Solving for Continuity in a Piecewise Function

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Homework Help Overview

The discussion revolves around finding values for A and B that ensure continuity in a piecewise function defined by different expressions over specified intervals. The function includes a rational expression for x < 2, a quadratic expression for 2 < x < 3, and a linear expression for x ≥ 3.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss limits of the function as x approaches 2 and 3 from both sides. There are attempts to set limits equal to each other to find values for a and b, with some participants expressing uncertainty about their methods and results.

Discussion Status

Participants are actively questioning their approaches and reasoning, particularly regarding the limits and continuity conditions at the transition points of the piecewise function. Some guidance has been offered on setting up equations based on limits, but there is no explicit consensus on the correct method or values.

Contextual Notes

There is mention of confusion regarding the values of a and b, with participants noting discrepancies between their findings and expected results. The discussion includes a focus on ensuring that left-side and right-side limits are equal at the transition points.

Econometricia
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1. Find the values of A and B that make the function continuous.
f(x) = (x2-4) /(x-2) When x < 2
f(x) = ax2-bx +3 When 2 < x < 3
f(x) = 2x - a + b When X is > or equal to 3








3. I took the limit of the equation and set it equal to the second to solve for a and b. After I wrote that in terms of b and tried to solve for a , but I got a= -(7/2) and b = 5.5 The correct solution is a=b=1/2

Thank You In Advance.
 
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What is the limit of f(x) as x --> 2 from the left?
What should be the limit of f(x) as x --> 2 from the right?

Same questions for x approaching 3.
 
The limit is 4 and should be for all correct?
 
Lim f(x) = 4, as x -->2, but I don't know what it is when x --> 3. However, you want the left-side and the right-side limits to be equal when x --> 3, so does that give you some idea of where to start?
 
No, the light bulb hasn't clicked yet.
 
Then go back and reread what I wrote. If you still have questions, ask them.
 
Well, what I did is I solved for the limit of the function as x-->2 from the left and got 4. I then set the second equation = to 4 and used x =2 to solve for a and b. After I solved for a and b I wrote b in terms of a and solved for the third equation to get the a. I know this is not the correct approach. Can you tell me where I began to go wrong?
 
Econometricia said:
Well, what I did is I solved for the limit of the function as x-->2 from the left and got 4. I then set the second equation = to 4 and used x =2 to solve for a and b.
Did you get 4a - 2b = 1?
Econometricia said:
After I solved for a and b I wrote b in terms of a and solved for the third equation to get the a.
This is where you went wrong. The problem is that at x = 3, you don't know what the function value should be. All you know is that the two formulas should give the same value when x = 3.
So, at x = 3, ax^2 -bx + 3 = 2x - a + b.
Substitute x in this equation, then you'll have a second equation in a and b.

Now with two equations in a and b, you should be able to solve them to find values for a and b.
Econometricia said:
I know this is not the correct approach. Can you tell me where I began to go wrong?
 

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