LIMITS, continuity piece defined function

[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)

 

[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.

 

[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.

 

[statdad]

One more point and one question. The function is

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

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.

 

[HallsofIvy]

Are you sure you have copied the problem correctly? There is NO way to make the function given continuous 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!