Let f and g be continuous functions on [ a , b ] such that f ( a ) ≥ g ( a ) and f ( b ) ≤ g ( b ). Prove f ( x 0 ) = g ( x 0 ) for at least one x 0 in [ a , b ].Here's what I have so far:Let h be a continuous function where h = f − gSince h ( b ) = f ( b ) − g ( b ) ≤ 0, and h ( a ) = f ( a ) − g ( a ) ≥ 0, then h ( b ) ≤ 0 ≤ h ( a ) is true.So, by IVT there exists some y such that...And that's where I need help. I read what the IVT is, but I could use some help explaining how it applies here, and why it finishes the proof. Thank you!

Question

Let   f and   g be continuous functions on   [  a  ,  b  ] such that   f  (  a  )  ≥  g  (  a  ) and   f  (  b  )  ≤  g  (  b  ). Prove   f  (      x    0    )  =  g  (      x    0    ) for at least one       x    0   in   [  a  ,  b  ].Here&#039;s what I have so far:Let   h be a continuous function where   h  =  f  −  gSince   h  (  b  )  =  f  (  b  )  −  g  (  b  )  ≤  0, and   h  (  a  )  =  f  (  a  )  −  g  (  a  )  ≥  0, then   h  (  b  )  ≤  0  ≤  h  (  a  ) is true.So, by IVT there exists some   y such that...And that&#039;s where I need help. I read what the IVT is, but I could use some help explaining how it applies here, and why it finishes the proof. Thank you!

beentjie8e · Accepted Answer

There exists some   y  ∈  (  a  ,  b  ) such that   h  (  y  )  =  0. So   f  (  y  )  =  g  (  y  ). The other case   a  =  b is trivial.

Let f and g be continuous functions on [a,b] such that f(a)>=g(a) and f(b)<=g(b). Prove f(x_0)=g(x_0) for at least one x_0 in [a,b].

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