anonymous
  • anonymous
What values of a and b makes the following question continuous?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
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DebbieG
  • DebbieG
To be continuous, you need the limit to exist at, and be = to, f(x) at each point. So for the "first chunk" of the domain, we have: \(f(x)=\dfrac{x^2-4}{x-2}~~~for~x<2\) So you need the limit at x=2 of that expression... can you tell me what that is? Now we'll call that L. The "next chunk" of the domain gives us: \(f(x)=ax^2-bx+1~~~for~2\le x<3\) That's a polynomial so its limit is just = it's value at x=2, so we need L from above to equal that value, e.g, you need: \(f(2)=ax^2-bx+1=L\) Then finally, you have \(f(x)=4x-a+b~~~for~x\ge 3\) Again this is a polynomial, so you just need the value at 3 from the middle chunk to equal the value at 3 here, e.g., you need: \(4(3)-a+b=a3^2-3b+1\) So I'm thinking that gives you (after you find L) a system of 2 equations in 2 unknowns a & b to solve. \(ax^2-bx+1=L\) \(12-a+b=9a-3b+1\)
anonymous
  • anonymous
so for that second chunk would it be 0= 4a-2b+1

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DebbieG
  • DebbieG
I'm not sure what you mean. Did you find L?
anonymous
  • anonymous
I found L to be 0/0 when using limit as x approaches 2 from the negative side
DebbieG
  • DebbieG
That's not a limit. YOu need to reduce, that rational function isn't in lowest form. Then find the limit.
DebbieG
  • DebbieG
I'm having trouble though... the solution is not working out for me, although the method seems sound.
anonymous
  • anonymous
ohhhh right oops :x thank you
DebbieG
  • DebbieG
...trying to figure out what's wrong....
anonymous
  • anonymous
so L is 4
DebbieG
  • DebbieG
yes
anonymous
  • anonymous
so then I would get 4a-2b=3 and then I'm completely lost lol
DebbieG
  • DebbieG
OK, I think I found my error.... Yes, that's right. Now that's one of your equations in 2 unknowns (that's what I should have put down at the bottom, where I re-capped the 2 equations).
DebbieG
  • DebbieG
But DO YOU UNDERSTAND WHY?? :) Do you understand how we came up with THAT equation?
DebbieG
  • DebbieG
The limits must be equal at the "crossover" points, e.g., when we cross from the first "chunk" to the middle "chunk" to the 3rd "chunk", the limits must be =
anonymous
  • anonymous
Yeah because youre using 2 for the first and second equation and 3 for second and third
DebbieG
  • DebbieG
well, yes... we have 2 unknowns, and 2 conditions: limits = at x=2 is one condition limits = at x=3 is the other condition Each condition gives us one equation, hence we get 2 equations in 2 unknowns. :)
anonymous
  • anonymous
right :)
anonymous
  • anonymous
and then do i like solve the second condition
DebbieG
  • DebbieG
Well, you have to solve the SYSTEM. 2 equations, 2 unknowns.
anonymous
  • anonymous
would that be 11= 10a-4b
anonymous
  • anonymous
i mean for the second part with the 3
DebbieG
  • DebbieG
Yes, that looks right
anonymous
  • anonymous
What would I do after this?
anonymous
  • anonymous
In relation of the two equation when using 2 a= 3+2b/4 and when using 3 a= 11+4b/10 should I set them equal to eachother
DebbieG
  • DebbieG
Well, you can't set them equal unless you solve both for one of the variables, first. I think elimination method would work better here. I'm so so sorry but I HAVE to go..... I do know that this method will work because I found a and b and can see from the graphs that they work.
anonymous
  • anonymous
its alrite thank you for your help :)

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