If f ( x ) = a x − x 3 1 + x 2 show that ∀ x [ f ′ ( x ) ≥ 0 iff a ≥ 9 8 ]Part of answer found online: f ′ ( x ) = a − 3 x 2 ( 1 + x 2 ) − 2 x ( x 3 ) ( 1 + 3 x 2 ) 2 = a + x 4 − 3 x 2 ( 1 + x 2 ) 2 My problem:I don't understand why: a − 3 x 2 ( 1 + x 2 ) − 2 x ( x 3 ) ( 1 + 3 x 2 ) 2 = a + x 4 − 3 x 2 ( 1 + x 2 ) 2 When i try to rearrange the left i get: a − 3 x 2 ( 1 + x 2 ) − 2 x ( x 3 ) ( 1 + 3 x 2 ) 2 = a − 3 x 4 + 3 x 2 − 2 x 4 ( 1 + 3 x 2 ) 2 = a − x 4 + 3 x 2 ( 1 + 3 x 2 ) 2 And from there: a − x 4 + 3 x 2 ( 1 + 3 x 2 ) 2 = a + − 1 1 ⋅ x 4 + 3 x 2 ( 1 + 3 x 2 ) 2 = a + − x 4 − 3 x 2 ( 1 + 3 x 2 ) 2 Am i making a simple mistake in the interpretation of the minus in front of a fraction?

Question

If   f  (  x  )  =  a  x  −            x      3              1      +              x        2             show that   ∀  x  [      f    ′    (  x  )  ≥  0 iff   a  ≥      9    8    ]Part of answer found online:      f    ′    (  x  )  =  a  −            3              x        2            (      1      +              x        2            )      −      2      x      (              x        3            )              (      1      +      3              x        2                    )        2              =  a  +                    x        4            −      3              x        2                    (      1      +              x        2                    )        2            My problem:I don&#039;t understand why:  a  −            3              x        2            (      1      +              x        2            )      −      2      x      (              x        3            )              (      1      +      3              x        2                    )        2              =  a  +                    x        4            −      3              x        2                    (      1      +              x        2                    )        2            When i try to rearrange the left i get:  a  −            3              x        2            (      1      +              x        2            )      −      2      x      (              x        3            )              (      1      +      3              x        2                    )        2              =  a  −            3              x        4            +      3              x        2            −      2              x        4                    (      1      +      3              x        2                    )        2              =  a  −                    x        4            +      3              x        2                    (      1      +      3              x        2                    )        2            And from there:  a  −                    x        4            +      3              x        2                    (      1      +      3              x        2                    )        2              =  a  +            −      1        1    ⋅                    x        4            +      3              x        2                    (      1      +      3              x        2                    )        2              =  a  +            −              x        4            −      3              x        2                    (      1      +      3              x        2                    )        2            Am i making a simple mistake in the interpretation of the minus in front of a fraction?

Bradley Forbes · Accepted Answer

It&#039;s not. There is a mistake in the denominator. It should be   (  1  +      x    2        )    2   not   (  1  +  3      x    2        )    2        f                          ′              (  x  )  =  a  −                    x        4            +      3              x        2                    (      1      +              x        2                    )        2

Landon Wolf · Accepted Answer

It should be       f    ′    (  x  )  =  a  −                    x        4            +      3              x        2                    (              x        2            +      1              )        2            We need to prove that      9    8    −                    x        4            +      3              x        2                    (              x        2            +      1              )        2              ≥  0  ,which is   (      x    2    −  3      )    2    ≥  0

Question: If f(x)=ax−(x^3)/(1+x^2) show that AAx[f′(x)>=0 iff a>=9/8] Part of answer found online: f'(x) = a-(3x^2(1+x^2)-2x(x^3))/((1+3x^2)^2) = a+(x^4 - 3x^2)/((1+x^2)^2)

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