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Homework Help: Need to find where the following function increase/decrease

  1. Aug 4, 2010 #1
    1. The problem statement, all variables and given/known data

    f(x)=(3x^2-6x)/(x^2-4x+3)

    3. The attempt at a solution
    I already have the intercepts: x= 0, x= 2, y= 0
    Horizontal Asymptote: y= 3
    [STRIKE]Vertical Asymptote: x=[/STRIKE]

    Derivative using Quotient Rule:
    f'(x)=(6x-6)/(x^(2)-4x+3)-((3x^(2)-6x)(2x-4))/((x^(2)-4x+3)^(2))

    I'm stuck in the process of the derivative test... got no idea how to perform it :confused:
     
  2. jcsd
  3. Aug 4, 2010 #2

    Mark44

    Staff: Mentor

    That should be
    f'(x)=[(6x-6)(x^2-4x+3) - (3x^2-6x)(2x-4)]/((x^2-4x+3)^2)

    The denominator is never negative, so you should be able to tell where this function is increasing (f'(x) > 0) and decreasing (f'(x) < 0) by determining the intervals where the numerator is positive or negative.

    Multiply the pairs of factors in the numerator and combine like terms. You should end up with a quadratic expression in the numerator that, with any luck, you can factor.



     
  4. Aug 4, 2010 #3
    Ok, if I multiply the pairs of factors and combine like terms I get something like this... correct me if I'm mistaken:

    f'(x)=[tex](-6x3+ 12x2 + 18x - 30)/ (x2- 4x + 3)2[/tex]
     
  5. Aug 4, 2010 #4

    Mark44

    Staff: Mentor

    Can you rewrite that so that it's easier to read? Don't mix [ sup] tags in with LaTeX.

    There shouldn't be an x3 term, so I think you made a mistake.
     
  6. Aug 4, 2010 #5
    (-6x^3 + 12x^2 + 18x + 3) / (x^2 - 4x + 3)^2

    that's what I got.. but I'm not sure
    I'll do the process again to see if I made a mistake
     
  7. Aug 4, 2010 #6
    ok... I checked the process... I found the mistake... now the equation stands like this:

    (-18x^2 + 54x - 42)/(x^2 - 4x + 3)^2
     
  8. Aug 4, 2010 #7

    Mark44

    Staff: Mentor

    That's not what I'm getting. The denominator is fine, but you have mistakes in the numerator.

    The numerator is
    6(x - 1)(x2 - 4x + 3) - 6x(x - 2)(x - 2)
    = 6(x3 - 5x2 + 7x - 3) - 6x(x2 - 4x + 4)

    If you expand those pairs of factors and collect like terms, what do you get?
     
  9. Aug 4, 2010 #8
    umm...
    6x^4 - 36x^3 + 24x^2 + 18x - 18
     
  10. Aug 4, 2010 #9

    Mark44

    Staff: Mentor

    No, that's not even close.
    Starting with 6(x - 1)(x2 - 4x + 3) - 6x(x - 2)(x - 2)
    = 6(x3 - 5x2 + 7x - 3) - 6x(x2 - 4x + 4)

    show me step by step what you're doing.
     
  11. Aug 4, 2010 #10
    Ok, after this, I did the following (hope I get it right now)

    = 6(x^3 - 5x^2 + 7x - 3) - 6x^3 + 24x^2 - 24x
    = 6x^3 - 30x^2 + 42x - 18 - 6x^3 + 24x^2 - 24x
    = -6x^2 -18x - 18
     
  12. Aug 4, 2010 #11

    Mark44

    Staff: Mentor

    That's better, but not perfect. It should be -6x^2 + 18x - 18, which can be factored to -6(x^2 - 3x + 3).

    So f'(x) = [-6(x^2 - 3x + 3)]/[(x^2 - 4x + 3)^2]

    The denominator is always >= 0, so the sign of f'(x) is controlled by -6(x^2 - 3x + 3). When is f'(x) > 0 and when is f'(x) < 0?
     
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