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BasiC MatH

  1. Sep 20, 2005 #1
    Alright this is confusing me a bit...

    Find k so that the following function is continuous on any interval?
    [tex] f(x)=kx [/tex] if [tex]0 \leq x \leq 2 [/tex] and [tex]f(x)=5x^2 [/tex] if [tex] 2\leq x [/tex]

    Alright I know the answer is 10, but I don't understand how you get there

    I mean I just doubled 5, because I took it off an example in the book that had answer

    but i would like be able to do eveutnally do

    If possible choose k so that the following function is continusous on any function

    [tex]f(z) = \left\{ \begin{array}{rcl}
    \frac{5x^3-10x^2}{x-2} & \mbox{ }
    & x\neq2 \\
    k & \mbox{ } & x=2
  2. jcsd
  3. Sep 20, 2005 #2
    You want the function to be continuous at x=2. Therefore, k*2 = 5*2^2. Solve for k
  4. Sep 21, 2005 #3
    So would tthe answer to the second one be false

  5. Sep 21, 2005 #4

    No the answer is true if you make k=20.

    You see, you can factor out 5x^2 from the top, and then the (x-2) will cancel. If you then sub in x=2 you get 20. Thus if you make k=20, the function becomes continous.
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