plot to see that.
– Moo
Nov 28 '18 at 5:41
I plotted it, the max value is going towards the infinite. X has to be a real value as well. How do we know whether question asks about local or global maximum?
– Sherlock Watson
Nov 28 '18 at 5:44
I plotted it, the max value is going towards the infinite. X has to be a real value as well. How do we know whether question asks about local or global maximum?
– Sherlock Watson
Nov 28 '18 at 5:44
That has to be given, generally, to avoid such ambiguities. I was actually writing a comment that hinted at how the question is poorly phrased for this reason.
– Eevee Trainer
Nov 28 '18 at 5:44
That has to be given, generally, to avoid such ambiguities. I was actually writing a comment that hinted at how the question is poorly phrased for this reason.
– Eevee Trainer
Nov 28 '18 at 5:44
1
1
Are you familiar with differentiation / the notion of maxima,minima of a differentiable function?
– астон вілла олоф мэллбэрг
Nov 28 '18 at 5:49
Are you familiar with differentiation / the notion of maxima,minima of a differentiable function?
– астон вілла олоф мэллбэрг
Nov 28 '18 at 5:49
|
show 2 more comments
3 Answers
3
active
oldest
votes
$p(x)= (8+x)^3(8-x)^4$ is a polynomial of degree $7$,
with leading term $x^7$.
1)No abs. maximum ,no abs. minimum . $(x rightarrow pm infty )$
2) $3$ zeroes at $x=-8$, $4$ zeroes at $x=8$.
3)$ -8 <x < 8$ : $p(x) >0$.
(Has a relative maximum).
answered Nov 28 '18 at 7:28
Peter SzilasPeter Szilas
10.9k2720
add a comment |
We have that
$$g(x)=(8+x)^3(8-x)^4implies g'(x)=3(8+x)^2(8-x)^4-4(8+x)^3(8-x)^3=0$$
$$3(8+x)^2(8-x)^4=4(8+x)^3(8-x)^3$$
which is true when
- $x=pm 8$
- $3(8-x)=4(8+x) implies 7x=-8 implies x=-frac87$
and the latter is a local maximum.
answered Nov 28 '18 at 11:23
gimusigimusi
1
add a comment |
You can use the wavy curve approach to find the range in which the local extremas and global extremas.
The wavy curve method is specifically used for polynomial equations.
Using this method you can easily come to know that there is no global maxima and global minima. However in the range (-8,8) there exists a local maxima. That point is x=-8/7
answered Nov 28 '18 at 11:43
Bhargav KaleBhargav Kale
313
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$p(x)= (8+x)^3(8-x)^4$ is a polynomial of degree $7$,
with leading term $x^7$.
1)No abs. maximum ,no abs. minimum . $(x rightarrow pm infty )$
2) $3$ zeroes at $x=-8$, $4$ zeroes at $x=8$.
3)$ -8 <x < 8$ : $p(x) >0$.
(Has a relative maximum).
answered Nov 28 '18 at 7:28
Peter SzilasPeter Szilas
10.9k2720
add a comment |
$p(x)= (8+x)^3(8-x)^4$ is a polynomial of degree $7$,
with leading term $x^7$.
1)No abs. maximum ,no abs. minimum . $(x rightarrow pm infty )$
2) $3$ zeroes at $x=-8$, $4$ zeroes at $x=8$.
3)$ -8 <x < 8$ : $p(x) >0$.
(Has a relative maximum).
answered Nov 28 '18 at 7:28
Peter SzilasPeter Szilas
10.9k2720
add a comment |
$p(x)= (8+x)^3(8-x)^4$ is a polynomial of degree $7$,
with leading term $x^7$.
1)No abs. maximum ,no abs. minimum . $(x rightarrow pm infty )$
2) $3$ zeroes at $x=-8$, $4$ zeroes at $x=8$.
3)$ -8 <x < 8$ : $p(x) >0$.
(Has a relative maximum).
answered Nov 28 '18 at 7:28
Peter SzilasPeter Szilas
10.9k2720
$p(x)= (8+x)^3(8-x)^4$ is a polynomial of degree $7$,
with leading term $x^7$.
1)No abs. maximum ,no abs. minimum . $(x rightarrow pm infty )$
2) $3$ zeroes at $x=-8$, $4$ zeroes at $x=8$.
3)$ -8 <x < 8$ : $p(x) >0$.
