# ContinuityWatch

Announcements
#1
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.
0
6 days ago
#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).
0
#3
(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).
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.
0
6 days ago
#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.
Since you've covered the IVT, consider the function f(x)-g(x)
0
6 days ago
#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.
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.
0
#6
(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.
Oh ****. Makes sense. Love for that bro
0
6 days ago
#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.
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.
Last edited by RDKGames; 6 days ago
0
6 days ago
#8
I suppose my method didn't have much rigour... xD
0
#9
(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.
Yeah I get what you mean. For me understanding the question is harder than doing the question so it at least helped spark something.
1
X

new posts Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### University open days

• University of Hertfordshire
Wed, 11 Dec '19
• University of Lincoln
Wed, 11 Dec '19
• Bournemouth University
Wed, 11 Dec '19

### Poll

Join the discussion

#### Do you work while at uni?

Yes I work at university (55)
31.61%
No I don't (84)
48.28%
I work during the holidays (35)
20.11%