Continuous functions/Limits and continuity

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August 30th 2007, 04:33 PM
Jane Doe

Continuous functions/Limits and continuity

Uhm, yeah, so I was studying for my test and realized that there's a lot I really didn't understand. This is one of those things:

Instructions:

Find a value for a so that the function is continuous.

1. f(x) = {2x + 3, x ≤ 2; ax + 1, x > 2}

2. f(x) = {4 - x^2, x < -1; ax^2 - 1, x >= 1}

3. f(x) = {x^2 + x + a, x < 1; x^3, x >= 1}

(This thing here: >= is a greater than or equals to sign since I couldn't figure out how to make it and I couldn't find anything to copy-paste off of the internet.)

The other problem I had was a little different.

Instructions:

a) Draw the graph of f.
b) At what points c in the domain of f does the limit exist?
c) At what points c does only the left-hand limit exist?
d) At what points c does only the right-hand limit exist?

Problem:

4. f(x) = {x, -1 ≤ x < 0, or 0 < x ≤ 1; 1, x=0; 0, x<-1, or x >1}

My problem here is drawing the graph. It's the 'or's' that throw me off. I'm fairly certain that if I were to figure out the graph, that I could easily do the rest of the problem. I may be wrong, of course...

(Also, if it's not too much trouble, and I'm sorry if it sounds a bit pretentious especially since I'm asking for help, but if you could maybe do part of it or explain how it goes and allow me to figure it out for myself? It is for a test, as I said, and I want to make sure that I perfectly understand it. But y'know, only if it's not too much trouble.)

Also, I was wondering another thing, though this is just a general question or comment. Does it break the rules that I'm putting so many problems/questions in one post or is that still okay? I'm unsure.
August 30th 2007, 05:05 PM
Plato

1 Attachment(s)

Here is some help on #3. The graph is for a=0.
The left-limit is 2+a at x=1; the right-limit is 1.
Now what should we make a in order to make them equal?
August 30th 2007, 05:24 PM
Jane Doe

Uhm, okay...so it looks like that graph would be continuous if the parabola were to move a little to the right. And if a=0 there, then should a=-1?

It seems to me that I could get the answer by trial and error with my calculator, is that right? If so, do you know if there's a way I could determine the answer algebraically? I'm not sure that we're allowed to use a calculator.

Though I remember, when hearing my classmates doing a similar problem, something about solving the top portion for x and then plugging x in? Or am I way off there?
August 30th 2007, 05:30 PM
Plato

Quote:

Originally Posted by Jane Doe

that graph would be continuous if the parabola were to move a little to the DOWN. And if a=0 there, then should a=-1?

CORRECT!
August 31st 2007, 01:40 AM
red_dog

Quote:

Originally Posted by Jane Doe

Problem:

4. f(x) = {x, -1 ≤ x < 0, or 0 < x ≤ 1; 1, x=0; 0, x<-1, or x >1}

My problem here is drawing the graph. It's the 'or's' that throw me off. I'm fairly certain that if I were to figure out the graph, that I could easily do the rest of the problem. I may be wrong, of course...

You can write the function in a more explicit way.
$f(x)=\left\{\begin{array}{cc} 0, & x<-1\\ x, & -1\leq x<0\\ 1, & x=0\\ x, & 0<x\leq 1\\ 0, & x>1 \end{array}\right.$

Now, can you draw the graph?