(Has a relative maximum).
answered Nov 28 '18 at 7:28
Peter SzilasPeter Szilas
10.9k2720
edited Nov 28 '18 at 7:44
answered Nov 28 '18 at 7:28
Peter SzilasPeter Szilas
10.9k2720
answered Nov 28 '18 at 7:28
Peter SzilasPeter Szilas
10.9k2720
answered Nov 28 '18 at 7:28
Peter SzilasPeter Szilas
10.9k2720
10.9k2720
add a comment |
add a comment |
We have that
$$g(x)=(8+x)^3(8-x)^4implies g'(x)=3(8+x)^2(8-x)^4-4(8+x)^3(8-x)^3=0$$
$$3(8+x)^2(8-x)^4=4(8+x)^3(8-x)^3$$
which is true when
- $x=pm 8$
- $3(8-x)=4(8+x) implies 7x=-8 implies x=-frac87$
and the latter is a local maximum.
answered Nov 28 '18 at 11:23
gimusigimusi
1
add a comment |
We have that
$$g(x)=(8+x)^3(8-x)^4implies g'(x)=3(8+x)^2(8-x)^4-4(8+x)^3(8-x)^3=0$$
$$3(8+x)^2(8-x)^4=4(8+x)^3(8-x)^3$$
which is true when
- $x=pm 8$
- $3(8-x)=4(8+x) implies 7x=-8 implies x=-frac87$
and the latter is a local maximum.
answered Nov 28 '18 at 11:23
gimusigimusi
1
add a comment |
We have that
$$g(x)=(8+x)^3(8-x)^4implies g'(x)=3(8+x)^2(8-x)^4-4(8+x)^3(8-x)^3=0$$
$$3(8+x)^2(8-x)^4=4(8+x)^3(8-x)^3$$
which is true when
- $x=pm 8$
- $3(8-x)=4(8+x) implies 7x=-8 implies x=-frac87$
and the latter is a local maximum.
answered Nov 28 '18 at 11:23
gimusigimusi
1
We have that
$$g(x)=(8+x)^3(8-x)^4implies g'(x)=3(8+x)^2(8-x)^4-4(8+x)^3(8-x)^3=0$$
$$3(8+x)^2(8-x)^4=4(8+x)^3(8-x)^3$$
which is true when
- $x=pm 8$
- $3(8-x)=4(8+x) implies 7x=-8 implies x=-frac87$
and the latter is a local maximum.
answered Nov 28 '18 at 11:23
gimusigimusi
1
answered Nov 28 '18 at 11:23
gimusigimusi
1
answered Nov 28 '18 at 11:23
gimusigimusi
1
answered Nov 28 '18 at 11:23
gimusigimusi
1
1
add a comment |
add a comment |
You can use the wavy curve approach to find the range in which the local extremas and global extremas.
The wavy curve method is specifically used for polynomial equations.
Using this method you can easily come to know that there is no global maxima and global minima. However in the range (-8,8) there exists a local maxima. That point is x=-8/7
answered Nov 28 '18 at 11:43
Bhargav KaleBhargav Kale
313
add a comment |
You can use the wavy curve approach to find the range in which the local extremas and global extremas.
The wavy curve method is specifically used for polynomial equations.
Using this method you can easily come to know that there is no global maxima and global minima. However in the range (-8,8) there exists a local maxima. That point is x=-8/7
answered Nov 28 '18 at 11:43
Bhargav KaleBhargav Kale
313
add a comment |
You can use the wavy curve approach to find the range in which the local extremas and global extremas.
The wavy curve method is specifically used for polynomial equations.
Using this method you can easily come to know that there is no global maxima and global minima. However in the range (-8,8) there exists a local maxima. That point is x=-8/7
answered Nov 28 '18 at 11:43
Bhargav KaleBhargav Kale
313
You can use the wavy curve approach to find the range in which the local extremas and global extremas.
The wavy curve method is specifically used for polynomial equations.
Using this method you can easily come to know that there is no global maxima and global minima. However in the range (-8,8) there exists a local maxima. That point is x=-8/7
answered Nov 28 '18 at 11:43
Bhargav KaleBhargav Kale
313
answered Nov 28 '18 at 11:43
Bhargav KaleBhargav Kale
313
answered Nov 28 '18 at 11:43
Bhargav KaleBhargav Kale
313
answered Nov 28 '18 at 11:43
Bhargav KaleBhargav Kale
313
313
add a comment |
add a comment |
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Yes Moo. Thank you very much.
– Sherlock Watson
Nov 28 '18 at 5:40
There is a local maximum and no global maximum - do a plot to see that.
– Moo
Nov 28 '18 at 5:41
I plotted it, the max value is going towards the infinite. X has to be a real value as well. How do we know whether question asks about local or global maximum?
– Sherlock Watson
Nov 28 '18 at 5:44
That has to be given, generally, to avoid such ambiguities. I was actually writing a comment that hinted at how the question is poorly phrased for this reason.
– Eevee Trainer
Nov 28 '18 at 5:44
1
Are you familiar with differentiation / the notion of maxima,minima of a differentiable function?
– астон вілла олоф мэллбэрг
Nov 28 '18 at 5:49