If f ( x ) = 1 π ( arcsin ⁡ x + arccos ⁡ x + arctan ⁡ x ) + x + 1 x 2 + 2 x + 10 , , Then max value of f(x)

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If       f    (    x    )    =          1      π              (      arcsin      ⁡      x      +      arccos      ⁡      x      +      arctan      ⁡      x      )        +                  x        +        1                              x          2                +        2        x        +        10                  ,  , Then max value of f(x)

behk0 · Accepted Answer

The domain should be   x  ∈  [  −  1  ,  1  ]We have  f  (  x  )  =      1    π        (          π      2        +    arctan    ⁡    x    )    +            x      +      1                      x        2            +      2      x      +      10      so      f    ′    (  x  )  =      1          π      (      1      +              x        2            )        +            9      −      (      x      +      1              )        2                    (              x        2            +      2      x      +      10              )        2            This is positive because of   (  x  +  1      )    2    ≤  4Since f(x) is increasing, the answer is   f  (  1  )  =      47          /        52

cooloicons62 · Accepted Answer

For   x  ∈  [  −  1  ,  1  ]      f    ′    (  x  )  =                    8        −        2        x        −                  x          2                            (        10        +        2        x        +                  x          2                          )          2                      +            1              π        (        1        +                  x          2                )              &amp;gt;  0So, f is increasing in   [  −  1  ,  1  ]. Thus, the maxima is taken at x=1. Plugging that in, the maxima is  f  (  1  )  =      1    π    (      π    2    +      π    4    )  +      2    13    =      3    4    +      2    13    =      47    52

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