1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook