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Proving a function is continuous

  1. Oct 21, 2005 #1
    I am working to prove that this function is continuous at [itex] x = 2 [/itex]

    [tex] {f(x) = 9x - 7} [/tex]

    To do this I know that I have to show that [itex] \vert f(x) – f(a) \vert < \epsilon [/itex] and that [itex] \vert x - a < \delta \vert [/itex]

    I tried to come up with a relationship between [itex] \vert x - 2 \vert [/itex] and [itex] \epsilon [/itex] so I could get an appropriate number to choose for [itex] \delta [/itex]

    This is as far as I got

    [tex] \vert {f(x) – f(a)} \vert < \epsilon [/tex]
    [tex] \vert {9x – 7} \vert < \epsilon [/tex]

    I’m stuck. All of the examples the text shows give equations where it is easy to factor out the [itex] \vert {x - a} \vert [/itex] term.

    A push in the right direction would be appreciated.
     
    Last edited: Oct 21, 2005
  2. jcsd
  3. Oct 21, 2005 #2

    quasar987

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    Wow. First of all, I suggest you use the more conventionnal sign "-" for minus, instead of 8211; :wink:.

    To show a function is continuous at 'a', you must how that the limit as x approaches a is f(a). Here, a = 2 and f(2) = 9(2)-7 = 11. So given e>0, we must find 'd' such that 0<|x-2|<d ==> |9x-7 - 11|=|9x-18|=9|x-2|<e.

    Mmmh.
     
  4. Oct 22, 2005 #3
    Thanks. Latex was being weird yesterday. I'm not sure why it put the 8211; in there in place of some - but not all. Weird.
     
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