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The discussion revolves around solving a piecewise function defined as f(x) with three segments: (x²-4)/(x-2) for x < 2, ax² - bx + 1 for 2 ≤ x < 3, and 4x - a + b for x ≥ 3. The user successfully factored the first segment and established that f(2) = 4, leading to the equation 4 = 4a - 2b + 1. The user then derived the equation 3 = 4a - 2b but struggled to find the values of a and b. Another user suggested applying continuity conditions at x = 3 to create a system of equations for solving a and b.

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Homework Statement



This is a piece wise function of course. f(x) =

(x2-4) / (x-2) if x is less than two.

ax2 - 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.

Homework Equations


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.
 
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SmittenWCalc said:

Homework Statement



This is a piece wise function of course. f(x) =

(x2-4) / (x-2) if x is less than two.

ax2 - 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.

Homework Equations


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.
Written in a more useful way, your equation is

4a - 2b = 3Now, 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.
 
SmittenWCalc said:

Homework Statement



This is a piece wise function of course. f(x) =

(x2-4) / (x-2) if x is less than two.

ax2 - 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.

Homework Equations


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.
Hello SmittenWCalc. Welcome to PF!

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

[itex]\displaystyle \lim_{x\to\,2+}f(x)=\lim_{x\to\,2+} (ax^2+bx+1)=4a+2b+1\ .[/itex]

Do something similar at x=3 .
 

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