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Removable continuity

  1. Dec 17, 2009 #1
    This is from my old exam

    f(x) = for x <1 (x-1)^2
    for 1 <= x <= 4 ax+b find a and b so that fx is continuous for all x
    for x <4 sqrt (2x+1)

    so i guess i start evaluating some limit.

    since the ax+b is define everywhere b/w x = 1 and x = 4, i guess i would use the third function f(x) = sqrt(2x+1) for limit goes to 4

    f(x) = limit (x goes to 4) sqrt(s2+1) = 3
    then i think i would let a = 3 ????
    3x+b = ???? what is y then????
     
  2. jcsd
  3. Dec 17, 2009 #2
    Where might this function fail to be continuous?
     
  4. Dec 17, 2009 #3
    x = 4 and x = 1
     
  5. Dec 17, 2009 #4
    Yes, so what does it mean for this function to be continuous at x = 1 and x = 4?
     
  6. Dec 17, 2009 #5
    oh so the ax +b = y
    since a = 3, for x = 4, y use the third equation which gives 9
    3x+b = 9
    am i correct
     
  7. Dec 17, 2009 #6
    Why does a = 3? The definition of continuity is: f is continuous at c if [tex]\lim_{ x \to c } f(x) = f(c)[/tex].
     
  8. Dec 17, 2009 #7
    Shouldn't it be :

    for x > 4 : sqrt(2x+1)
     
  9. Dec 18, 2009 #8

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Yes, as jeques says, you must mean that [itex]f(x)= \sqrt{2x+1}[/itex] for x> 4.

    What is the limit of f as x goes to 1 from below?
    What is the limit of f as x goes to 1 from above?

    For f to be continuous there, those two limits must be the same.

    What is the limit of f as x goes to 4 from below?
    What is the limit of f as x goes to 4 from above?

    For f to be continous there, those two limits must be the same.

    You cannot determine a or b from either of those alone. Each gives an equation in both a and b.
     
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