How to make this piecewise continuous (multiple variables, lines)

**skinsdomination09** · September 13th 2012, 03:54 PM

f(x) = (x-3)(x-2)(x+1), .....................AND x doesn't equal (equal with the line going across it) -1, 3
.........____________
..........(x+1) (x-3)

4px-2 ....x=-1

5mx² - 2x .....x=3

This is no calculator and the answer isn't as important as knowing the steps to do it. I'm confused as to which number (-1 or 3) I set x equal to before setting the functions equal to eachother.

**johnsomeone** · September 13th 2012, 04:26 PM

$f(x) = \frac{(x-3)(x-2)(x+1)}{(x+1)(x-3)}, x \notin \{-1, 3 \}$ already is peicewise continuous. I assume the problem is actually asking how to make it continuous.

Here are some questions for you. I'm going to define a function on the same domain. Let $g(x) = x-2, x \notin \{-1, 3 \}$ .

First Question: What's the relationship, if any, between $f$ and $g$ ?

Second Question: Can you think of a continuous function defined on all of $\mathbb{R}$ that equals $g$ on $g$ 's domain of ( $x \notin \{-1, 3 \})$ ?

In other words, find a function $\tilde{g}: \mathbb{R} \rightarrow \mathbb{R}$ such that $\tilde{g}$ is continuous everywhere, and $\tilde{g}(x) = g(x)$ for all $x \notin \{-1, 3 \}$ .

What you're dealing with is something called a "removeable discontinuity", for reasons you'll hopefully soon understand. For now, classify these problems under "So Simple They're Confusing".

**skinsdomination09** · September 13th 2012, 04:35 PM

Originally Posted by johnsomeone

$f(x) = \frac{(x-3)(x-2)(x+1)}{(x+1)(x-3)}, x \notin \{-1, 3 \}$ already is peicewise continuous. I assume the problem is actually asking how to make it continuous.

Here are some questions for you. I'm going to define a function on the same domain. Let $g(x) = x-2, x \notin \{-1, 3 \}$ .

First Question: What's the relationship, if any, between $f$ and $g$ ?

Second Question: Can you think of a continuous function defined on all of $\mathbb{R}$ that equals $g$ on $g$ 's domain of $x \notin \{-1, 3 \})$ ?

In other words, find a function $\tilde{g}: \mathbb{R} \rightarrow \mathbb{R}$ such that $\tilde{g}$ is continuous everywhere, and $\tilde{g}(x) = g(x)$ for all $x \notin \{-1, 3 \}$ .

What you're dealing with is something called a "removeable discontinuity", for reasons you'll hopefully soon understand. For now, classify these problems under "So Simple They're Confusing".

I gotta be honest with you, I have no idea what you're talking about lol.

It asks " Find the value of f k,m that will make f(x) continuous everywhere."

I figure you'd have to set the functions equal to eachother and insert an x value to solve for p and then solve for m after setting the first and third functions equal to eachother.

Am I wrong? How is this done. I have a test tommorow and I just found out there might be a question on this tommorow. (Although most will be straightfoward limits)

**johnsomeone** · September 13th 2012, 04:43 PM

OK, I thought you had 3 separate questions there. What you wrote didn't make much sense - I thought I'd deciphered it, but it seems I did so incorrectly.
If you're going to get help, people will have to understand your problem. For instance, you said the problem was posed as "Find the value of f k,m that will make f(x) continuous everywhere." How can I help you find k when your post doesn't even have a k?
Do you think you understand the problem you're asking about? - Not how to solve it, but what it even means? If you could clarify what the problem actually is, maybe we could try again.
-----
?? Might the problem be asking this:

$f(x) = 4px - 2$ when $x < -1$

$f(-1) = \ ?$

$f(x) = \frac{(x-3)(x-2)(x+1)}{(x+1)(x-3)}$ when $-1 < x < 3$

$f(3) = \ ?$

$f(x) = 5mx^2 - 2x$ when $x > 3$

Find the values of $m$ and $p$ , and values of $f(-1)$ and $f(3)$ , so that $f$ is continuous. Could that be the problem?

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