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Function range

  1. Apr 25, 2016 #1
    1. The problem statement, all variables and given/known data
    For real ##x##, the function ##\dfrac{(x-a)(x-b)}{x-c}## will assume all real values provided

    (A) ##a > b > c##
    (B) ##a < b < c##
    (C) ##a > c > b##
    (D) ##a < c < b##
    2. Relevant equations


    3. The attempt at a solution
    I'm really stuck on this. I know that ##x = c## is a vertical asymptote. I also know that I could substitute in values for ##a, b## and ##c## and graph it, but I don't want to do that. Again, I'm really not sure how to go about it. Please help.
     
  2. jcsd
  3. Apr 25, 2016 #2

    SammyS

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    Where are the x-intercepts?
     
  4. Apr 26, 2016 #3
    a and b.
     
  5. Apr 26, 2016 #4

    PeroK

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    Isn't the question will assume positive real values?
     
  6. Apr 26, 2016 #5

    SammyS

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    No. It makes sense as written.

    It's saying that the range should include all real numbers, (-∞, ∞) .


    @erisedk ,
    What is the is the behavior of the function as x → ± ∞ ?
     
    Last edited: Apr 26, 2016
  7. Apr 26, 2016 #6
    As x → +∞ f(x) → ∞
    and as x → -∞ f(x) → -∞
     
  8. Apr 26, 2016 #7

    SammyS

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    Good.

    Now starting with a negative at the left, in what order do you want two intercepts and a vertical asymptote ?
     
  9. Apr 26, 2016 #8
    1. There are multiple answers to this problem.

    2. I understand what you're saying, we'll get these sort of hyperbola type graphs on both sides of the vertical asymptote, and on opposite sides of an asymptotes, the infinities that a function approaches are opposite (i.e + and - ∞)

    3. I expanded the function and differentiated it to find the points of maxima/minima. I got a quadratic for the roots of f'(x) that has the discriminant ##4(c^2 - ac - bc + ab)##. Now, I'm not sure if I'm supposed to figure out what sign this is for the number of roots, because it seems complicated to work out for different sets of ##a, b## and ##c##.

    4. Other than that, assuming that the discriminant is positive(for no reason, just assuming) so two roots, we'll have to figure out the maximum value of the function on the left side of ##c## and minimum value to the right of ##c## and compare them. Finding the roots of the quadratic for the maxima and minima seems rather arduous.

    I don't really know what to do now.
     
  10. Apr 26, 2016 #9

    PeroK

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    The substitution ##y = x-c## might be a good start, with the aim of transforming to as simple a function as possible.
     
  11. Apr 26, 2016 #10
    This is taking too much time, I suppose I should abandon the question at this point, especially since I've at least understood the key ideas.
     
  12. Apr 26, 2016 #11
    Thank you everyone for your help!
     
  13. Apr 26, 2016 #12

    SammyS

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    Use a sign graph.

    There are three critical points: x = a, x = b, and x = c. The rational function in question will change sign at each of these (none are equal in the choices) as x increases.

    Trace the function from left to right (increasing x.) There are three possibilities. What does the graph look like for each?

    1. x - intercept, x - intercept, vertical asymptote
    2. x - intercept, vertical asymptote, x - intercept
    3. vertical asymptote, intercept, x - intercept
     
  14. Apr 27, 2016 #13
    Yeah, I did that. It doesn't solve the complete problem because there still is the question of whether the maxima on the left and minima on the right of the asymptote have some space between them. Furthermore, in case 2, we wouldn't get maximas or minimas, we'd get inflection points(or perhaps not even inflection points, not exactly sure there, that would happen if the discriminant of the derivative is negative).
     
  15. Apr 27, 2016 #14

    SammyS

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    In the first and third cases, you can't be sure regarding there being a gap in the range, without more analysis, which I know you're trying to avoid.

    However you can be sure that in case 2, the entire range is covered.

    There are not inflection points. That may be difficult to see. The second derivative is messy, but we are in the pre-calculus forum so ...
     
  16. Apr 27, 2016 #15
    I think I figured it out!!!!
    ##y = \dfrac{(x-a)(x-b)}{x-c} ##
    ##(x - c) y = x^2 - (a+b)x + ab##
    ## x^2 - (a+b+y)x + ab + cy = 0##
    Since x is real
    D ≥ 0
    ## (a+b+y)^2 - 4(ab + cy) ≥ 0##
    ##y^2 + 2y(a+b-2c) + (a-b)^2 ≥ 0##
    So, now discriminant of the quadratic in y should be negative as this quadratic is always positive
    ##4(a+b-2c)^2 - 4(a-b)^2 ≤ 0##
    ## (a + b - 2c + a - b)(a+b -2c - a +b)≤0##
    ##4(a-c)(b-c)≤0##
    Ignoring the equality in ≤
    ## a - c<0 ## and ## b - c > 0## OR ## a - c > 0 ## and ## b -c < 0##
    ## a < c < b ## OR ## a > c > b##
     
  17. Apr 27, 2016 #16

    SammyS

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    That's quite interesting !

    A few picky issues remain such as the = of the ≥ , etc.
     
  18. Apr 27, 2016 #17
    Yeah, I just chose to ignore it.
     
  19. Apr 27, 2016 #18

    SammyS

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    You could analyze this function in terms of its slant asymptote.

    Using long division or some other method, such as Mark44's decomposition, we get that
    ## \dfrac{(x-a)(x-b)}{x-c} = x + c-(a+b)+(c-a)(c-b)\dfrac{1}{x-c}##

    ## \quad = x + G+H\dfrac1{x-c}##​
    for constants, G and H. Which have values:
    ##H=(c-a)(c-b)\ ##: Very important
    ##G=c-(a+b)\ ##: Not so important for this​

    We can view our function as the sum of two basic functions; a linear function with slope of 1, and a reciprocal function, shifted horizontally, and having a coefficient, H, which could be either positive or negative. It's the sign of H that is crucial for the problem in this thread.
     
    Last edited: Apr 27, 2016
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