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Continuity Question

  1. Dec 26, 2011 #1
    A function f is defined as follows:

    f(x) = 2cos(x) if x≤c,
    = ax^2 + b if x > c .

    Where a,b, and c are constants. If b and c are given. find all values of a for which f is continous at the point x = c

    a = (2cos(c) - b)/c^2 if c ≠ 0 ; if c = 0 there is no solution unless b = 2.

    I dont understand how if c = 0 and b = 2 there is a solution at c = 0

    For instance... (2cos(c) - 2)/c^2 = (2/c)( Cos(c) - 1 )/c = 2/0*0 as c → 0 which is still a discontinuity.
  2. jcsd
  3. Dec 26, 2011 #2


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    In order for f(x) to be continuous at x = 0, f(0) must be defined and:

    [tex]\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)[/tex]

    Looking at the limit part of this, calculate [itex]\lim_{x \to 0^-} f(x)[/itex]. What do you get? Now look at [itex]\lim_{x \to 0^+} f(x)[/itex]. What is the only value of b that we can have to get the right limit?
  4. Dec 26, 2011 #3


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    Generalizing on what gb7nash said: f is continuous at at x=c if and only if
    [itex]\displaystyle \lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)[/itex]​
    Of course since cos(x) is continuous on ℝ, [itex]\displaystyle \lim_{x \to c^-}f(x)=f(c)[/itex] is true for all real c.

    [itex]\displaystyle \lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x)[/itex] leads directly to the equation
    [itex]\displaystyle 2\cos(c)=ac^2+b[/itex]​
    Use that last equation to solve the two cases:
    (1) c = 0

    (2) c ≠ 0​
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