Mixing
“Switching†of a continuous function
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Let $f, g: [0,1] rightarrow mathbbR$ be continuous functions and consider the continuous function $h(x) = maxf(x), g(x)$. Suppose that at $0$, $f$ is "active" and at $1$, $g$ is active. That is $h(0) = f(0) neq g(0)$, and similarly at $x = 1$. Is it true that there must exist some point $t in [0,1]$ such that $h$ "switches" from $f$ to $g$? If not I think this should violate continuity but I don't see exactly how.
continuity
asked Aug 6 at 19:59
user395788
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Let $f, g: [0,1] rightarrow mathbbR$ be continuous functions and consider the continuous function $h(x) = maxf(x), g(x)$. Suppose that at $0$, $f$ is "active" and at $1$, $g$ is active. That is $h(0) = f(0) neq g(0)$, and similarly at $x = 1$. Is it true that there must exist some point $t in [0,1]$ such that $h$ "switches" from $f$ to $g$? If not I think this should violate continuity but I don't see exactly how.
continuity
asked Aug 6 at 19:59
user395788
284
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $f, g: [0,1] rightarrow mathbbR$ be continuous functions and consider the continuous function $h(x) = maxf(x), g(x)$. Suppose that at $0$, $f$ is "active" and at $1$, $g$ is active. That is $h(0) = f(0) neq g(0)$, and similarly at $x = 1$. Is it true that there must exist some point $t in [0,1]$ such that $h$ "switches" from $f$ to $g$? If not I think this should violate continuity but I don't see exactly how.
continuity
asked Aug 6 at 19:59
user395788
284
Let $f, g: [0,1] rightarrow mathbbR$ be continuous functions and consider the continuous function $h(x) = maxf(x), g(x)$. Suppose that at $0$, $f$ is "active" and at $1$, $g$ is active. That is $h(0) = f(0) neq g(0)$, and similarly at $x = 1$. Is it true that there must exist some point $t in [0,1]$ such that $h$ "switches" from $f$ to $g$? If not I think this should violate continuity but I don't see exactly how.
continuity
asked Aug 6 at 19:59
user395788
284
asked Aug 6 at 19:59
user395788
284
asked Aug 6 at 19:59
user395788
284
asked Aug 6 at 19:59
user395788
284
284
add a comment |Â
add a comment |Â
1 Answer
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Consider $u=g-f$. $u$ takes a negative value at $0$, a positive one at $1$, and is continuous, so...
answered Aug 6 at 20:01
Nicolas FRANCOIS
3,2501415
Damn, I figured there was probably a standard trick to show this easily. Thanks.
– user395788
Aug 6 at 20:03
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Consider $u=g-f$. $u$ takes a negative value at $0$, a positive one at $1$, and is continuous, so...
answered Aug 6 at 20:01
Nicolas FRANCOIS
3,2501415
Damn, I figured there was probably a standard trick to show this easily. Thanks.
– user395788
Aug 6 at 20:03
add a comment |Â
up vote
1
down vote
accepted
Consider $u=g-f$. $u$ takes a negative value at $0$, a positive one at $1$, and is continuous, so...
answered Aug 6 at 20:01
Nicolas FRANCOIS
3,2501415
Damn, I figured there was probably a standard trick to show this easily. Thanks.
– user395788
Aug 6 at 20:03
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Consider $u=g-f$. $u$ takes a negative value at $0$, a positive one at $1$, and is continuous, so...
answered Aug 6 at 20:01
Nicolas FRANCOIS
3,2501415
Consider $u=g-f$. $u$ takes a negative value at $0$, a positive one at $1$, and is continuous, so...
answered Aug 6 at 20:01
Nicolas FRANCOIS
3,2501415
answered Aug 6 at 20:01
Nicolas FRANCOIS
3,2501415
answered Aug 6 at 20:01
Nicolas FRANCOIS
3,2501415
answered Aug 6 at 20:01
Nicolas FRANCOIS
3,2501415
3,2501415
Damn, I figured there was probably a standard trick to show this easily. Thanks.
– user395788
Aug 6 at 20:03
add a comment |Â
Damn, I figured there was probably a standard trick to show this easily. Thanks.
– user395788
Aug 6 at 20:03
Damn, I figured there was probably a standard trick to show this easily. Thanks.
– user395788
Aug 6 at 20:03
Damn, I figured there was probably a standard trick to show this easily. Thanks.
– user395788
Aug 6 at 20:03
add a comment |Â
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