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Caitlyn Cole 2022-04-30 Answered
Suppose the function f : R R is continuous. For a natural number k, let x 1 , x 2 , . . . , x k be points in [ a , b ]. Prove there is a point z in [ a , b ] at which
f ( z ) = ( f ( x 1 ) + f ( x 2 ) + . . . + f ( x k ) ) / k
So I'm thinking about applying the intermediate value theorem:

If
a < x 1 < b , a < x 2 < b , . . . , a < x k < b
then
f ( a ) < f ( x 1 ) < f ( b ) , . . . , f ( a ) < f ( x k ) < f ( b )
or
k . f ( a ) < f ( x 1 ) + . . . + f ( x k ) < k . f ( b )
f ( a ) < ( f ( x 1 ) + f ( x 2 ) + . . . + f ( x k ) ) / k < f ( b )
But I couldn't think of any way to prove that f ( a ) < f ( x k ) < f ( b ) or is it even true?

EDIT: Thanks everyone for your effort.
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Answers (2)

Bethany Mills
Answered 2022-05-01 Author has 14 answers
Hint:

Take z { x 1 , x k } such that f ( z ) f ( x j ) for any x j .

Take w { x 1 , x k } such that f ( w ) f ( x j ) for any x j .

Then
f ( z ) f ( x 1 ) + + f ( x k ) k f ( w )
Now, use the intermediate value theorem.
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tswe0uk
Answered 2022-05-02 Author has 19 answers
m 1 k i = 1 k f ( x i ) M where M = max f ( x ) , m = min f ( x ) on [ a , b ], Now apply IVT
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