Find the values of a and b that make f continuous everywhere?

[jordan123]

Find the values of a and b that make f continuous everywhere??

1. Find the values of a and b that make f continuous everywhere??

Homework Equations

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

The Attempt at a Solution

I don't think it matters what value a and b are, because it is discontinuous at 2, hence it will never be a completely continuous function. No? Shed some light if I am wrong.

 

[benorin]

What is the $$\lim_{x\to 2^-}f(x)$$?

 

[HallsofIvy]

1. Find the values of a and b that make f continuous everywhere??

Homework Equations

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

The Attempt at a Solution

I don't think it matters what value a and b are, because it is discontinuous at 2, hence it will never be a completely continuous function. No? Shed some light if I am wrong.

Why do you say it is discontinuous at 2? As benorin says, look at
$$\lim_{x\rightarrow 2^-} f(x)= \lim_{x\rightarrow 2}\frac{x^2- 4}{x-2}$$

You will also want to look at
$$\lim_{x\rightarrow 2^+} f(x)= \lim_{x\rightarrow 2} ax^2- bx+ 3$$
which, of course, depends on a and b.

 

[jordan123]

Why do you say it is discontinuous at 2? As benorin says, look at
$$\lim_{x\rightarrow 2^-} f(x)= \lim_{x\rightarrow 2}\frac{x^2- 4}{x-2}$$

Right, the limit of that would be 4. But its not continuous at 4 because the piecewise function is x<2, hence not equal at 2.

 

[jordan123]

I know that there are limits from the left side. And the right I am guessing as well. But the piecewise function describes x that for both the left and right functions as not equal to zero. From the left it is x<2 and from the right it is 2<x<3, so both funtions will never have a value at f(2), right? So there will be a hole at f(2) making it impossible to make the function continuous everywhere. no? Explain please if I am wrong.

 

[HallsofIvy]

Right, the limit of that would be 4. But its not continuous at 4 because the piecewise function is x<2, hence not equal at 2.

Finally, it dawns on me. Yes, if f(x) is (x²-4)/(x-2) only for x less than 2 and f(x) is x²- ax- b only for x greater than 2, no values of a and b will give it a value at x= 2 and so it cannot be continuous. If the first were "$x\le 2$" or the second "$2\le x$" then it would be doable but the way it is written, no values of a and b will make the function continuous at x= 2.

 

[jordan123]

Finally, it dawns on me. Yes, if f(x) is (x²-4)/(x-2) only for x less than 2 and f(x) is x²- ax- b only for x greater than 2, no values of a and b will give it a value at x= 2 and so it cannot be continuous. If the first were "$x\le 2$" or the second "$2\le x$" then it would be doable but the way it is written, no values of a and b will make the function continuous at x= 2.

Exactly, ok, thanks! I don't know if its a typo in the book, but thanks for clarifying.

 

[smith007]

I hate to bump up an older thread but I just ran into this same question. We must have the same text or something.

The question is to find a and b so that f is continuous everywhere.

So can we not take lim->2 of the first equation for 4. The lim->2 of equation 2 for 4a -2b+3.

Then 4a-2b+3=4 so 4a-2b=1.


Next we do the same at Lim->3 for equation 2 and 3.

Getting 6-a+b and 9a-3b+3. Take 6-a+b=9a-3b+3 so b = (8a-3)/4


I am not really sure if I am working towards an answer here or just spining my wheels. could anyone comment?

 

[Nick89]

The problem, as identified before, is that the function is simply not defined on x = 2.
Look at the first two 'if' conditions:

The first specifies the function for x smaller than 2 and the second for x larger than 2 (not larger/smaller than or equal to). The third specifies the function for x larger than 3. There is no specification for x = 2. So the function does not exist at x = 2 so it can never be continuous by merely changing the constants. It can only be made continuous if you would re-define the piecewise conditions.

Note that the graph does not 'jump' or go to infinity or something near x = 2 (provided that you choose the right constants a and b) but it merely has a 'hole' in the graph at x = 2.

 

[laralds]

I know that this is coming almost 2 years late but here it goes for what its worth.

Yes, there is a hole on 2 and the first function equals 4...this is fixable! a little tricky but you can do it making the other functions to start at 2 (where the hole is) and making it continuous from there...my superman math tutor Brad showed me how:
1) function= 4a-2b+4=4 -------> 4a-2b= 0
then at x= 3 lim as x---> 3- (ax to the sq - bx+4) = 9a - 3b +4
lim as x---> 3+ (2x-a+b)= 6-a+b

so,
9a-3b+4 = 6-a+b --------> 10a-4b=2

2) 4a-2b= 0
10a-4b=2
----------
comes to a= 1 and b= 2

 

[Mark44]

I know that this is coming almost 2 years late but here it goes for what its worth.

