Question

I need help with this concept of Continuity and Discontinuity.
Determine whether each function is continuous at the given x value. Justify your answer using the continuity test.
y=x^3-4; x=1

Answer 1

Also i would gladly appreciate it if someone could explain the Continuity Test

Answer 2

\(f(x)\) is continuous at \(x=a\) if and only if\[~~~~~~~~~\lim_{x\to a}f(x)=f(a)\]

Answer 3

that is, is the limit at that point the same as the value of the function at that point?

Answer 4

what is\[\lim_{x\to1}x^2-4\]?

Answer 5

Um..im sorry, but i havent learned lim. Would you mind explaining what that means?

Answer 6

oh, that is kinda hard. I'm not sure what kind of continuity test you are referring to then.

Answer 7

what is the definition you have?

Answer 8

Iam in precalculus. I just started the course.

Answer 9

well the limit is the value that the function approaches as x gets close to the specified number, in this case as x approaches 1

Answer 10

We have infinite disconituty, Jump Discontinuity, Point Discontinuity, and Everywhere Discontious.

Answer 11

yeah you need limits for all that, and I can't explain all the subtleties of the concept that quickly. I suggest you read something like this:
http://tutorial.math.lamar.edu/Classes/CalcI/LimitsIntro.aspx
and come back with any questions you may have after.

Answer 12

Im looking at my textbook right now, and we have nothing about limits. Thats whats confusing me.

Answer 13

well if you want to be lazy about it you can simply say
"is f(x) ever divided by zero, does it ever have a square root of a negative number, or the logarithm of a negative number or zero?"
if the answer is no, and it is not a piecewise function, then f(x) is continuous everywhere

Answer 14

Hahaha, im not lazy about it. I was just confused. Our teacher didnt explain it well.

Answer 15

No I was just kidding, didn't mean to call you lazy. It's just an incomplete approach.

Answer 16

Okay..I will research on that topic a little later. How would you describe the end behavior of a function then? Like if you had y=x^3 +2x^2+x-1

Answer 17

Does that involve limits too?

Answer 18

yes, but this is a little easier to ignore the limit bit
in the ends the largest power of x takes over, in this case x^3 will win out

Answer 19

so as x gets large and positive, x^3 gets large and positive, going off to infinity

Answer 20

in the notation of limits we would write\[\lim_{x\to\infty}x^3+2x^2+x-1=\infty\]that is "the limit as x goes to infinity of f(x) is (positive) infinity

Answer 21

what about when x is large and negative?

Answer 22

\[\lim_{x\to-\infty}x^3+2x^2+x-1=?\]

Answer 23

Im gonna find out how to use limits. Because im not understanding this at all. I appreciate your help. I will find out how to do it, and come back to this

Answer 24

okay, good luck