# LIMITS, continuity piece defined function

1. Sep 21, 2008

### susan__t

The question asks to find a value for a and b that makes f continuous everywhere.

f(x)=
$$\frac{x - 4}{x-2}$$ , where x<2

ax2 - bx + 3 , where 2<x<3

2x - a +b , where x > or = 3

I know that in order for a function to be continuous the limit as x approaches 2 must be equal from the left and right and f(2) must be defined. I don't know how I can possibly make it continuous if x never is equal to 2, then f(x) would not be possible and it would be a removable discontinuity.
There's no answer available at the back of my text or in the solutions manual so it seems a bit fishy to me
Any help solving this would be great (if there even is a solution)

2. Sep 21, 2008

### JG89

In order for a function to be continuous as a point p, the function must be defined at f(p) and the limit of f(x) as x approaches p must be equal to f(p).

In order for this function we are considering here to be continuous, one of the conditions is that it must be continuous at x = 2. Also notice that when x approaches 2, the function increases beyond all positive bounds. This should help you out.

Last edited: Sep 21, 2008
3. Sep 22, 2008

### slider142

Most likely they meant make f continuous everywhere on its domain, as x=2 is not part of its domain. Then your only concern is to make sure the line and the quadratic meet at their endpoints.

4. Sep 22, 2008

One more point and one question. The function is

$$f(x) = \begin{cases} \frac{x-4}{x-2} & \text{ if } x < 2\\ ax^2 - bx + 3 & \text{ if } 2 < x < 3\\ 2x-a + b & \text{ if } x \ge 3 \end{cases}$$

As slider142 mentioned, part of the work is going to involve making sure that the quadratic and the linear portions agree when you look at the limits at $$x = 3$$. However, this is only one condition, and you have two unknown constants, $$a, b$$. The other condition will involve the right-hand limit of the quadratic at $$x = 2$$, so you will
need to decide on how

$$\lim_{x \to 2^+} f(x)$$

should bd defined, and this will give you the second condition on $$a, b$$.

Now for my question: Is the end of the linear portion $$-a + b$$ or is it $$-(a+b)$$? It doesn't make any difference for us, but it can for your answer.

5. Sep 22, 2008

### HallsofIvy

Staff Emeritus
Are you sure you have copied the problem correctly? There is NO way to make the function given continous at x= 2 because its limit, as x approaches 2 from below, does not exist.

If, however, the function were defined as (x2- 4)/(x- 2) for x less than 2, then it would be possible.

If, in fact, the problem is to find a and b such that
$$f(x) = \begin{cases}\frac{x^2-4}{x-2} & \text{ if } x < 2\\ax^2 - bx + 3 & \text{ if } 2 < x < 3\\2x-a + b & \text{ if } x \ge 3\end{cases}$$
then, because, for x not equal to 2, (x2- 4)/(x-2)= (x-2)(x+2)/(x-2), which, for x not equal to 2, is just x+ 2 and has limit 4, from below, at x= 2.

Now, you want to find a and b such that the limit of ax2- bx+ 3, as x approaches 2 from above, is 4. That's easy because that is a polynomial and so its limit is 4a- 2b+ 3. You must have 4a- 2b+ 3= 4.

As x approaches 3, from below, you have 9a- 3b+ 3 and, from above, again because 2x- a+ b is a polynomial, the limit is 6-a+ b.

Assuming that the problem really had (x2-4)/(x-2) rather than(x-4)/(x-2), you must solve the two equations 4a- 2b+ 3= 4 and 9a- 3b+ 3= 6- a+ b for a and b.

If the problem really is what you say, then there is no solution!