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Math Help - Change in x and differentials

  1. #1
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    Change in x and differentials

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  2. #2
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    f ' (x)= ( f(x)-f(xo) )/ (x-xo)

    this implies..

    f(x)-f(xo) = f ' (x) (x-xo)
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  3. #3
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    Hello, zaurm!

    Use differentials to approximate the change in
    f(x) \:=\:4x^5 - 6x^4 + 3x^2 - 5 .if x changes from 1 to 1.02
    We have: . x = 1,\:dx \:=\:0.02

    . . And: . df \;=\;(20x^4 - 24x^3 + 6x)\,dx

    Therefore: . df \;=\;\bigg[20(1^4) - 24(1^3) + 6(1) - 5\bigg]\,(0.02) \;=\;\boxed{0.04}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    I like to check my answer by calculating the actual change.

    . . f(1) \;=\;4(1^5) - 6(1^4) + 3(1^2) - 5 \;=\;-4

    . . f(1.02) \;=\;4(1.02^5) - 6(1.02^4) + 3(1.02^2) - 5 \;=\;-3.957069747

    The actual change is: . \Delta f \;=\;-3.957069747 - (-4) \;=\;{\color{blue}0.04}2930253

    . . . . Close enough!

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