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LIMITS, continuity piece defined function

  1. Sep 21, 2008 #1
    The question asks to find a value for a and b that makes f continuous everywhere.

    [tex]\frac{x - 4}{x-2}[/tex] , where x<2

    ax2 - bx + 3 , where 2<x<3

    2x - a +b , where x > or = 3

    I know that in order for a function to be continuous the limit as x approaches 2 must be equal from the left and right and f(2) must be defined. I don't know how I can possibly make it continuous if x never is equal to 2, then f(x) would not be possible and it would be a removable discontinuity.
    There's no answer available at the back of my text or in the solutions manual so it seems a bit fishy to me
    Any help solving this would be great (if there even is a solution)
  2. jcsd
  3. Sep 21, 2008 #2
    In order for a function to be continuous as a point p, the function must be defined at f(p) and the limit of f(x) as x approaches p must be equal to f(p).

    In order for this function we are considering here to be continuous, one of the conditions is that it must be continuous at x = 2. Also notice that when x approaches 2, the function increases beyond all positive bounds. This should help you out.
    Last edited: Sep 21, 2008
  4. Sep 22, 2008 #3
    Most likely they meant make f continuous everywhere on its domain, as x=2 is not part of its domain. Then your only concern is to make sure the line and the quadratic meet at their endpoints.
  5. Sep 22, 2008 #4


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    Homework Helper

    One more point and one question. The function is

    f(x) = \begin{cases}
    \frac{x-4}{x-2} & \text{ if } x < 2\\
    ax^2 - bx + 3 & \text{ if } 2 < x < 3\\
    2x-a + b & \text{ if } x \ge 3

    As slider142 mentioned, part of the work is going to involve making sure that the quadratic and the linear portions agree when you look at the limits at [tex] x = 3 [/tex]. However, this is only one condition, and you have two unknown constants, [tex] a, b [/tex]. The other condition will involve the right-hand limit of the quadratic at [tex] x = 2 [/tex], so you will
    need to decide on how

    \lim_{x \to 2^+} f(x)

    should bd defined, and this will give you the second condition on [tex] a, b [/tex].

    Now for my question: Is the end of the linear portion [tex] -a + b[/tex] or is it [tex] -(a+b) [/tex]? It doesn't make any difference for us, but it can for your answer.
  6. Sep 22, 2008 #5


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    Staff Emeritus
    Science Advisor

    Are you sure you have copied the problem correctly? There is NO way to make the function given continous at x= 2 because its limit, as x approaches 2 from below, does not exist.

    If, however, the function were defined as (x2- 4)/(x- 2) for x less than 2, then it would be possible.

    If, in fact, the problem is to find a and b such that
    [tex]f(x) = \begin{cases}\frac{x^2-4}{x-2} & \text{ if } x < 2\\ax^2 - bx + 3 & \text{ if } 2 < x < 3\\2x-a + b & \text{ if } x \ge 3\end{cases}[/tex]
    then, because, for x not equal to 2, (x2- 4)/(x-2)= (x-2)(x+2)/(x-2), which, for x not equal to 2, is just x+ 2 and has limit 4, from below, at x= 2.

    Now, you want to find a and b such that the limit of ax2- bx+ 3, as x approaches 2 from above, is 4. That's easy because that is a polynomial and so its limit is 4a- 2b+ 3. You must have 4a- 2b+ 3= 4.

    As x approaches 3, from below, you have 9a- 3b+ 3 and, from above, again because 2x- a+ b is a polynomial, the limit is 6-a+ b.

    Assuming that the problem really had (x2-4)/(x-2) rather than(x-4)/(x-2), you must solve the two equations 4a- 2b+ 3= 4 and 9a- 3b+ 3= 6- a+ b for a and b.

    If the problem really is what you say, then there is no solution!
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