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Math Help - Continuity

  1. #1
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    Continuity

    Hey everyone, I can't figure this problem out:

    Find the constants a and b that will make the following function continuous everywhere.



    f(x)=
    • a cosx if x<0
    • sin \left(\frac{pi}{2}x+3) if 0<=x<=2
    • x^2-x+b if x>2
    Thanks in advance!
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  2. #2
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    Quote Originally Posted by bodel View Post
    Hey everyone, I can't figure this problem out:

    Find the constants a and b that will make the following function continuous everywhere.



    f(x)=
    • a cosx if x<0
    • sin \left(\frac{pi}{2}x+3) if 0<=x<=2
    • x^2-x+b if x>2
    Thanks in advance!
    I assume you are aware of the definition of continuity at a point ...

    \displaystyle \lim_{x \to 0^+} f(x) = \sin(3)<br />

    f(0) = \sin(3)

    therefore \displaystyle \lim_{x \to 0^-} f(x) = \sin(3) = a\cos(0)

    a = ?


    \displaystyle \lim_{x \to 2^-} f(x) = \sin(\pi+3) = -\sin(3)

    f(2) = -\sin(3)

    therefore, \displaystyle \lim_{x \to 2^+} f(x) = 2^2 - 2 + b = -\sin(3)

    b = ?
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  3. #3
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    Thanks skeeter!! I actually got the same answers for both sides of the limit of x=0 and x=2. I was getting confused because I knew they had to equal each other, but I didn't actually set the answers equal and solve for the constant.

    Would a = sin(3) and b = -sin(3)-2
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  4. #4
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    Quote Originally Posted by bodel View Post
    I actually got the same answers for both sides of the limit of x=0 and x=2. Would a = sin(3) and b = -sin(3)-2
    The the value of a is correct.

    b=-\sin(3)-2. WHY?
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  5. #5
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    \displaystyle \lim_{x \to 2^+} f(x) = 2^2 - 2 + b = -\sin(3)

    b = ?

    1. 4-2+b = -sin(3)
    2. 2+b = -sin(3)
    3. b = -sin(3) - 2?

    I'm prone to dumb mistakes so please point them out if you find some.
    Quote Originally Posted by Plato View Post
    The the value of a is correct.

    b=-\sin(3)-2. WHY?
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  6. #6
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    b = -\sin(3) - 2

    works for me.
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