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Question involving concept of $max.min.$
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In a book I’m been reading and solving from, I encountered a problem that is troubling me for too long as I can’t understand the concept behind the $max, min$ calls in the question and have been wondering if it’s a typographical error but either way I’d want to know for sure.
Let $f(x)=f_1(x) - 2f_2(x)$
where $f_1(x)= begincasesmin, & text$ \
max,& text$
endcases$
Therefore, $ f_1(x) = x^2$, $xin R$
and $ f_2(x) = begincasesmin, & text$ \
max,& text$
endcases$
Therefore, $f_2(x) = |x|, xin R$
Till here I understood the max, min calls.
Now, the next function is what I’m not able to understand-
and let $g(x) = begincasesminf(t):-3 le t le x,& text$-3 le x<0$ \
maxf(t): 0 le t le x,& text$0le x le 3$
endcases$
So what exactly does $g(x)$ mean? Does the above function $g(x)$ actually signify something or is there something missing in the space?
I checked the solution as I couldn’t go further in the question without solving this doubt. In the solution provided, it says
$g(x) =begincases
f(x), & text$-3 le x < -1$ \
-1, & text$-1 le x < 0$\
0, & text$0le x le 2$\
f(x), & text$2 < x le 3$\
endcases$
$g(x) =begincases
x^2 + 2x, & text$-3 le x < -1$ \
-1, & text$-1 le x < 0$\ tag1
0, & text$0le x le 2$\
x^2 - 2x, & text$2 < x le 3$\
endcases$
How to handle the $t$ along with the $x$ and prove $g(x) = (1)$? Or is there something missing in the definition of $g(x)$ that makes $g(x) = (1)$? Any help would be appreciated.
functions optimization
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In a book I’m been reading and solving from, I encountered a problem that is troubling me for too long as I can’t understand the concept behind the $max, min$ calls in the question and have been wondering if it’s a typographical error but either way I’d want to know for sure.
Let $f(x)=f_1(x) - 2f_2(x)$
where $f_1(x)= begincasesmin, & text$ \
max,& text$
endcases$
Therefore, $ f_1(x) = x^2$, $xin R$
and $ f_2(x) = begincasesmin, & text$ \
max,& text$
endcases$
Therefore, $f_2(x) = |x|, xin R$
Till here I understood the max, min calls.
Now, the next function is what I’m not able to understand-
and let $g(x) = begincasesminf(t):-3 le t le x,& text$-3 le x<0$ \
maxf(t): 0 le t le x,& text$0le x le 3$
endcases$
So what exactly does $g(x)$ mean? Does the above function $g(x)$ actually signify something or is there something missing in the space?
I checked the solution as I couldn’t go further in the question without solving this doubt. In the solution provided, it says
$g(x) =begincases
f(x), & text$-3 le x < -1$ \
-1, & text$-1 le x < 0$\
0, & text$0le x le 2$\
f(x), & text$2 < x le 3$\
endcases$
$g(x) =begincases
x^2 + 2x, & text$-3 le x < -1$ \
-1, & text$-1 le x < 0$\ tag1
0, & text$0le x le 2$\
x^2 - 2x, & text$2 < x le 3$\
endcases$
How to handle the $t$ along with the $x$ and prove $g(x) = (1)$? Or is there something missing in the definition of $g(x)$ that makes $g(x) = (1)$? Any help would be appreciated.
functions optimization
asked yesterday
Hola
304
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
In a book I’m been reading and solving from, I encountered a problem that is troubling me for too long as I can’t understand the concept behind the $max, min$ calls in the question and have been wondering if it’s a typographical error but either way I’d want to know for sure.
Let $f(x)=f_1(x) - 2f_2(x)$
where $f_1(x)= begincasesmin, & text$ \
max,& text$
endcases$
Therefore, $ f_1(x) = x^2$, $xin R$
and $ f_2(x) = begincasesmin, & text$ \
max,& text$
endcases$
Therefore, $f_2(x) = |x|, xin R$
Till here I understood the max, min calls.
