Let f be a function such that f : U ⊆ R n → R . U is open in R n and path connected and f is continuous. Let x 1 , x 2 ∈ U. Proof that for all c ∈ [ f ( x 1 ) , f ( x 2 ) ] there exists an x ∈ U such that f ( x ) = c. I'm supposed to use one-dimensional intermediate value theorem to proof this. There is a hint stating that I should look out for a function φ : [ 0 , 1 ] → U such that we use a "useful" composition of f and φ. I really don't know how to do this proof I would appreciate help a lot!

Question

Let   f be a function such that   f  :  U  ⊆            R        n    →      R  .   U is open in             R        n   and path connected and   f is continuous. Let       x    1    ,      x    2    ∈  U. Proof that for all   c  ∈  [  f  (      x    1    )  ,  f  (      x    2    )  ] there exists an   x  ∈  U such that   f  (  x  )  =  c. I&#039;m supposed to use one-dimensional intermediate value theorem to proof this. There is a hint stating that I should look out for a function   φ  :  [  0  ,  1  ]  →  U such that we use a &quot;useful&quot; composition of   f and   φ. I really don&#039;t know how to do this proof I would appreciate help a lot!

Alexis Fields · Accepted Answer

You could let   ϕ be a continuous function satisfying   ϕ  (  0  )  =      x    1   and   ϕ  (  1  )  =      x    2  . Its existence is guaranteed by path-connectedness. The composition   f  ∘  ϕ is continuous, maps   [  0  ,  1  ] to       R  , and satisfies   f  ∘  ϕ  (  0  )  =  f  (      x    1    ) and   f  ∘  ϕ  (  1  )  =  f  (      x    2    ). There must exist   t  ∈  [  0  ,  1  ] with   f  ∘  ϕ  (  t  )  =  c: what can you say about   ϕ  (  t  )?

Sonia Ayers · Accepted Answer

you need to find a point   x  ∈  U satisfying   f  (  x  )  =  c. So far there is a point   t  ∈  [  0  ,  1  ], guess i`m right

Let f be a function such that f : U ⊆ <mrow class="MJX-

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