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Linear approx verifying?

  1. Jun 15, 2014 #1
    1. The problem statement, all variables and given/known data
    Verify the given linear approximation at
    a = 0.
    Then determine the values of x for which the linear approximation is accurate to within 0.1. (Enter your answer using interval notation. Round your answers to three decimal places.)

    1/(1 + 3x)^3 ≈ 1 − 9x
    2. Relevant equations

    I have no idea how to find the x values

    3. The attempt at a solution

    (1/(1+3x)^3)-0.1<1-9x<(1/(1+3x)^3)+0.1 is how my textbook sets up the problem and then all of a sudden they seem to compute the x values that is unknown to me. is it simply just solving the inequality for x (subtract 1 and then divide by -9 having x=0) thanks and any help will be appreciated.
     
  2. jcsd
  3. Jun 15, 2014 #2

    haruspex

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    Yes, it is just a matter of solving each inequality for x, but not setting x = 0 anywhere.
    E.g. for (1+3x)-3-0.1<1-9x, solve the equation (1+3x)-3-0.1=1-9x to find the end of the interval. Of course, solving a cubic is non-trivial, so you should use a suitable approximation.
    Solving each cubic will give you one or three real values. (Two is theoretically possible, but most unlikely.) You then need to figure out which ones are relevant and which side of them satisfies the inequality.
     
  4. Jun 16, 2014 #3

    Ray Vickson

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    Actually, you get quartics, because
    [tex] \frac{1}{(1+3x)^3}\pm 0.1 = 1-9x \Longrightarrow 1 \pm 0.1(1+3x)^3 = (1+3x)^3(1-9x)[/tex]
     
  5. Jun 16, 2014 #4

    haruspex

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    How true. Same advice, though.
     
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