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Continuity for piecewise

phrox

New member
Sep 7, 2013
20
I don't really understand this question...

I'm given a graph with x approaching and hitting 1, making y=2 (filled dot). Then there's a discontinuity jump at (1,3) which is the empty dot, then there's 2 other points on this small curve with empty dots at (3,4.5) and at (5,4), then another jump to a filled dot at (5,1).

For the first question being find point c that the function has jump discontinuity but is left-continuous which I know would be c=5.

It's the second question that doesn't make sense to me. It is:
What value should you assign to g(c) to make f right - continuous at x=c?

This question just doesn't make sense to me. What is g(c) on the graph? The empty dotted curve? Would by "f" he meant the whole f(x) function which has discontinuity?

Thanks!!
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Can you include an image of the graph?
 

phrox

New member
Sep 7, 2013
20
Sure thing,

 

Plato

Well-known member
MHB Math Helper
Jan 27, 2012
196
Then there's a discontinuity jump at (1,3) which is the empty dot, then there's 2 other points on this small curve with empty dots at (3,4.5) and at (5,4), then another jump to a filled dot at (5,1).
For the first question being find point c that the function has jump discontinuity but is left-continuous which I know would be c=5.
It's the second question that doesn't make sense to me. It is:
What value should you assign to g(c) to make f right - continuous at x=c?
This question just doesn't make sense to me. What is g(c) on the graph? The empty dotted curve? Would by "f" he meant the whole f(x) function which has discontinuity?
I do not really follow some of this question, but the graph helps.
The function \(\displaystyle f\) has a jump discontinuity at \(\displaystyle x=1\) but it is left-continuous there.
The function \(\displaystyle f\) has a jump discontinuity at \(\displaystyle x=5\) but it is lright-continuous there.
See this web page.

The function \(\displaystyle f\) has an discontinuity at \(\displaystyle x=3\) that some call a removable discontinuity .
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
If g(x) = f(x) except for at points of discontinuity of f, then to make g right-continuous at c = 1 (which is the only place where f has a jump discontinuity but is left-continuous, your answer of c = 5 is wrong) we must choose g(1) = 3.

This is just a conjecture on my part, you haven't told us what "g" is. A summary of the discontinuities at f:

x = 1:

f is left-continuous, and has a jump discontinuity

x = 3

f is neither left-continuous nor right-continuous and has a jump discontinuity. if we were to replace the value f(3) = 2.5 (approx.) with f(3) = 4.5 (approx., it is REALLY hard to tell from your graph) f would be continuous at 3 (which is why the continuity is called "removable")

x = 5

f is right-continuous and has a jump discontinuity.

Neither the discontinuity at 1, nor at 5 can be fixed by assigning some "other" value for f(x), the resulting function will still be discontinuous there (sometimes these are called "essential" discontinuities).
 

phrox

New member
Sep 7, 2013
20
I do agree with that. But for the second question, it doesn't make sense because there has to be an answer according to my prof. He said if it doesn't exist, put DNE. But that doesn't work.

Also, 1) is correct at c=5 because it just simply says it's correct when I have it right.
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
It? What is this "it"?

If by "it" you mean some automated software on a computer-graded test/problem set, I can assure you "it" most certainly can be wrong:

1) The program could have an internal error.
2) Whichever human supplied the "answer data file" could have made a mistake.

In general, mathematical statements are not true because someone or something says they are, they are true because they are logically consistent with, and dependent upon other facts or axioms given before-hand. "Proof by authority" is NOT accepted by the mathematical community at large.

IF (and this is actually a big "if") you have portrayed the graph of f(x) reasonably accurately (in particular, if the holes are solid or "hollow" at the "special points" as you depict it) then f is NOT left-continuous at c = 5.

I state the formal definition of left-continuous here:

A function $f:\Bbb R \to \Bbb R$ is said to be left-continuous at $a$ if:

$\displaystyle \lim_{x \to a-} f(x) = f(a)$.

From the graph you have supplied, we see that:

$\displaystyle \lim_{x \to 5-} f(x) = 4$

but $f(5) = 1$

As 4 does not equal 1, the matter is settled, and anyone who says differently is just wrong.

Perhaps we are missing some crucial information here, but it sounds like someone has confused left with right.