Now, the next function is what I’m not able to understand-
and let $g(x) = begincasesminf(t):-3 le t le x,& text$-3 le x<0$ \
maxf(t): 0 le t le x,& text$0le x le 3$
endcases$
So what exactly does $g(x)$ mean? Does the above function $g(x)$ actually signify something or is there something missing in the space?
I checked the solution as I couldn’t go further in the question without solving this doubt. In the solution provided, it says
$g(x) =begincases
f(x), & text$-3 le x < -1$ \
-1, & text$-1 le x < 0$\
0, & text$0le x le 2$\
f(x), & text$2 < x le 3$\
endcases$
$g(x) =begincases
x^2 + 2x, & text$-3 le x < -1$ \
-1, & text$-1 le x < 0$\ tag1
0, & text$0le x le 2$\
x^2 - 2x, & text$2 < x le 3$\
endcases$
How to handle the $t$ along with the $x$ and prove $g(x) = (1)$? Or is there something missing in the definition of $g(x)$ that makes $g(x) = (1)$? Any help would be appreciated.
functions optimization
asked yesterday
Hola
304
In a book I’m been reading and solving from, I encountered a problem that is troubling me for too long as I can’t understand the concept behind the $max, min$ calls in the question and have been wondering if it’s a typographical error but either way I’d want to know for sure.
Let $f(x)=f_1(x) - 2f_2(x)$
where $f_1(x)= begincasesmin, & text$ \
max,& text$
endcases$
Therefore, $ f_1(x) = x^2$, $xin R$
and $ f_2(x) = begincasesmin, & text$ \
max,& text$
endcases$
Therefore, $f_2(x) = |x|, xin R$
Till here I understood the max, min calls.
Now, the next function is what I’m not able to understand-
and let $g(x) = begincasesminf(t):-3 le t le x,& text$-3 le x<0$ \
maxf(t): 0 le t le x,& text$0le x le 3$
endcases$
So what exactly does $g(x)$ mean? Does the above function $g(x)$ actually signify something or is there something missing in the space?
I checked the solution as I couldn’t go further in the question without solving this doubt. In the solution provided, it says
$g(x) =begincases
f(x), & text$-3 le x < -1$ \
-1, & text$-1 le x < 0$\
0, & text$0le x le 2$\
f(x), & text$2 < x le 3$\
endcases$
$g(x) =begincases
x^2 + 2x, & text$-3 le x < -1$ \
-1, & text$-1 le x < 0$\ tag1
0, & text$0le x le 2$\
x^2 - 2x, & text$2 < x le 3$\
endcases$
How to handle the $t$ along with the $x$ and prove $g(x) = (1)$? Or is there something missing in the definition of $g(x)$ that makes $g(x) = (1)$? Any help would be appreciated.
functions optimization
asked yesterday
Hola
304
asked yesterday
Hola
304
asked yesterday
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304
asked yesterday
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304
304
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add a comment |Â
2 Answers
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You have concluded that $f_1(x)=x^2$ and $f_2(x)=|x|$.
Given, $f(x)=f_1(x)-2f_2(x)$
When, $x>0$
$$f(x)=x^2-2x$$
When $x<0$
$$f(x)=x^2+2x$$
You can easily calculate that $f(x)$ achieves its local mininum $x=+1,-1$ as $f(x)=-1$ and local maximum at $x=0$ as $f(x)=0$.
You are given that, $g(x) = begincasesminf(t):-3 le t le x,& text$-3 le x<0$ \
maxf(t): 0 le t le x,& text$0le x le 3$
endcases$
CASE I
As long as $xlt-1$.
$f(x)$ i.e. the closing element of the set will correspond to the functions minimum value.
But as $xgeq-1$
There will always lie a $t=-1$ in $-3leqtleqx$ such that $f(t)$ can achieve its minimum value, i.e. $f(t)=-1$ at $t=-1$.
