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Continuity question

  1. Aug 17, 2013 #1
    1. The problem statement, all variables and given/known data

    For which values of a E ℝ, is the function given by

    f(x) = pieceise function
    x^2+4x-4, x<a
    cos((x-a)/2) , x ≥ a.

    continuous at x=a


    2. Relevant equations

    I'm getting stuck on the algebra part to be honest.

    3. The attempt at a solution
    lim x→a f(x)= f(a) to be continuous

    lim x→a- = x^2+4x-4 and this must be equal to lim x→a+ = cos((x-a)/2)

    I think its 1 because cos(1-1/2)=1 and 1+4-4=1

    I'm just confused on the working out part, how to I algebraically manipulate the equations to show this? or am I completely wrong.

    thanks in advance.
     
  2. jcsd
  3. Aug 17, 2013 #2

    vela

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    You seem to have the basic idea. You need
    $$\lim_{x \to a^-} x^2 + 4x - 4 = \lim_{x \to a^+} \cos\frac{x-a}{2}.$$ So what are the two sides equal to?
     
  4. Aug 17, 2013 #3
    Thanks for the reply, I'm guessing the value of (a) in the equation has to be 1 for both sides of the equation to equal 1.
    I'm almost certain a=1 for the function to be continuous, just having a hard time showing the working out besides me just putting a=1 into the limit equation.

    Thanks again.
     
  5. Aug 17, 2013 #4

    vela

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    Don't plug any value in for a. Just work out what the limits equal in terms of a.
     
  6. Aug 17, 2013 #5
    Thank you for your replies, but I literally have no idea of how to solve a, i think its the cosine function that is throwing me off. Any tips for trying to solve it?
     
  7. Aug 17, 2013 #6

    vela

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    Yes, do what I've already suggested twice.
     
  8. Aug 17, 2013 #7

    pasmith

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    You are given that if [itex]x \geq a[/itex] then [itex]f(x) = \cos((x - a)/2)[/itex]. Therefore [itex]f(a) = \cos((a-a)/2) = \cos(0) = 1[/itex].
     
  9. Aug 17, 2013 #8

    Evo

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    Quoted for the record.

    Please remember to ALWAYS quote the OP's posts.
     
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