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Range Question

  1. Jul 25, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the range of

    y = (3x-1)/(2x^2 + x - 6)

    I was going to take the derivative of the numerator and denomenator and then use the quotent rule to find the derivative of the function and find the critical values but I ran into the imaginary number during the proess and am woundering if I did nothing wrong... if I did nothing wrong then what exactly does this mean?

    2. Relevant equations



    3. The attempt at a solution

    y = (3x-1)/(2x^2 + x - 6) = h(x)
    f(x) = (3x-1)
    g(x)= 2x^2 + x - 6

    dy/dx f(x) = ( 3(x + a) -1 -3x + 1 ) / a = (3x + 3a -1 - 3x +1)/a = (3a)/a = 3

    dy/dx f(g) = ( 2(x + a)^2 + x + a - 6 -2x^2 -x +6)/a = (2(x^2 + a^2 + 2ax) + a -2x^2)/a = (2x^2 + 2a^2 + 4ax + a -2x^2)/a = (2a^2 + 4ax +a)/a = 2a + 4x + 1 = 4x + 1

    dy/dx h(x) = ( (2x^2 + x -6)3-(3x - 1)(4x +1) )/ (2x^2 + x - 6)^2 = ( 6x^2 + 3x - 18 -(12x^2 + 3x - 4x -1) )/(2x^2 + x - 6)^2 = ( 6x^2 + 3x - 18 -(12x^2 -x - 1) )/(2x^2 + x - 6)^2 = ( 6x^2 + 3x - 18 -12x^2 + x + 1)/(2x^2 + x - 6)^2 = ( -6x^2 + 4x -17)/(2x^2 + x - 6)^2

    At this point I proceded by setting the numerator to zero and solving which is were I got stuck at least I think I don't see what I have done any where even while typing this thing up charecter by charecter, normally this is a good way to catch mistakes but I have found nothing wrong with my work...

    -6x^2 + 4x -17 = 0

    (-4 +/- sqrt( 4^2 - 4(-6)(-17) ) )/(2(-6)) = -4/-12 +/- (1/-12)sqrt( 4 - (-6)(-17) ) = 1/3 +/- (-1/6)sqrt(2(2 - (-3)(-17)) = 1/3 +/- (-1/6)sqrt(2(2-51)) = 1/3 +/- (-1/6)sqrt(2(-47) = 1/3 +/- (-7/6)sqrt(-2) = 1/3 +/- (-7i/6)sqrt(2)

    Is this correct or did I just simply do something wrong here? Don't see were or how...
     
  2. jcsd
  3. Jul 25, 2010 #2

    Mentallic

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    Homework Helper

    It's really hard to follow, especially when you're taking the derivative.

    Simply use the quotient rule?

    Anyway, if you're running into imaginary numbers when setting the derivative to zero means the derivative doesn't equal zero anywhere which means the function doesn't have any turning points!
     
  4. Jul 25, 2010 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    I cannot see any reason to use Calculus at all here. The numerator is 0 only at x= 1/3, the denominator is 0 at x= -2 and x= 3/2. There are vertical asymptotes at x= -2 and x= 3/2. The horizontal asympote is y= 0. For x< -2, each factor is negative and, since there are 3 factors, the entire fraction is negative. The value of y goes to negative infinity as x goes to -2 from below.

    For x between -2 and 1/3, x+ 2 is positive but the other two factors are still negative so the fraction is positive. y goes to positive infinity as x goes to -2 from above. For x between 1/3 and 3/2, x+ 2 and 3x- 1 are both positive while 2x- 3 is still negative. The fraction is negative and y goes to negative infinity as x goes to 3/2 from above.

    That alone is enough to show that y takes on all values between -2 and 3/2 and give the range.
     
  5. Jul 25, 2010 #4
    SO I put the equation into a derivative calculator and got the same results so there are no turning points? I'm sure that there has to be on no? Thye are there they are just complex correct?
     
  6. Jul 25, 2010 #5
    Well at x = -2 has a well defined value of -.28 there is no asymptote at x=-2 no? and at 3/2 it's about .11570248 there is no vertical asymptote at this value???

    Using the calculate max on the graphing calculator I get a turning point at x = .999999894 with a local max of 1/8

    so the range is (1/8, -infinity) now how do I do this without a graphing calculator and cheating setting the derivitive to zero gave me imaginary number... please help me as I even put the funciton into a derivitive calculator and got the same derivitive I found... What do I do
     
    Last edited: Jul 25, 2010
  7. Jul 25, 2010 #6
    It actually crosses and equals one at x=1/3 and becomes positive... the range is not (-infinity, zero)...
     
  8. Jul 25, 2010 #7
    Well, it doesn't have a well defined value at -2, since the denominator is zero. It is not even a removable singularity, since the numerator is nonzero there. Where did the -.28 came from?!
     
  9. Jul 25, 2010 #8
    lol no idea sorry about taht but ya can you help me find the range pleaes?
     
  10. Jul 25, 2010 #9
    As x goes to -2, f(x) goes to...?
    As x goes to 1/3, f(x) goes to...?
    This facts + continuity between -2 and 1/3 give the range immediately.
    Good luck!
     
  11. Jul 25, 2010 #10
    hmmm so the way I solved using calculus gave me a value that includes I the equation turns around at infinity no? So does that suggest that infinity can be calculated?
     
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