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

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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  1. Answer:

    [tex]f(126) \approx 5.01333[/tex]

    Step-by-step explanation:

    Given

    [tex]\sqrt[3]{126}[/tex]

    Required

    Solve using differentials

    In differentiation:

    [tex]f(x+\triangle x) \approx f(x) + \triangle x \cdot f'(x)[/tex]

    Express 126 as 125 + 1;

    i.e.

    [tex]x = 125; \triangle x = 1[/tex]

    So, we have:

    [tex]f(125+1) \approx f(125) + 1 \cdot f'(125)[/tex]

    [tex]f(126) \approx f(125) + 1 \cdot f'(125)[/tex]

    To calculate f(125), we have:

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

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

    [tex]f(125) = 5[/tex]

    So:

    [tex]f(126) \approx f(125) + 1 \cdot f'(125)[/tex]

    [tex]f(126) \approx 5 + 1 \cdot f'(125)[/tex]

    [tex]f(126) \approx 5 + f'(125)[/tex]

    Also:

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

    Rewrite as:

    [tex]f(x) = x^\frac{1}{3}[/tex]

    Differentiate

    [tex]f'(x) = \frac{1}{3}x^{\frac{1}{3} – 1}\\[/tex]

    Using law of indices, we have:

    [tex]f'(x) = \frac{x^\frac{1}{3}}{3x}[/tex]

    So:

    [tex]f'(125) = \frac{125^\frac{1}{3}}{3*125}[/tex]

    [tex]f'(125) = \frac{5}{375}[/tex]

    [tex]f'(125) = \frac{1}{75}[/tex]

    So, we have:

    [tex]f(126) \approx 5 + f'(125)[/tex]

    [tex]f(126) \approx 5 + \frac{1}{75}[/tex]

    [tex]f(126) \approx 5 + 0.01333[/tex]

    [tex]f(126) \approx 5.01333[/tex]

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