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Trying to find limit involving continuity concept

  1. Jan 27, 2013 #1
    1. The problem statement, all variables and given/known data
    Find c such that the function f(x) { x^2-9 while x≤ c and 6x-18 x > c } is continuous everywhere.


    2. Relevant equations
    Given above. Basic algebra.


    3. The attempt at a solution

    I made a number line. Showing that x^2-9 is approaching from the left side and 6x-18 is approaching from the right side given the designation of the inequality.

    If there were numbers, I could easily do it, but the "c" is throwing me off.

    An attempt to a similar problem with numbers would simply be chug and plug and knowing which side it comes from left or right, and if there was a variable "a" in there as well you can substitute.

    I have no idea how to start the problem, would it become c^2-9 and 6c-18? and try to solve for c?
     
  2. jcsd
  3. Jan 27, 2013 #2
    ##f(x)=\begin{cases} x^2-9 & \mbox{if } x \leq c\\ 6x-18 & \mbox{if } x>c\end{cases}##

    For ##f(x)## to be continuous at ##c##, you must satisfy
    ##\displaystyle\lim_{x\to c^{+}} f(x) = \displaystyle\lim_{x\to c^{-}} f(x)##

    That is,
    ##\displaystyle\lim_{x\to c^{+}} x^2-9 = \displaystyle\lim_{x\to c^{-}} 6x-18##

    That's basically what (it seems) you've done already, but this is the reasoning behind the step you've taken.
     
  4. Jan 27, 2013 #3

    SammyS

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    IMO: It's more important to know why a method works than to know a method without understanding it.

    So, here are some questions:

    What is necessary in order that f(x) be continuous at x = c ?

    What is [itex]\displaystyle \lim_{x\to c^-}\,x^2-9\ ?[/itex]

    What is [itex]\displaystyle \lim_{x\to c^+}\,6x-18\ ?[/itex]
     
  5. Jan 27, 2013 #4
    But I don't know how to move on from there.. I end up getting something like x/x = √-3
     
  6. Jan 27, 2013 #5
    do you have to set the two limits equal?
     
  7. Jan 27, 2013 #6

    SammyS

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    If you solve c2-9 =6c-18 for c,

    how do you get x/x = √-3 ?

    There is no x in c2-9 =6c-18 , and x/x = 1 .
     
  8. Jan 27, 2013 #7

    SammyS

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    Yes, which should be clear if you answer the question about the function being continuous at x = c .
     
  9. Jan 27, 2013 #8
    I put x because the other poster. But, essentially wouldn't it be:

    c^2-9 = 6c-18
    c^2=6c-9
    c^2/c = -3
    ..idk what to do ... if c was a number in the inequality then i could easily do this problem

    alright hold on..
     
  10. Jan 27, 2013 #9
    bump

    nvm got it.

    Since f is continuous everywhere, it is continuous at x = c.
    So the left-hand limit must equal the right-hand limit. Thus:
    c^2 - 9 = 6c - 18
    c^2 - 6c + 9 = 0
    (c - 3)^2 = 0
    c = 3

    thank you!
     
    Last edited: Jan 27, 2013
  11. Jan 27, 2013 #10
    Yes, that's what I was suggesting.
     
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