minimum value of f ( x ) = ( 1 + x ) 0.8 1 + x 0.8 without derivativeminimum value of f ( x ) = ( 1 + x ) 0.8 1 + x 0.8 , x ≥ 0 without derivativeBinomial expansion of ( 1 + x ) 0.8 = 1 + 0.8 x − ( 0.8 ⋅ ( 0.8 − 1 ) x 2 ) 2 + ⋯ but from above does not get anythingcould some help me with this

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minimum value of       f    (    x    )    =                  (        1        +        x                  )                      0.8                                      1        +                  x                      0.8                               without derivativeminimum value of       f    (    x    )    =                  (        1        +        x                  )                      0.8                                      1        +                  x                      0.8                                ,    x    ≥    0   without derivativeBinomial expansion of       (    1    +    x          )              0.8              =    1    +    0.8    x    −                  (        0.8        ⋅        (        0.8        −        1        )                  x          2                )            2        +    ⋯  but from above does not get anythingcould some help me with this

helsedel1v · Accepted Answer

If you set   x  =      e    u   you get:  f  (  x  )  =            (      1      +              e        u                    )                  4                      /                    5                            1      +              e                  4          u                      /                    5                      =      2          −      1              /            5        ⋅            cosh      ⁡      (      u              /            2              )                  4                      /                    5                            cosh      ⁡      (      2      u              /            5      )      that is an even function of   u. It follows that the minimum of   f is achieved at   u  =  0, i.e. at   x  =  1It is enough to show that   log  ⁡  cosh  ⁡            4      x        5    ≤      4    5    log  ⁡  cosh  ⁡  x for any   x  ≥  0. That follows from   tanh  ⁡            4      x        5    ≤  tanh  ⁡  x, that is trivial, through termwise integration.

mastegotgd · Accepted Answer

Raise your function to the power of five:  (  f  (  x  )      )    5    =            (                        (          1          +          x                      )                          0.8                                                1          +                      x                          0.8                                          )        5    (  f  (  x  )      )    5    =            (      1      +      x              )        4                    1      +      5              x                  0.8                    +      10              x                  1.6                    +      10              x                  2.4                    +      5              x                  3.2                    +              x        4              (  f  (  x  )      )    5    =            1      +      4      x      +      6              x        2            +      4              x        3            +              x        4                    1      +      5              x                  0.8                    +      10              x                  1.6                    +      10              x                  2.4                    +      5              x                  3.2                    +              x        4              (  f  (  x  )      )    5    =                    x                  −          2                    +      4              x                  −          1                    +      6      +      4      x      +              x        2                            x                  −          2                    +      5              x                  −          1.2                    +      10              x                  −          0.4                    +      10              x                  0.4                    +      5              x                  1.2                    +              x        2            As   (  f  (  x  )      )    5    =  (  f  (      x          −      1        )      )    5   then you will have a minimum (or maximum) when   x  =      x          −      1      , i.e. when   x  =  1 (ignoring negative values as its unclear which root you would want for   x  &amp;lt;  0).

minimum value of f(x)=((1+x)^(0.8))/(1+x^(0.8)) without derivative Binomial expansion of (1+x)^(0.8)=1+0.8 x-((0.8*(0.8-1)x^2))/(2)+...but from above does not get anything could some help me with this

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