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Math approximation problem

  1. Oct 28, 2008 #1
    Let f(x) = x^(1/3). The equation of the tangent line to f(x) at x = 8 can be written in the form y = mx+b where m is: and where b is:
    Using this, we find our approximation for 8.1^(1.3) is:

    I found the slope to be 1/12
    I found b to be 1.3333333333333333333
    I still can't get the answer for the approximation for 8.1^(1/3).
    I plugged it correctly in the mx+b equation but it won't work.
    Is there another way to do this? Please help.

  2. jcsd
  3. Oct 29, 2008 #2


    Staff: Mentor

    Re: Approximation

    The exact equation for your tangent line at (8, 2) is
    y = 1/12 * x + 4/3

    When x = 8.1, what is the value of y on the tangent line? That's your approximation for (8.1)^(1/3).
  4. Oct 29, 2008 #3


    User Avatar
    Staff Emeritus
    Science Advisor

    Re: Approximation

    Perhaps simpler: y= (1/12)(x- 8)+ 2.

    Edited thanks to Mark44.
    Last edited: Oct 29, 2008
  5. Oct 29, 2008 #4


    Staff: Mentor

    Re: Approximation

    The line has to pass through (8, 2), not (4, 2). You might have overlooked the fact that we're dealing with the cube root function, not the square root function.
  6. Oct 29, 2008 #5
    Re: Approximation

    Thanks alot for the help. I was able to get the answer.

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