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Solving absolute function

  1. Sep 3, 2008 #1
    1. The problem statement, all variables and given/known data

    a. Find the value of x such as fx < gx where fx = |2x -1| and gx = x(2-x)
    b. evaluate [tex] \displaystyle\int^1_0 [gx - fx]\,dx [/tex]

    2. Relevant equations

    3. The attempt at a solution
    For question a I make it into 2 equation to 2x-1 = 2x-x^2 and 1 - 2x = 2x - x^2. I solve it and find the value of x = 1, 0.2679 and 3.73. The problem is, which interval should i choose if there is no graph to be sketched? And how do i get 0.2679 = 2 - sqrt of 3?

    I have no idea for question b.

    Thanks :)
  2. jcsd
  3. Sep 3, 2008 #2


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    Homework Helper

    You should always sketch a graph, whether they ask you to or not. It will make your life much easier. |2x-1|=2x-1 for x>=1/2 and |2x-1|=1-2x for x<=1/2. Work separately on those two intervals. There are only two points of intersection. And to get 0.2679...=2-sqrt(3) exactly, use the quadratic formula to solve the equation. Don't just put it into a calculator and get the numerical results.
  4. Sep 3, 2008 #3


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    Problem 1:
    I haven't checked your numbers yet, but you shouldn't find three solutions, only 2. Once you have found the values of [tex] x [/tex] where [tex] f(x) = g(x) [/tex], those are the only places the two functions can be equal. Call the two locations of equality [tex] a, b [/tex], and for purposes of my notes assume that [tex] a < b [tex].

    Pick three numbers [tex] x_1 < a [/tex], [tex] a < x_2 < b [/tex], and [tex] b < x_3 [/tex]. Compare the function values at each [tex] x [/tex]. If [tex] f < g [/tex] at your choice, it will be for all other values in that interval. (The same if [tex] g > f [/tex]).

    For problem 2: I assume you want

    \int_0^1 |g(x) - f(x)| \, dx

    To do this you'll need to split this integral into pieces, depending on where [tex] g(x) > f(x) [/tex] and [tex] g(x) < f(x) [/tex]. But that's exactly why you solved problem 1.
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