Therefore, $g(x) = begincases f(x);;;; -3leqxlt-1 \
-1 ;;;;;; -1leqxleq0
endcases$
CASE II
As $0leqxleq2$,
There will always lie a $t=0$ in $0leqtlt2$, such that $f(t)$ achieves its maximum as $f(t)=0$ at $t=0$.
As $2ltxleq3$
$f(t)$ is strictly increasing, therefore, its maximum value corresponds to the end point of the set, i.e. $x$ achieving maximum $f(x)$.
Therefore, $g(x) = begincases; 0;;;;;;; 0leqxleq2 \
-1 ;;;;;; 2lexleq3
endcases$
CONCLUSION
$g(x) =begincases
x^2 + 2x, & text$-3 le x < -1$ \
-1, & text$-1 le x < 0$\ tag1
0, & text$0le x le 2$\
x^2 - 2x, & text$2 < x le 3$\
endcases$
answered yesterday
prog_SAHIL
732217
add a comment |Â
up vote
3
down vote
$g(x) = begincasesminf(t):-3 le t le x,& text$-3 le x<0$ \
maxf(t): 0 le t le x,& text$0le x le 3$
endcases$ means, in Layman terms,
$g(x)$ is the minimum value the function $y = f(x)$ takes in the interval $[-3, x]$, when $x$ is in $[-3, 0]$;
and maximum value the function $y = f(x)$ takes in the interval $[0, x]$, when $x$ is in $[0, 3].$
answered yesterday
ab123
1,274318
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You have concluded that $f_1(x)=x^2$ and $f_2(x)=|x|$.
Given, $f(x)=f_1(x)-2f_2(x)$
When, $x>0$
$$f(x)=x^2-2x$$
When $x<0$
$$f(x)=x^2+2x$$
You can easily calculate that $f(x)$ achieves its local mininum $x=+1,-1$ as $f(x)=-1$ and local maximum at $x=0$ as $f(x)=0$.
You are given that, $g(x) = begincasesminf(t):-3 le t le x,& text$-3 le x<0$ \
maxf(t): 0 le t le x,& text$0le x le 3$
endcases$
CASE I
As long as $xlt-1$.
$f(x)$ i.e. the closing element of the set will correspond to the functions minimum value.
But as $xgeq-1$
There will always lie a $t=-1$ in $-3leqtleqx$ such that $f(t)$ can achieve its minimum value, i.e. $f(t)=-1$ at $t=-1$.
Therefore, $g(x) = begincases f(x);;;; -3leqxlt-1 \
-1 ;;;;;; -1leqxleq0
endcases$
CASE II
As $0leqxleq2$,
There will always lie a $t=0$ in $0leqtlt2$, such that $f(t)$ achieves its maximum as $f(t)=0$ at $t=0$.
As $2ltxleq3$
$f(t)$ is strictly increasing, therefore, its maximum value corresponds to the end point of the set, i.e. $x$ achieving maximum $f(x)$.
Therefore, $g(x) = begincases; 0;;;;;;; 0leqxleq2 \
-1 ;;;;;; 2lexleq3
endcases$
CONCLUSION
$g(x) =begincases
x^2 + 2x, & text$-3 le x < -1$ \
-1, & text$-1 le x < 0$\ tag1
0, & text$0le x le 2$\
x^2 - 2x, & text$2 < x le 3$\
endcases$
answered yesterday
prog_SAHIL
732217
add a comment |Â
up vote
1
down vote
accepted
You have concluded that $f_1(x)=x^2$ and $f_2(x)=|x|$.
Given, $f(x)=f_1(x)-2f_2(x)$
When, $x>0$
$$f(x)=x^2-2x$$
When $x<0$
$$f(x)=x^2+2x$$
You can easily calculate that $f(x)$ achieves its local mininum $x=+1,-1$ as $f(x)=-1$ and local maximum at $x=0$ as $f(x)=0$.
