question on function and image
$begingroup$
Consider $f: R^2-> R$ given by
$f(v,w)=(v+w)/2 +|v-w|/2$
We also had to show that $g=f$ where $g: R^2-> R$ given by
g(v,w)= v if v>=w and w if v<0
Derive with proof the image of [0,1] times [-1,1] under f.
Quite stuck with this question i think i have figured out that the image is [0,1] but not sure. Then you have to prove this so something to do with showing f([0,1] times [-1,1]) is a subset of [0,1] and show the other way but no idea reall yhow to do this. Thanks for help in advance
im ok with the g=f part just need help with the image part.
functions
asked Dec 3 '18 at 23:38
HarryHarry
253
$endgroup$
add a comment |
$begingroup$
Consider $f: R^2-> R$ given by
$f(v,w)=(v+w)/2 +|v-w|/2$
We also had to show that $g=f$ where $g: R^2-> R$ given by
g(v,w)= v if v>=w and w if v<0
Derive with proof the image of [0,1] times [-1,1] under f.
Quite stuck with this question i think i have figured out that the image is [0,1] but not sure. Then you have to prove this so something to do with showing f([0,1] times [-1,1]) is a subset of [0,1] and show the other way but no idea reall yhow to do this. Thanks for help in advance
im ok with the g=f part just need help with the image part.
functions
asked Dec 3 '18 at 23:38
HarryHarry
253
$endgroup$
add a comment |
$begingroup$
Consider $f: R^2-> R$ given by
$f(v,w)=(v+w)/2 +|v-w|/2$
We also had to show that $g=f$ where $g: R^2-> R$ given by
g(v,w)= v if v>=w and w if v<0
Derive with proof the image of [0,1] times [-1,1] under f.
Quite stuck with this question i think i have figured out that the image is [0,1] but not sure. Then you have to prove this so something to do with showing f([0,1] times [-1,1]) is a subset of [0,1] and show the other way but no idea reall yhow to do this. Thanks for help in advance
im ok with the g=f part just need help with the image part.
functions
asked Dec 3 '18 at 23:38
HarryHarry
253
$endgroup$
Consider $f: R^2-> R$ given by
$f(v,w)=(v+w)/2 +|v-w|/2$
We also had to show that $g=f$ where $g: R^2-> R$ given by
g(v,w)= v if v>=w and w if v<0
Derive with proof the image of [0,1] times [-1,1] under f.
Quite stuck with this question i think i have figured out that the image is [0,1] but not sure. Then you have to prove this so something to do with showing f([0,1] times [-1,1]) is a subset of [0,1] and show the other way but no idea reall yhow to do this. Thanks for help in advance
im ok with the g=f part just need help with the image part.
functions
functions
asked Dec 3 '18 at 23:38
HarryHarry
253
asked Dec 3 '18 at 23:38
HarryHarry
253
edited Dec 3 '18 at 23:44
Harry
asked Dec 3 '18 at 23:38
HarryHarry
253
asked Dec 3 '18 at 23:38
HarryHarry
253
asked Dec 3 '18 at 23:38
HarryHarry
253
253
add a comment |
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
The definition of $f(v,w)$ suggests that it is just the maximum of $v$ and $w$
The image is $[0,1] $ because you can simply pick $v=0$ and run $w$ over $[0,1]$
answered Dec 3 '18 at 23:57
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
$endgroup$
add a comment |
$begingroup$
Consider the cases $v geq w$ and $v<w$. In the first case $|v-w|=v-w$ so $f(v,w)=v$ and in the second case $|v-w|=w-v$ so $f(v,w)=w$. This proves that $f=g$. Note that $|x|+x geq 0$ for any real number $x$. Hence $f$ takes only non-negative values. From the form of $g$ it follows that the image is contained in $[0,1]$. To show that the image is all of $[0,1]$ just note that $g(t,0)=t$ for all $t in [0,1]$.
