Use linear approximation i.e the tangent line ...?

I am very confused how to solve this, please help!

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4 Answers

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  • ted s
    Lv 7
    4 weeks ago

    y - f(a) = f ' ( a ) ( x - a ).......f ' = ( 1/4) x^(-3/4)...f '(81) = 1/108...a = 81& f(a) = 3

  • Ian H
    Lv 7
    4 weeks ago

    f(81) = 81^(1/4) = 3

    f'(x) = (1/4)x^(-3/4)

    f'(81) = (1/4)*81^(-3/4) = 1/(4*27) = 1/108

    L(x) = f’(81)(x – 81) + f(81)

    L(x) = (1/108)(x – 81) + 3

    L(81.04) = 3 + (0.04/108) ~ 3.000370370370

    Compare ... 81.04 ^(1/4) = 3.000370301803 calculator

  • TomV
    Lv 7
    4 weeks ago

    Let f(x) = x^(1/4)

    f'(x) = (1/4)x^(-3/4)

    For values of x near a:

    f'(a) ≈ [f(x) - f(a)]/(x-a)

    f(x) ≈ f(a) + f'(a)(x-a)

    Let a = 81. Then f(81.04) = f(81) + f'(81)(0.04)

    f(81) = 81^(1/4) = 3

    f'(81) = (1/4)/27 = 1/108

    f(81.04) = (81.04)^(1/4)  ≈ 3 + 0.04/108 ≈ 3.000370370_

    (Actual value,  3.0003703018... differs from the approximate value in the 8th decimal place)

  • 4 weeks ago

    You are asked to find L(81.04). The tangent line approximates the function in a close neighborhood around the tangency point. I don't see how this can be confusing if you spent at least 5 minutes studying these things.

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