Calculus 2: Partial fractions problem. Finding the value of a constantI encountered the following problem.Let f(x) be a quadratic function such that f ( 0 ) = − 6 and ∫ f ( x ) x 2 ( x − 3 ) 8 d xis a rational function.Determine the value of f′(0)Here's what I tried. I decomposed the fraction integrand below f ( x ) x 2 ( x − 3 ) 8 = A 1 x + A 2 x 2 + ∑ i = 1 8 B i ( x − 3 ) i By finding a common denominator, I determined f ( x ) = A 1 x ( x − 3 ) 8 + A 2 ( x − 3 ) 8 + x 2 ∑ j = 1 8 [ B j ( x − 3 ) 8 − j ] f ′ ( x ) = A 1 ( x − 3 ) 8 + 8 A 2 ( x − 3 ) 7 + D ( x )where D(x) is a function such that D(0)=0. (These are all the remaining terms that go away when we plug 0 into f′(0)).I used the information that D ( 0 ) = 0 to get the equation f ( 0 ) = A 2 ( − 3 ) 8 = − 6 → A 2 = − 2 3 7 This leaves us with f ′ ( 0 ) = A 1 ( − 3 ) 8 + 8 ⋅ − 2 3 7 ⋅ ( − 3 ) 7 f ′ ( 0 ) = 3 8 ⋅ A 1 + 16I was told f ′ ( 0 ) = 16, however I cannot convince myself the value of A 1 . I would think that the fact that f(x) is a quadratic function should come into play here. Please let me know what you think.

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Calculus 2: Partial fractions problem. Finding the value of a constantI encountered the following problem.Let f(x) be a quadratic function such that   f  (  0  )  =  −  6 and  ∫            f      (      x      )                      x        2            (      x      −      3              )        8              d  xis a rational function.Determine the value of f′(0)Here&#039;s what I tried. I decomposed the fraction integrand below            f      (      x      )                      x        2            (      x      −      3              )        8              =            A      1        x    +            A      2              x      2        +      ∑          i      =      1        8              B      i              (      x      −      3              )        i            By finding a common denominator, I determined  f  (  x  )  =      A    1    x  (  x  −  3      )    8    +      A    2    (  x  −  3      )    8    +      x    2        ∑          j      =      1        8    [      B    j    (  x  −  3      )          8      −      j        ]      f    ′    (  x  )  =      A    1    (  x  −  3      )    8    +  8      A    2    (  x  −  3      )    7    +  D  (  x  )where D(x) is a function such that D(0)=0. (These are all the remaining terms that go away when we plug 0 into f′(0)).I used the information that   D  (  0  )  =  0 to get the equation  f  (  0  )  =      A    2    (  −  3      )    8    =  −  6  →      A    2    =            −      2              3      7      This leaves us with      f    ′    (  0  )  =      A    1    (  −  3      )    8    +  8  ⋅            −      2              3      7        ⋅  (  −  3      )    7        f    ′    (  0  )  =      3    8    ⋅      A    1    +  16I was told       f    ′    (  0  )  =  16, however I cannot convince myself the value of       A    1  . I would think that the fact that f(x) is a quadratic function should come into play here. Please let me know what you think.

bargeolonakc · Accepted Answer

Step 1Note that since f is quadratic with   f  (  0  )  =  0, we can write f(x) as  f  (  x  )  =  −  6  +      f    ′    (  0  )  x  +      1    2        f    ″    (  0  )      x    2  Then, the integrand becomes                    (1)                                          f            (            x            )                                              x              2                        (            x            −            3                          )              8                                      =                              −            6                                              x              2                        (            x            −            3                          )              8                                      +                                            f              ′                        (            0            )                                x            (            x            −            3                          )              8                                      +                                            1              2                                      f              ″                        (            0            )                                (            x            −            3                          )              8                                          The last term on the right-hand side of (1) integrates to the rational function,   ∫                    1        2                    f        ″            (      0      )              (      x      −      3              )        8                d  x  =  −      1    2        f    ″    (  0  )      1          7      (      x      −      3              )        7              +  C and is not implicated , therefore, in the ensuing analysis.Step 2Using partial fraction expansion, we can write the first term on the right-hand side of (1) as                    (2)                                          −            6                                              x              2                        (            x            −            3                          )              8                                      =                              −            16                                6561                        x                          +                  16                      6561                        (            x            −            3            )                          +                  (          other terms that integrate to rational functions          )                    and the second term on the right-hand side of (1) as                    (3)                                                        f              ′                        (            0            )                                x            (            x            −            3                          )              8                                      =                                            f              ′                        (            0            )                                6561                        x                          −                                            f              ′                        (            0            )                                6561                        (            x            −            3            )                          +                  (          other terms that integrate to rational functions          )                    Therefore, to annihilate the terms that integrate to log(x) and   log  ⁡  (  x  −  3  ) in (2) and (3), we must have      f    ′    (  0  )  =  16

Chloe Arnold · Accepted Answer

Step 1Consider that your partial fraction decomposition is equal to             A      1        x    +            B      1              x      −      3        +  h  (  x  )  , where h(x) is the sum of the other terms.Step 2The anti-derivative of each of those other terms is a rational function. But   ∫  (            A      1        x    +            B      1              x      −      3        )    d  x  =      A    1    ln  ⁡      |    x      |    +      B    1    ln  ⁡      |    x  −  3      |   is not a rational function unless       A    1    =      B    1    =  0. So       A    1    =  0 and you are finished.

Calculus 2: Partial fractions problem. Finding the value of a constant. Let f(x) be a quadratic function such that f(0)=-6 and int (f(x))/(x^2(x-3)^8)dx

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