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Showing continuity

  1. Oct 11, 2004 #1
    let f be a function defined by f(x)=4-x^2 when x=<1 and k+x when x>1
    What value of k will f be continuous at x=1?

    I know the answer is k=2, however, I dont know how to show to correct work. I got 2 when I sketched a graph of 4-x^2 and plugged in some numbers but I dont know how to show it algebraically/using an actual method.
     
  2. jcsd
  3. Oct 11, 2004 #2

    NateTG

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    Are you familiar with epsilon-delta proofs of continuity?
     
  4. Oct 11, 2004 #3

    BobG

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    To be continous, the limit has to exist (in this case, you have two functions on either side of 1 - they should both have the same limit as they approach 1) and f(x) has to equal the limit.

    Find the limit of the first function as x approaches 1 from the left.

    [tex]
    \lim_{x\rightarrow 1^-} 4-x^2
    [/tex]

    The limit of the function, f(x)=k+1, as x approaches 1 from the right must equal the limit of the first function.

    [tex]
    \lim_{x\rightarrow 1^+} k+1 = \lim_{x\rightarrow 1^-} 4-x^2
    [/tex]

    Substitute the value for k that makes this true.

    In other words,

    [tex]
    \lim_{x\rightarrow 1^+} k+1 = \lim_{x\rightarrow 1^+} k+\lim_{x\rightarrow 1^+}1
    [/tex]
    k and 1 are both constants, so the limit of k = k and the limit of 1 = 1.
    [tex]
    \lim_{x\rightarrow 1^+} k+\lim_{x\rightarrow 1^+}1 = k+1 = 3
    [/tex]

    so, subtracting 1 from both sides, k=2
     
    Last edited: Oct 11, 2004
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