Yes, there is a hole on 2 and the first function equals 4...this is fixable! a little tricky but you can do it making the other functions to start at 2 (where the hole is) and making it continuous from there...my superman math tutor Brad showed me how:
1) function= 4a-2b+4=4 -------> 4a-2b= 0
then at x= 3 lim as x---> 3- (ax to the sq - bx+4) = 9a - 3b +4
lim as x---> 3+ (2x-a+b)= 6-a+b

so,
9a-3b+4 = 6-a+b --------> 10a-4b=2

2) 4a-2b= 0
10a-4b=2
----------
comes to a= 1 and b= 2

You're new to the forum, so probably haven't had a chance to read the rules, part of which is shown below.

On helping with questions: Any and all assistance given to homework assignments or textbook style exercises should be given only after the questioner has shown some effort in solving the problem. If no attempt is made then the questioner should be asked to provide one before any assistance is given. Under no circumstances should complete solutions be provided to a questioner, whether or not an attempt has been made.​

Since the thread is nearly two years old, and the OP seems to have gotten it, there's probably no harm done, but please take note of the rules.

 

[HallsofIvy]

But the crucial point (which I think I missed when I first responded to this thread- ah, well, I was younger then!) is that the first line defines the function for "x< 2" and the second line defines it for "2< x< 3". Yes, you can set a and b so that the limit as x goes to 2 exists but since the function has NO value at x= 2, that does not make it continuous.

 

[Ryker]

But couldn't you say the function is continuous everywhere where it's defined? If it isn't defined in point x = 2, wouldn't that then mean x = 2 isn't even a part of the function? A logarithmic function is said to be continuous even though it is normally only defined for x > 0. Why would it not be the same in this case? I'd say there was probably an error in the problem being given out and that it didn't in fact shoot to omit the defining of the function in x = 2, but still.

I'm just thinking aloud, though, not saying I'm right :)

 

[Gib Z]

But couldn't you say the function is continuous everywhere where it's defined? If it isn't defined in point x = 2, wouldn't that then mean x = 2 isn't even a part of the function? A logarithmic function is said to be continuous even though it is normally only defined for x > 0. Why would it not be the same in this case? I'd say there was probably an error in the problem being given out and that it didn't in fact shoot to omit the defining of the function in x = 2, but still.

I'm just thinking aloud, though, not saying I'm right :)

This is true. Many state that for continuity of a function f:D to C at (a, f(a) ), we require for all $\epsilon > 0$ there exists $\delta>0$ such that for all x with $$|x-a|<\delta$$ we have $$|f(x) - f(a)|< \epsilon$$.

Unless it is made clear beforehand, it should really say "..for all x in D with...".

 

[SocialEng]

Ok. Similar problem.

Except, that the three intervals for the function are:
X<2
2$$\leq$$X<3
X$$\geq$$3

Which means that the function has the possibility of being continuous, given proper A and B values.

My question is, is there an algebraic way to assault this? It seems like there would be a better way to find A and B than just plugging in random numbers.

Once again, I'm trying to find A and B such that the graph is continuous on all x, given the three intervals above with the three functions:

(x$$^{2}$$-4)/(x-2) on (-inf,2)
ax$$^{2}$$ - bx + 1 on [2,3)
2x-a+b on [3,inf)

Thoughts?

 

[Nick89]

What you need to do is simply make sure that the functions meet at the same value at the boundaries of the intervals.

Let's take the first two functions:
$$\frac{x^2-4}{x-2}$$
$$ax^2-bx+1$$
The first one is defined up to x = 2 while the second one is defined only between x = 2 and x = 3. So the problem is at x = 2, you need to ensure that the values of the two functions are equal at x = 2.

Now, you cannot simply plug in x = 2 in the first function because you'll get 0/0, so you'll have to take the limit to x = 2. That gives you the value that that function will go to in the limit that x goes to 2. Let's call this value $c$ (because it is a constant).
On the other hand, you can simply plug in x = 2 in the second function and you will know what value it has in x = 2. Of course that value will depend on a and b. Let's call that value $f(a,b)$.

Your problem now is to determine for which values of a and b you have
$$c = f(a,b)$$

Of course there are two unknowns (a and b) and only one equation so at this moment this is impossible (you can only determine a as a function of b or vice versa), but you have a third function which gives you the second equation that you need to determine both a and b.