calculus

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use differentials to approximate the value of the squareroot of 4.3

  • calculus -

    Let
    f(x)=sqrt(x)=x^(1/2)
    f'(x)=1/(2sqrt(x))
    and
    f(x0+Δx)=f(x0)+f'(x0)*Δx (approx.)
    or
    Δx=(f(x0+Δx)-f(x0))/f'(x0) approx.
    Knowing that 2^2=4, and 2.1^2=4.41
    Try x0=2
    Δx=(4.3-2^2)/(1/(2*2)
    =0.3/(4)
    =0.075
    or
    Approximately, x+Δx=2+.075=2.075
    (check: 2.075^2=4.305625)

    Try x0=2.1, x0^2=4.41...
    to get a still better approximation.

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