answered Dec 3 '18 at 23:57
Kavi Rama MurthyKavi Rama Murthy
55.7k42158
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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$begingroup$
The definition of $f(v,w)$ suggests that it is just the maximum of $v$ and $w$
The image is $[0,1] $ because you can simply pick $v=0$ and run $w$ over $[0,1]$
answered Dec 3 '18 at 23:57
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
$endgroup$
add a comment |
$begingroup$
The definition of $f(v,w)$ suggests that it is just the maximum of $v$ and $w$
The image is $[0,1] $ because you can simply pick $v=0$ and run $w$ over $[0,1]$
answered Dec 3 '18 at 23:57
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
$endgroup$
add a comment |
$begingroup$
The definition of $f(v,w)$ suggests that it is just the maximum of $v$ and $w$
The image is $[0,1] $ because you can simply pick $v=0$ and run $w$ over $[0,1]$
answered Dec 3 '18 at 23:57
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
$endgroup$
The definition of $f(v,w)$ suggests that it is just the maximum of $v$ and $w$
The image is $[0,1] $ because you can simply pick $v=0$ and run $w$ over $[0,1]$
answered Dec 3 '18 at 23:57
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
answered Dec 3 '18 at 23:57
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
answered Dec 3 '18 at 23:57
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
answered Dec 3 '18 at 23:57
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
41.5k42061
add a comment |
add a comment |
$begingroup$
Consider the cases $v geq w$ and $v<w$. In the first case $|v-w|=v-w$ so $f(v,w)=v$ and in the second case $|v-w|=w-v$ so $f(v,w)=w$. This proves that $f=g$. Note that $|x|+x geq 0$ for any real number $x$. Hence $f$ takes only non-negative values. From the form of $g$ it follows that the image is contained in $[0,1]$. To show that the image is all of $[0,1]$ just note that $g(t,0)=t$ for all $t in [0,1]$.
answered Dec 3 '18 at 23:57
Kavi Rama MurthyKavi Rama Murthy
55.7k42158
$endgroup$
add a comment |
$begingroup$
Consider the cases $v geq w$ and $v<w$. In the first case $|v-w|=v-w$ so $f(v,w)=v$ and in the second case $|v-w|=w-v$ so $f(v,w)=w$. This proves that $f=g$. Note that $|x|+x geq 0$ for any real number $x$. Hence $f$ takes only non-negative values. From the form of $g$ it follows that the image is contained in $[0,1]$. To show that the image is all of $[0,1]$ just note that $g(t,0)=t$ for all $t in [0,1]$.
answered Dec 3 '18 at 23:57
Kavi Rama MurthyKavi Rama Murthy
55.7k42158
$endgroup$
add a comment |
$begingroup$
Consider the cases $v geq w$ and $v<w$. In the first case $|v-w|=v-w$ so $f(v,w)=v$ and in the second case $|v-w|=w-v$ so $f(v,w)=w$. This proves that $f=g$. Note that $|x|+x geq 0$ for any real number $x$. Hence $f$ takes only non-negative values. From the form of $g$ it follows that the image is contained in $[0,1]$. To show that the image is all of $[0,1]$ just note that $g(t,0)=t$ for all $t in [0,1]$.
answered Dec 3 '18 at 23:57
Kavi Rama MurthyKavi Rama Murthy
55.7k42158
$endgroup$
Consider the cases $v geq w$ and $v<w$. In the first case $|v-w|=v-w$ so $f(v,w)=v$ and in the second case $|v-w|=w-v$ so $f(v,w)=w$. This proves that $f=g$. Note that $|x|+x geq 0$ for any real number $x$. Hence $f$ takes only non-negative values. From the form of $g$ it follows that the image is contained in $[0,1]$. To show that the image is all of $[0,1]$ just note that $g(t,0)=t$ for all $t in [0,1]$.
answered Dec 3 '18 at 23:57
Kavi Rama MurthyKavi Rama Murthy
55.7k42158
answered Dec 3 '18 at 23:57
Kavi Rama MurthyKavi Rama Murthy
55.7k42158
answered Dec 3 '18 at 23:57
Kavi Rama MurthyKavi Rama Murthy
55.7k42158
answered Dec 3 '18 at 23:57
Kavi Rama MurthyKavi Rama Murthy
55.7k42158
55.7k42158
add a comment |
add a comment |
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