SCALCET8 3.10.025. Use a linear approximation (or differentials) to estimate the given number. (Round your answer to five decimal places.) 3

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SCALCET8 3.10.025. Use a linear approximation (or differentials) to estimate the given number. (Round your answer to five decimal places.) 3 126

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Kiệt Gia 3 years 2021-08-03T06:31:01+00:00 1 Answers 285 views 0

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  1. quangkhai
    0
    2021-08-03T06:32:51+00:00
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    Answer:

    f(126) \approx 5.01333

    Step-by-step explanation:

    Given

    \sqrt[3]{126}

    Required

    Solve using differentials

    In differentiation:

    f(x+\triangle x) \approx f(x) + \triangle x \cdot f'(x)

    Express 126 as 125 + 1;

    i.e.

    x = 125; \triangle x = 1

    So, we have:

    f(125+1) \approx f(125) + 1 \cdot f'(125)

    f(126) \approx f(125) + 1 \cdot f'(125)

    To calculate f(125), we have:

    f(x) = \sqrt[3]{x}

    f(125) = \sqrt[3]{125}

    f(125) = 5

    So:

    f(126) \approx f(125) + 1 \cdot f'(125)

    f(126) \approx 5 + 1 \cdot f'(125)

    f(126) \approx 5 + f'(125)

    Also:

    f(x) = \sqrt[3]{x}

    Rewrite as:

    f(x) = x^\frac{1}{3}

    Differentiate

    f'(x) = \frac{1}{3}x^{\frac{1}{3} - 1}\\

    Using law of indices, we have:

    f'(x) = \frac{x^\frac{1}{3}}{3x}

    So:

    f'(125) = \frac{125^\frac{1}{3}}{3*125}

    f'(125) = \frac{5}{375}

    f'(125) = \frac{1}{75}

    So, we have:

    f(126) \approx 5 + f'(125)

    f(126) \approx 5 + \frac{1}{75}

    f(126) \approx 5 + 0.01333

    f(126) \approx 5.01333

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