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Homework Help: Making a piecewise function continuous everywhere

  1. Oct 25, 2014 #1
    1. The problem statement, all variables and given/known data

    Given the piecewise function

    f(x) = \left\{
    \frac{(2-x)^2-p}{x} &: x < q\\
    r(x+6) &: q \leq x <2 \\
    x^3-p &: x \geq 2

    Find the values of p,q,r such that f(x) is continuous everywhere and f(2) = p

    3. The attempt at a solution
    Since f(2) = p,
    f(2) = (2)^3 - p = p\\

    If f(x) is continuous everywhere, then

    \lim_{x \to 2} f(x) = f(2)

    But f(2)=p, and r(x+6) is always smooth, hence

    r(2+6) = 4 \\
    r = \frac{1}{4}

    My difficulty is with q. Again from (assumed) continuity, at x=q,

    \lim_{x \to q} f(x) = f(q) \\
    \lim_{x \to q} \frac{(2-x)^2-4}{x} = \frac{1}{2}(q+6) \\
    \lim_{x \to q} x-4 = \frac{1}{2}(q+6) \\
    2q-8 = q+6 \\

    but from the original function, q <2. The answer given is supposedly q=0, but I do not see then how the function could be continuous.

    Are my calculations wrong? I tried assuming that q =0, hence instead of factorising the limit I used l'hopital but no dice.

    I am including the original picture as a comparison.
  2. jcsd
  3. Oct 25, 2014 #2


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    There's a bit of a error there, I'm sorry to say.
  4. Oct 25, 2014 #3
    My mistake. I wrote that incorrectly, it should be [itex] r=\frac{1}{2}[/itex].

    However in the following working, I indeed used [itex] r=\frac{1}{2}[/itex].
  5. Oct 25, 2014 #4


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    You are correct. In order that f(2)= 2, p must be 4 and in order that the function be continuous at x= 2, r must be 1/2 (you say "1/4" but I presume that is a typo since you later use "1/2"). But in that case, there is NO value of q, less than 2, such that [itex][(2- q)^2- 4]/q[/itex] is equal to (1/2)(q+ 6) so there is no value of q, less than 2, such that f(x) is continuous at x= q.
    Last edited by a moderator: Oct 25, 2014
  6. Oct 25, 2014 #5


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    The question looks wrong. Also, it doesn't cover the case where x = 0, so I assume there is a typo somewhere.
  7. Oct 25, 2014 #6
    Thanks for the replies. I'll just chalk this one up as one of those incorrect questions.
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