You are given that, $g(x) = begincasesminf(t):-3 le t le x,& text$-3 le x<0$ \
maxf(t): 0 le t le x,& text$0le x le 3$
endcases$
CASE I
As long as $xlt-1$.
$f(x)$ i.e. the closing element of the set will correspond to the functions minimum value.
But as $xgeq-1$
There will always lie a $t=-1$ in $-3leqtleqx$ such that $f(t)$ can achieve its minimum value, i.e. $f(t)=-1$ at $t=-1$.
Therefore, $g(x) = begincases f(x);;;; -3leqxlt-1 \
-1 ;;;;;; -1leqxleq0
endcases$
CASE II
As $0leqxleq2$,
There will always lie a $t=0$ in $0leqtlt2$, such that $f(t)$ achieves its maximum as $f(t)=0$ at $t=0$.
As $2ltxleq3$
$f(t)$ is strictly increasing, therefore, its maximum value corresponds to the end point of the set, i.e. $x$ achieving maximum $f(x)$.
Therefore, $g(x) = begincases; 0;;;;;;; 0leqxleq2 \
-1 ;;;;;; 2lexleq3
endcases$
CONCLUSION
$g(x) =begincases
x^2 + 2x, & text$-3 le x < -1$ \
-1, & text$-1 le x < 0$\ tag1
0, & text$0le x le 2$\
x^2 - 2x, & text$2 < x le 3$\
endcases$
answered yesterday
prog_SAHIL
732217
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You have concluded that $f_1(x)=x^2$ and $f_2(x)=|x|$.
Given, $f(x)=f_1(x)-2f_2(x)$
When, $x>0$
$$f(x)=x^2-2x$$
When $x<0$
$$f(x)=x^2+2x$$
You can easily calculate that $f(x)$ achieves its local mininum $x=+1,-1$ as $f(x)=-1$ and local maximum at $x=0$ as $f(x)=0$.
You are given that, $g(x) = begincasesminf(t):-3 le t le x,& text$-3 le x<0$ \
maxf(t): 0 le t le x,& text$0le x le 3$
endcases$
CASE I
As long as $xlt-1$.
$f(x)$ i.e. the closing element of the set will correspond to the functions minimum value.
But as $xgeq-1$
There will always lie a $t=-1$ in $-3leqtleqx$ such that $f(t)$ can achieve its minimum value, i.e. $f(t)=-1$ at $t=-1$.
Therefore, $g(x) = begincases f(x);;;; -3leqxlt-1 \
-1 ;;;;;; -1leqxleq0
endcases$
CASE II
As $0leqxleq2$,
There will always lie a $t=0$ in $0leqtlt2$, such that $f(t)$ achieves its maximum as $f(t)=0$ at $t=0$.
As $2ltxleq3$
$f(t)$ is strictly increasing, therefore, its maximum value corresponds to the end point of the set, i.e. $x$ achieving maximum $f(x)$.
Therefore, $g(x) = begincases; 0;;;;;;; 0leqxleq2 \
-1 ;;;;;; 2lexleq3
endcases$
CONCLUSION
$g(x) =begincases
x^2 + 2x, & text$-3 le x < -1$ \
-1, & text$-1 le x < 0$\ tag1
0, & text$0le x le 2$\
x^2 - 2x, & text$2 < x le 3$\
endcases$
answered yesterday
prog_SAHIL
732217
You have concluded that $f_1(x)=x^2$ and $f_2(x)=|x|$.
Given, $f(x)=f_1(x)-2f_2(x)$
When, $x>0$
$$f(x)=x^2-2x$$
When $x<0$
$$f(x)=x^2+2x$$
You can easily calculate that $f(x)$ achieves its local mininum $x=+1,-1$ as $f(x)=-1$ and local maximum at $x=0$ as $f(x)=0$.
