Continuity Tests

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September 20th 2009, 07:49 PM
VitaX

Continuity Tests

At which points are is the function continuous?

y = (3x - 1)^(1/4)

The function is continuous at the interval [1/3, infinity). Discontinuous at (-infinity,1/3] because a negative answer is obtained.
So to apply the continuity tests, do I just substitute in any internal point and prove what I said above is correct?
Also did I use the brackets and parenthesis correctly in writing an interval or does it not matter?
September 20th 2009, 08:14 PM
Kasper

Quote:

Originally Posted by VitaX

At which points are is the function continuous?

y = (3x - 1)^(1/4)

The function is continuous at the interval [1/3, infinity). Discontinuous at (-infinity,1/3] because a negative answer is obtained.
So to apply the continuity tests, do I just substitute in any internal point and prove what I said above is correct?
Also did I use the brackets and parenthesis correctly in writing an interval or does it not matter?

Right off the top, just to clarify; the function is not just discontinuous per se below $\frac{1}{3}$ but it is undefined. Also, I would change the interval, to being that the function is continuous (and defined) over the interval $[\frac{1}{3}, \infty)$ . If you want to look at it from the undefined side, the interval would be $(-\infty,\frac{1}{3})$ , because the function IS defined at $\frac{1}{3}$ . Can't have closed brackets on both intervals. I usually like to write that the function is continuous over it's domain where $x \geq \frac{1}{3}$ . Take a sample point on either side to verify.
September 20th 2009, 08:21 PM
VitaX

Quote:

Originally Posted by Kasper

Right off the top, just to clarify; the function is not discontinuous per se below $\frac{1}{3}$ but it is undefined. Also, I would change the interval, to being that the function is continuous (and defined) over the interval $[\frac{1}{3}, \infty)$ . If you want to look at it from the undefined side, the interval would be $(-\infty,\frac{1}{3})$ , because the function IS defined at $\frac{1}{3}$ . Can't have closed brackets on both intervals.

So when writing the interval in which all points are continuous use a bracket at the point in which the limit is 0. [1/3,infinity) The function is continuous at all the points but not defined at 1/3?
September 20th 2009, 08:29 PM
Kasper

Quote:

Originally Posted by VitaX

So when writing the interval in which all points are continuous use a bracket at the point in which the limit is 0. [1/3,infinity) The function is continuous at all the points but not defined at 1/3?

Close, the interval $[\frac{1}{3},\infty)$ is domain of this function, and this function is continuous at all points on its domain. So this interval, while the functions domain is the also interval of continuity for the function.

Keep in mind that the function is indeed defined at $x=\frac{1}{3}$ . Roots are undefined when the radicand is negative right? It is still defined at $x=\frac{1}{3}$ ! It just gives a value of 0. On top of that, we know that basic root functions are continuous over their domains.

Here is a quick way to find the domain AND interval of continuity.

We have the function $f(x)=(3x-1)^\frac{1}{4}$

We know that the radicand must be greater than zero right? It also can be equal to 0, because the $f(x)=0\ at\ x=\frac{1}{3}$ .

So the domain and interval of continuity lies in the inequality;

$3x-1 \geq 0$

$3x \geq 1$

$x \geq \frac{1}{3}$

Just to add on interval notation, we use the closed bracket on \frac{1}{3} because the point is included in the interval.

$[\frac{1}{3}, \infty)$ is the same as $x \geq \frac{1}{3}$

$(\frac{1}{3}, \infty)$ is the same as $x > \frac{1}{3}$

Does this help clarify?
September 20th 2009, 08:36 PM
VitaX

Yah its clearer now. So when I find the domain of the function, I should pick an internal point within the domain and substitute that point for x in the function to prove my point?
September 20th 2009, 08:44 PM
Kasper

Quote:

Originally Posted by VitaX

Yah its clearer now. So when I find the domain of the function, I should pick an internal point within the domain and substitute that point for x in the function to prove my point?

Yeah, you can pick a point within the domain to show continuity in that interval, and I would take a point in the undefined interval to show that the function is undefined there. These aren't "proofs" per se as much as they are verification. I'm not sure if you've touched epsilon-delta proofs? There are better ways of proving where a function is continuous, but it depends on what level of math you are at. Follow your course's example on that note.

When you are looking for domains or continuity, always look for places that would mess up the graph of the function, like zero-denominators, possible negative radicands, etc. These things can hack your domain in half, like the root in your problem restricted the domain.

For example $f(x)=\frac{1}{x}$ , would have the domain $(-\infty, 0)\cup(0,\infty)$ with open brackets because 0 is not defined.
September 20th 2009, 08:47 PM
VitaX

Quote:

Originally Posted by Kasper

Yeah, you can pick a point within the domain, to show continuity in that interval, and I would take a point in the undefined interval to show that the function is undefined there. These aren't "proofs" per se as much as they are verification. I'm not sure if you've touched epsilon-delta proofs? There are better ways of proving where a function is continuous, but it depends on what level of math you are at. Follow your course's example on that note.

When you are looking for domains or continuity, always look for places that would mess up the graph of the function, like zero-denominators, possible negative radicands, etc. These things can hack your domain in half, like the root here restricted the domain.

For example $f(x)=\frac{1}{x}$ , would have the domain $(-\infty, 0)\cup(0,\infty)$ with open brackets because 0 is not defined.

Alright thats easy to understand.