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 ) ) / kSo I'm thinking about applying the intermediate value theorem:If a &lt; x 1 &lt; b , a &lt; x 2 &lt; b , . . . , a &lt; x k &lt; bthen f ( a ) &lt; f ( x 1 ) &lt; f ( b ) , . . . , f ( a ) &lt; f ( x k ) &lt; f ( b )or k . f ( a ) &lt; f ( x 1 ) + . . . + f ( x k ) &lt; k . f ( b ) f ( a ) &lt; ( f ( x 1 ) + f ( x 2 ) + . . . + f ( x k ) ) / k &lt; f ( b )But I couldn't think of any way to prove that f ( a ) &lt; f ( x k ) &lt; f ( b ) or is it even true?EDIT: Thanks everyone for your effort.

Question

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    )  )      /    kSo I&#039;m thinking about applying the intermediate value theorem:If  a  &amp;lt;      x    1    &amp;lt;  b  ,  a  &amp;lt;      x    2    &amp;lt;  b  ,  .  .  .  ,  a  &amp;lt;      x    k    &amp;lt;  bthen  f  (  a  )  &amp;lt;  f  (      x    1    )  &amp;lt;  f  (  b  )  ,  .  .  .  ,  f  (  a  )  &amp;lt;  f  (      x    k    )  &amp;lt;  f  (  b  )or  k  .  f  (  a  )  &amp;lt;  f  (      x    1    )  +  .  .  .  +  f  (      x    k    )  &amp;lt;  k  .  f  (  b  )  f  (  a  )  &amp;lt;  (  f  (      x    1    )  +  f  (      x    2    )  +  .  .  .  +  f  (      x    k    )  )      /    k  &amp;lt;  f  (  b  )But I couldn&#039;t think of any way to prove that   f  (  a  )  &amp;lt;  f  (      x    k    )  &amp;lt;  f  (  b  ) or is it even true?EDIT: Thanks everyone for your effort.

Bethany Mills · Accepted Answer

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.

tswe0uk · Accepted Answer

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

Suppose the function f : <mrow class="MJX-TeXAtom-ORD"> <mi mathvariant="double-struc

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