# Continuity of a Function

1. Feb 3, 2010

### Econometricia

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

2. Feb 3, 2010

### Staff: Mentor

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.

3. Feb 3, 2010

### Econometricia

The limit is 4 and should be for all correct?

4. Feb 3, 2010

### Staff: Mentor

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?

5. Feb 3, 2010

### Econometricia

No, the light bulb hasn't clicked yet.

6. Feb 3, 2010

### Staff: Mentor

Then go back and reread what I wrote. If you still have questions, ask them.

7. Feb 3, 2010

### Econometricia

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?

8. Feb 4, 2010

### Staff: Mentor

Did you get 4a - 2b = 1?
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.