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Calculus Problem

  1. Nov 14, 2004 #1
    f(x) = (ax+b)/(x-c) has the following properties

    i) the graph of f is symmetric with respect to the y-axis
    ii) lim x->2^+ f(x) = + infinite
    iii) f^prime (1) = -2

    a) Determine the values of a, b and c.

    i think c = 2

    derivative of f(x) = (-ca - b)/(x-c)^2

    so f^prime(1) = -2

    -ca-b = (1-c)^2 * -2

    if c = 2 then b = -2a + 2

    I'm stuck from here. I have no idea how to do this. I tried using this equation
    (ax+b)/(x-2) = (-ax + b)/(-x-2)

    since symmetric over y axis and i plugged in b, but it turned out to be useless. Please help. thanks
  2. jcsd
  3. Nov 14, 2004 #2
    The equation will be competely determined if you have 3 equations for these three unknowns (a,b,c).

    Your equations should be:

    1) f(x) = f(-x) for the first condition (an even function)
    2) as x goes to 2^+(f(x)) the funciton blows up. the only way the function can blow up is if the numerator goes to infinity or the denominator goes to zero. The numerator will still be finite in this limit so the denominator has to go to zero. so you should have 2^f(x) -c =0 take the natural log of both sides and you should get f(x) = ln c / ln 2 (look up your identities in a math table if this doesn't seem familar)
    3) f'(x=1) = -2
  4. Nov 15, 2004 #3


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    tiger-striped-cat: I suspect that "ii) lim x->2^+" was meant as "limit as x goes to 2 from above, not 2f(x). In that case, you just have to note that the denominator, x-c, must be 0 when x= 2: 2- c= 0. c= 2 as in the initial post.

    Saying that "the graph of f is symmetric with respect to the y-axis" means, as tiger-striped-cat said, that f(x)= f(-x). Instead of doing that in general, just look at a few values of x: if x= 1, f(1)= (a+b)/(1-2)= (-a+b)/(-1-2)= f(-1); if x= 3, f(3)= (3a+b)/(3-2)= (-3a+b)/(-3-1)= f(-3). You may not even need the derivative!
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