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Finding y=f(x)

  1. Apr 29, 2015 #1
    1. The problem statement, all variables and given/known data
    Given two curves y=f(x) passing through (0,1) and ##g(x)=\int\limits_{-\infty}^xf(t)dt## passing through (0,1/n). The tangents drawn to both curves at the points with equal abscissae intersect on the x-axis. Find y=f(x).

    2. Relevant equations
    None

    3. The attempt at a solution
    g(0)=##\int\limits_{-\infty}^0f(x)dx##=1/n
    let the abscissae be x.
    The tangent to y=f(x) is y=xf'(x)+c
    Can I directly differentiate g(x) to get slope?
     
  2. jcsd
  3. Apr 29, 2015 #2

    HallsofIvy

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    By the "fundamental theorem of Calculus", [itex]g'(x)= f(x)[/itex]
     
  4. Apr 29, 2015 #3
    What about the lower limit? So differentiating ##\int\limits_a^xf(t)dt## with respect to x gives the same value for any a?
     
  5. Apr 29, 2015 #4
    Ok. Let integral of f(t) from a to x be F(x)-F(a). Since F(a) is a constant, g'(x) is f(x). Thank you for helping.
     
  6. Apr 29, 2015 #5

    HallsofIvy

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    Yes. For any a and b, [itex]\int_a^x f(x)dx= \int_b^x f(x)dx+ \int_a^b f(x)dx[/itex] and [itex]\int_a^b f(x)dx[/itex] is a constant.
     
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