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Math Help - Coordinate Geometry

  1. #1
    Member classicstrings's Avatar
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    Coordinate Geometry

    Hello again! I'm in need of some help with these two parts.

    Find the value(s) of k for which the equation x + 1 + (1/x) - k = 0, x not equal to 0, has no real solutions.

    A second function is defined as g(x) = b/(f(x)), b greater than 0, where f(x) = (x^2 + x + 1)/(x), x not equal to 0. Determine the range of g(x), leaving the answer in terms of b, and clearly explain your reasoning.
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  2. #2
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    For the first one.
    Because x is not zero, multiply through by x: x^2 +(1-k)x +1 =0.
    Make the discriminant less than zero.
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  3. #3
    Member classicstrings's Avatar
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    Thanks for the help, still stuck on the 2nd part.
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  4. #4
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by classicstrings View Post
    Hello again! I'm in need of some help with these two parts.

    Find the value(s) of k for which the equation x + 1 + (1/x) - k = 0, x not equal to 0, has no real solutions.

    A second function is defined as g(x) = b/(f(x)), b greater than 0, where f(x) = (x^2 + x + 1)/(x), x not equal to 0. Determine the range of g(x), leaving the answer in terms of b, and clearly explain your reasoning.
    For the first problem, I would subtract 1/x then multiply both sides by x:
    x(x + 1 - k) = -1
    x^2 + (1 - k)x + 1 = 0

    Using the quadratic equation:
    x = {-(1 - k) +/- sqrt[(1 - k)^2 - 4]}/2

    Now we need to find where this will have no real values, which means we only need to see where sqrt[(1 - k)^2 - 4] does not exist, thus were (1 - k)^2 - 4 < 0.

    (1 - k)^2 - 4 < 0
    (1 - k)^2 < 4
    |1 - k| < 2

    This gives us two sets of problems:
    1 - k < 2
    k > -1

    1 - k > -2
    k < 3

    Therefore, for k is any value in the range: (-1,3), the function above has no real roots.
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  5. #5
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by classicstrings View Post
    Hello again! I'm in need of some help with these two parts.

    Find the value(s) of k for which the equation x + 1 + (1/x) - k = 0, x not equal to 0, has no real solutions.

    A second function is defined as g(x) = b/(f(x)), b greater than 0, where f(x) = (x^2 + x + 1)/(x), x not equal to 0. Determine the range of g(x), leaving the answer in terms of b, and clearly explain your reasoning.
    g(x) = (bx)/(x^2 + x + 1) for x != 0 and b > 0.

    Well, to find it's range, we need to find it max/min values. Take it's derivative and set it equal to 0 and also determine where it's undefined. Determine if it's increasing/decreasing on either side of this critical point and find the value of the function at this critical point. I think I can safely guarantee that range will be (-inf,0)U(0,inf) no matter what b is, but that's something you'll have to check.
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  6. #6
    Member classicstrings's Avatar
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    I get the derivative to be
    EDIT - made error in calculation.

    But the question wants the range in terms of b.
    Last edited by classicstrings; May 3rd 2007 at 08:31 AM.
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  7. #7
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by classicstrings View Post
    I get the derivative to be

    (bx)/((x^2 + x +1)^2)

    But the question wants the range in terms of b.
    g(x) = (bx)/(x^2 + x + 1)
    g'(x) = [b(x^2 + x + 1) - bx(2x + 1)]/(x^2 + x + 1)^2
    g'(x) = [bx^2 + bx + b - 2bx^2 - bx]/(x^2 + x + 1)^2
    g'(x) = [-bx^2 + b]/(x^2 + x + 1)^2

    Let g'(x) = 0:
    -bx^2 + b = 0 --> x^2 = 1 --> x = 1, -1

    Let g'(x) be undefined:
    (x^2 + x + 1)^2 = 0
    x^2 + x + 1 = 0
    This results in imaginary solutions, so we don't need to deal with this.

    Now using the points x = 1, -1 in g(x) we get:
    g(1) = b/3
    g(-1) = -b

    Now, we need to see if g is increasing/decreasing on the intervals (-inf,-1), (-1,0), (0,1), (1,inf).

    In g'(x):
    (-inf,-1) --> g'(x) --> {negative}
    (-1,0) --> g'(x) --> {positive}
    (0,1) --> g'(x) --> {positive}
    (1,inf) --> g'(x) --> {negative}

    x = -1 is a minimum
    x = 1 is a maximum

    My conclusions after this point are wrong.
    ------------------------------------------------------------------------------------

    Therefore, there are two parts to this graph, one is always above g(-1) = b/3, the other is always below g(1) = -b.

    The range is: (-inf,-b) U (b/3,inf)
    Last edited by ecMathGeek; May 3rd 2007 at 09:14 AM. Reason: To indicate the mistake I made.
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  8. #8
    Senior Member ecMathGeek's Avatar
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    It appears I was wrong about my original prediction that the range would be from (-inf,0) U (0,inf).
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  9. #9
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    Hello, classicstrings!

    A second function is defined as: .g(x) .= .b/(f(x)), .b > 0,
    . . where f(x) .= .(x + x + 1)/(x), .x ≠0.

    Determine the range of g(x), leaving the answer in terms of b,
    and clearly explain your reasoning.
    I will assume this is a Calculus problem and derivatives are allowed.


    We have: .g(x) .= .bx/(x + x + 1)

    . . . . . . . . . . . . . . . . . . . .b(x + x + 1) - bx(2x + 1)
    The derivative is: .g'(x) .= .--------------------------------
    . . . . . . . . . . . . . . . . . . . . . . . (x + x + 1)

    . . . . . . . . . . . . . . . . . . . . . . . . -b(x - 1)
    . . which simplifies to: .g'(x) .= .---------------- . = . 0
    . . . . . . . . . . . . . . . . . . . . . . .(x + x + 1)

    . . and has critical values: .x .= .1

    The corresponding y-values are: .y .= .b/3, -b


    We see that the x-axis is a horizontal asymptote.

    Hence: (1, b/3) is a maximum of the graph
    . . and: (-1, -b) is a minimum of the graph.


    The range of the function is: . -b .< .g(x) .< .b/3

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  10. #10
    Member classicstrings's Avatar
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    Thanks Soroban,

    I'm still a bit blurry, how do you realise that y = 0 is an asymptote and deduce that (1,b/3) is the max and (-1,-b) is min?

    -> For f(x) = (x^2 + x + 1)/(x), the asymptotes are y = x +1 and x = 0 right?
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  11. #11
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by classicstrings View Post
    Thanks Soroban,

    I'm still a bit blurry, how do you realise that y = 0 is an asymptote and deduce that (1,b/3) is the max and (-1,-b) is min?

    -> For f(x) = (x^2 + x + 1)/(x), the asymptotes are y = x +1 and x = 0 right?
    From g(x) = bx/(x^2 + x + 1)

    Lim{x->inf} g(x) = 0
    Lim{x->-inf} g(x) = 0

    It's the g(x) function from which we determine the horizontal asymptotes.

    And if you look back at the work I did, even though my conclusion was wrong (I did it quickly just before having to head off to school), I showed that at x = -1, g(x) has a relative minimum, at x = 1, g(x) has a relative maximum.
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