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March 30, 2017

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Find the linearization of f(x)= sqrtx at x=25 and use the linearization to approximate sqrt25.2

This is my math: 5+ 1/10 * (25.2-25) but it's wrong.

  • Calculus - ,

    The McLaurin series expansion of that function about x=25 is
    f(x) = f(25) + (x-25)[f'(x) at x=25)
    f'(x) = (1/2)*x^(-1/2)
    f'(25) = (1/2)/5 = 1/10
    Therefore
    f(x) = 5 + (x-25)/10
    f(25.2) = 5+ .02 = 5.02
    There is nothing wrong with your answer. The correct value is
    5.01996016...
    Maybe they wanted you to do the last step and compute an actual numerical approximation, 5.02.

  • Calculus - ,

    I typed in 5.01999999999 and it took it. Weird.

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