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Find the values of a and b that make f continuous everywhere 
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#1
Sep1112, 05:32 PM

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1. The problem statement, all variables and given/known data
This is a piece wise function of course. f(x) = (x^{2}4) / (x2) if x is less than two. ax^{2}  bx + 1 if x is greater than or equal to 2, or less than three. 4x  a + b if x is greater than or equal to three. 2. Relevant equations 3. The attempt at a solution Alright, I know enough to factor the top of the fist equation and get x+2. That means when x is two, f(x) is four. We can use f(x) in this case because we are making the function continuous. I've gotten as far as plugging in this value in the second equation and getting 4 = a4  b2 + 1 but I don't know what to do from here, or how to get the values of a and b. I think I subtract one from the right and get 3 = a4  2b Now I am definitely stuck. 


#2
Sep1112, 06:26 PM

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4a  2b = 3 Now, what about at x = 3? You want the function to be continuous there, as well, right? What needs to happen for f to be continuous at x = 3? That should give you another equation so that you have a system of two equations in the unknowns a and b. 


#3
Sep1112, 11:41 PM

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So, you have the following: [itex]\displaystyle \lim_{x\to\,2}f(x)=\lim_{x\to\,2}(x+2)=4\ .[/itex]Do something similar at x=3 . 


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