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Values of M so 2 curves don't intersect.

  1. Jan 11, 2013 #1


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    1. The problem statement, all variables and given/known data
    2 curves f(x) and g(x) don't intersect, find the range of values of m can be.

    2. Relevant equations
    $$f(x)=3x^2 - 2$$
    $$g(x) = mx-5$$

    3. The attempt at a solution

    Could I work out the 2 values for m that mean g(x) is a tangent to f(x) at some point and then the range will be between them?

    PS. I've already solved it via another approach but I was just wondering if this would work as it was my first initial idea on solving it.
    Last edited: Jan 11, 2013
  2. jcsd
  3. Jan 11, 2013 #2


    Staff: Mentor

    You're on the right track, the limiting values for m making the g(x) line a tangent to f(x) is correct.

    have you plotted 3x^2 - 2 ? that would give you a hint of what the values of m are.

    for g(x) you know one one point for certain: the y-intercept
  4. Jan 11, 2013 #3


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    I tried it again and I must have made a mistake when I tried this the first time as I definitely didn't get this.

    For g(x) to be a tangent of f(x) their gradients must be the same:
    $$f'(x)=6x \\
    g(x)=6x(x)-5 \\

    And they also must intersect eachother:$$
    Then just work out the values of m for the values of x:$$

  5. Jan 11, 2013 #4


    Staff: Mentor

    and you can check this by plotting the two lines and the original f(x).
  6. Jan 11, 2013 #5


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    I'm not sure you should focus on the slope. The two graphs will intersect when [itex]3x^2- 2= mx- 5[/itex] which is the same as [itex]3x^2- mx+ 3= 0[/itex].

    The "discriminant" of that is [tex](-m)^2- 4(3)(3)= m^2- 36= 0[/tex]. That is a parabola that is negative for x between -6 and 6. Since a quadratic equation has no real solutions if the discriminant is negative, the graphs do not intersect for m between -6 and 6.
  7. Jan 11, 2013 #6


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    That was the method I used in class to solve it after I tried working out the tangents but made a mistake copying something.

    Is there any advantage to that method, ie. would the tangents not work for some similar problems?
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