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Linear Approximation

  1. Nov 9, 2004 #1
    I need to find thelocal approximation of f(x)=x^(1/3) at x=26.6, knowing f(27)=3.

    Here's what I did, dont know if I did it right:

    f '(x)=(1/3)x^(-2/3)=[x^(-2/3)]/3

    slope of the tanget = slope of the secant
    [x^(-2/3)]/3=(y-y1)/(x-x1)
    [26.6^(-2/3)]/3=(y-3)/(x-27)

    now I sub in X and solve for y:
    [26.6^(-2/3)]/3=(y-3)/(26.6-27)
    y=2.985

    I dont think that's the answer at all, could someone check it?
     
  2. jcsd
  3. Nov 9, 2004 #2
    your answer is reasonable. why? it's just under the cube root of 27, which is what you're looking at. cube that number to check your results -- 26.597 pretty close.

    as i recall

    the formula is L(x) = f(a) + f'(a)(x-a)

    f(x) = x^(1/3)
    f'(x) = 1/(3x^(2/3))

    i arrive at the same answer plugging those values in.

    f(x) = x^(1/3)
     
  4. Nov 9, 2004 #3
    However, I'm not supposed to use a calculator to solve it...is there an easier way?
     
  5. Nov 9, 2004 #4
    well subsitute the values in fractional form

    3 + 1/3(27)^2/3(26.6-27)

    3+(1/27)(-4/10)

    3-4/270 = 3-2/135 = 2.985

    the fractional answer is correct, the decimal is simply an approximation of the fractional expressoin. (both are approximations, keep that in mind)
     
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