Let f be continuous on [a,b] and differential on (a,b). Prove that if a &gt;= 0 there are x1, x2, x3 ∈ (a,b) such that f ′ ( x 1 ) = ( b + a ) f ′ ( x 2 ) 2 x 2 = ( b 2 + b a + a 2 ) f ′ ( x 3 ) 3 x 3 2 I think this problem will use the generalized mean value theorem to solve. However, I don't know how to apply it. Can you suggest a solution? Thank you very much!

Question

Let f be continuous on [a,b] and differential on (a,b). Prove that if a &amp;gt;= 0 there are x1, x2, x3 ∈ (a,b) such that      f    ′    (      x    1    )  =  (  b  +  a  )                              f          ′                (                  x          2                )                    2                  x          2                      =  (      b    2    +  b  a  +      a    2    )                              f          ′                (                  x          3                )                    3                  x          3          2                    I think this problem will use the generalized mean value theorem to solve. However, I don&#039;t know how to apply it. Can you suggest a solution? Thank you very much!

iskakanjulc · Accepted Answer

By the mean value theorem,                    (1)                    ∃                  x          1                ∈        (        a        ,        b        )                :                                      f            (            b            )            −            f            (            a            )                                b            −            a                          =                  f          ′                (                  x          1                )                .            Consider   g  :  [      a    2    ,      b    2    ]  →            R       defined by   g  (  x  )  =  f  (      x    ). Applying the mean value theorem to   g, we infer that there is some   t  ∈  (      a    2    ,      b    2    ) such that            g      (              b        2            )      −      g      (              a        2            )                      b        2            −              a        2              =      g    ′    (  t  )    ,or equivalently, writing       x    2    :=      t  ,                    (2)                                          f            (            b            )            −            f            (            a            )                                (            b            −            a            )            (            b            +            a            )                          =                                            f              ′                        (                          t                        )                                2                          t                                      =                                            f              ′                        (                          x              2                        )                                2                          x              2                                              .            Similarly, apply the mean value theorem to   h  :  [      a    3    ,      b    3    ]  →            R       defined by   h  (  x  )  =  f  (      x          1              /            3        ). We deduce that there is some   s  ∈  (      a    3    ,      b    3    ) such that            h      (              b        3            )      −      h      (              a        3            )                      b        3            −              a        3              =      h    ′    (  s  )    ,or equivalently, writing       x    3    :=            s              1                  /                3            ,                    (3)                                          f            (            b            )            −            f            (            a            )                                (            b            −            a            )            (                          b              2                        +            b            a            +                          a              2                        )                          =                                            f              ′                        (                          s                              1                                  /                                3                                      )                                3                          s                              2                                  /                                3                                                    =                                            f              ′                        (                          x              3                        )                                3                          x              3              2                                              .

Let f be continuous on [a,b] and differential on (a,b). Prove that if a >= 0 there are x1, x2, x3 ∈

Answered question

Answer & Explanation

New Questions in Differential Calculus