You are given that, $g(x) = begincasesminf(t):-3 le t le x,& text$-3 le x<0$ \
maxf(t): 0 le t le x,& text$0le x le 3$
endcases$
CASE I
As long as $xlt-1$.
$f(x)$ i.e. the closing element of the set will correspond to the functions minimum value.
But as $xgeq-1$
There will always lie a $t=-1$ in $-3leqtleqx$ such that $f(t)$ can achieve its minimum value, i.e. $f(t)=-1$ at $t=-1$.
Therefore, $g(x) = begincases f(x);;;; -3leqxlt-1 \
-1 ;;;;;; -1leqxleq0
endcases$
CASE II
As $0leqxleq2$,
There will always lie a $t=0$ in $0leqtlt2$, such that $f(t)$ achieves its maximum as $f(t)=0$ at $t=0$.
As $2ltxleq3$
$f(t)$ is strictly increasing, therefore, its maximum value corresponds to the end point of the set, i.e. $x$ achieving maximum $f(x)$.
Therefore, $g(x) = begincases; 0;;;;;;; 0leqxleq2 \
-1 ;;;;;; 2lexleq3
endcases$
CONCLUSION
$g(x) =begincases
x^2 + 2x, & text$-3 le x < -1$ \
-1, & text$-1 le x < 0$\ tag1
0, & text$0le x le 2$\
x^2 - 2x, & text$2 < x le 3$\
endcases$
answered yesterday
prog_SAHIL
732217
edited 21 hours ago
answered yesterday
prog_SAHIL
732217
answered yesterday
prog_SAHIL
732217
answered yesterday
prog_SAHIL
732217
732217
add a comment |Â
add a comment |Â
up vote
3
down vote
$g(x) = begincasesminf(t):-3 le t le x,& text$-3 le x<0$ \
maxf(t): 0 le t le x,& text$0le x le 3$
endcases$ means, in Layman terms,
$g(x)$ is the minimum value the function $y = f(x)$ takes in the interval $[-3, x]$, when $x$ is in $[-3, 0]$;
and maximum value the function $y = f(x)$ takes in the interval $[0, x]$, when $x$ is in $[0, 3].$
answered yesterday
ab123
1,274318
add a comment |Â
up vote
3
down vote
$g(x) = begincasesminf(t):-3 le t le x,& text$-3 le x<0$ \
maxf(t): 0 le t le x,& text$0le x le 3$
endcases$ means, in Layman terms,
$g(x)$ is the minimum value the function $y = f(x)$ takes in the interval $[-3, x]$, when $x$ is in $[-3, 0]$;
and maximum value the function $y = f(x)$ takes in the interval $[0, x]$, when $x$ is in $[0, 3].$
answered yesterday
ab123
1,274318
add a comment |Â
up vote
3
down vote
up vote
3
down vote
$g(x) = begincasesminf(t):-3 le t le x,& text$-3 le x<0$ \
maxf(t): 0 le t le x,& text$0le x le 3$
endcases$ means, in Layman terms,
$g(x)$ is the minimum value the function $y = f(x)$ takes in the interval $[-3, x]$, when $x$ is in $[-3, 0]$;
and maximum value the function $y = f(x)$ takes in the interval $[0, x]$, when $x$ is in $[0, 3].$
answered yesterday
ab123
1,274318
$g(x) = begincasesminf(t):-3 le t le x,& text$-3 le x<0$ \
maxf(t): 0 le t le x,& text$0le x le 3$
endcases$ means, in Layman terms,
$g(x)$ is the minimum value the function $y = f(x)$ takes in the interval $[-3, x]$, when $x$ is in $[-3, 0]$;
and maximum value the function $y = f(x)$ takes in the interval $[0, x]$, when $x$ is in $[0, 3].$
answered yesterday
ab123
1,274318
answered yesterday
ab123
1,274318
answered yesterday
ab123
1,274318
answered yesterday
ab123
1,274318
1,274318
add a comment |Â
add a comment |Â
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