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Let f : [0, 1] → [0, 1] be continuous and let g : [0, 1] → R be continuous with
g(0) = 0, g(1) = 1.
Prove that we can find a ξ ∈ [0, 1] with f(ξ) = g(ξ)
Don't know how to get going. Any help would be appreciated.
g(0) = 0, g(1) = 1.
Prove that we can find a ξ ∈ [0, 1] with f(ξ) = g(ξ)
Don't know how to get going. Any help would be appreciated.
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#2
Not too sure about this as I am still in 6th form, but here is my attempt:
f is continuous and defined for all x and f(x) in the range [0,1]
g is continuous, and g(0) = 0, and g(1) = 1, meaning that it is also defined for all f(x) in the range [0,1]
Therefore, there must be some value ξ in the range [0,1] where f(ξ) = g(ξ).
f is continuous and defined for all x and f(x) in the range [0,1]
g is continuous, and g(0) = 0, and g(1) = 1, meaning that it is also defined for all f(x) in the range [0,1]
Therefore, there must be some value ξ in the range [0,1] where f(ξ) = g(ξ).
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(Original post by Mrepic Foulger)
Not too sure about this as I am still in 6th form, but here is my attempt:
f is continuous and defined for all x and f(x) in the range [0,1]
g is continuous, and g(0) = 0, and g(1) = 1, meaning that it is also defined for all f(x) in the range [0,1]
Therefore, there must be some value ξ in the range [0,1] where f(ξ) = g(ξ).
Not too sure about this as I am still in 6th form, but here is my attempt:
f is continuous and defined for all x and f(x) in the range [0,1]
g is continuous, and g(0) = 0, and g(1) = 1, meaning that it is also defined for all f(x) in the range [0,1]
Therefore, there must be some value ξ in the range [0,1] where f(ξ) = g(ξ).
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#4
(Original post by abz456)
Damn, what a lad. Enjoy sixth form lol. And yeah, thats what I thought when I attempted this. I used the intermediate value theorem to try and justify it.
Damn, what a lad. Enjoy sixth form lol. And yeah, thats what I thought when I attempted this. I used the intermediate value theorem to try and justify it.
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#5
(Original post by abz456)
Let f : [0, 1] → [0, 1] be continuous and let g : [0, 1] → R be continuous with
g(0) = 0, g(1) = 1.
Prove that we can find a ξ ∈ [0, 1] with f(ξ) = g(ξ)
Don't know how to get going. Any help would be appreciated.
Let f : [0, 1] → [0, 1] be continuous and let g : [0, 1] → R be continuous with
g(0) = 0, g(1) = 1.
Prove that we can find a ξ ∈ [0, 1] with f(ξ) = g(ξ)
Don't know how to get going. Any help would be appreciated.
Give a reason why it's continuous on the interval.
Show that h(0) = - f(0) which is less than or equal to 0.
and h(1) = 1 - f(1) which is greater than or equal to 0.
Then apply the intermediate value theorem to 0.
Show that there must be ξ for which h(ξ) = 0.
The result follows.
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(Original post by MarkFromWales)
Consider h(x) = g(x) - f(x).
Give a reason why it's continuous on the interval.
Show that h(0) = - f(0) which is less than or equal to 0.
and h(1) = 1 - f(1) which is greater than or equal to 0.
Then apply the intermediate value theorem to 0.
Show that there must be ξ for which h(ξ) = 0.
The result follows.
Consider h(x) = g(x) - f(x).
Give a reason why it's continuous on the interval.
Show that h(0) = - f(0) which is less than or equal to 0.
and h(1) = 1 - f(1) which is greater than or equal to 0.
Then apply the intermediate value theorem to 0.
Show that there must be ξ for which h(ξ) = 0.
The result follows.
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#7
(Original post by abz456)
Damn, what a lad. Enjoy sixth form lol. And yeah, thats what I thought when I attempted this. I used the intermediate value theorem to try and justify it.
Damn, what a lad. Enjoy sixth form lol. And yeah, thats what I thought when I attempted this. I used the intermediate value theorem to try and justify it.
Last edited by RDKGames; 6 days ago
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(Original post by RDKGames)
All they have done was rephrase the question and then answer it in a sense of "well, it's obvious!" and it might be once you understand it, but I hope you don't try doing that in an Analysis exam because you need to be rigorous about it (by using IVT) and not so hand-wavey.
All they have done was rephrase the question and then answer it in a sense of "well, it's obvious!" and it might be once you understand it, but I hope you don't try doing that in an Analysis exam because you need to be rigorous about it (by using IVT) and not so hand-wavey.
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