How to find the linearization of f ( x ) = x 1 2 at x=25?

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How to find the linearization of             f              (        x        )              =          x                        1          2                     at x=25?

rygynguo0rk · Accepted Answer

f              (        x        )              =          x                        1          2                      =          x      At       x    =    25  , we have       y    =          f              (        25        )              =    5        f    ′          (      x      )        =          1              2                  x                     so at       x    =    25  , the slope of the tangent line is       m    =    f    ′          (      25      )        =          1      10      Equation of a tangent line in the form of a point-slope:        y    -    5    =          1      10              (      x      -      25      )      Linearization at a=1 (in a form I am used to):      L          (      x      )        =    5    +          1      10              (      x      -      25      )      Free illustration of the linearization process            28        =          f              (        28        )                        f              (        28        )              ≈    L          (      28      )        =    5    +          1      10              (      28      -      25      )        =    5    +          1      10              (      3      )        =    5    +    0.3    =    5.3  (The point on the tangent line with       x  -coordinate       28   has       y  -coordinate       5.3  )

How to find the linearization of f(x) = x^(1/2) at x=25?

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