Limits Conceptual Question

Integrate

Integrate

Junior Member
Joined
May 17, 2018
Messages
107
F(X).png

I am having a hard time really understanding the question, but what I am gathering is if the question had said "f(x)=3 for all x GREATER THAN 7" then we would be able to determine f(7) because it is right continuous until it gets to 3? Which 3 would be our value for f(7) in the new condition I created?
 
If f(x) is left-continuous then f(7)=3. If f(x) is right-continuous we have no idea what f(7) equals.
Draw the graph and look at the possibilities.

Exactly what part of the question is not clear? There can be a jump discontinuity at x=3. Please post back with all the drawings you think are possible and state what questions/concerns you have about each one. This is a hard problem to help a student with if the tutoring is not face to face. However with your help it can happen but we need to see those graphs!
 
IMG_5915.jpg
I'm guessing it looks something like this

All values less than and NOT equal to 7 are equal 3

Then it jumps to another value (or it could be three. We have been told it can't be three from the right hand of the function) and is continuous from there on.
 
The problem contains 2 questions. I don't quite get the answer No.
 
You are not thinking about this clearly.

Just because I say that f(x)=3 for x<7 does NOT mean that f(7) can't be 3. Where does it say that? In fact, there was nothing said about what happens at x=7. f(x) is right continuous for x>7 does not say that f(x) is not right continuous from x>=7 (your own picture shows this).

You wrote We have been told it can't be three from the right hand of the function) and is continuous from there on. What do you mean by it?
Are you saying that f(7) can't be 3? Why not?

Your graph would have been a straight line right through (7,3) even including (7,3). The graph could be continuous!
 
Jomo said:
If f(x) is left-continuous then f(7)=3. If f(x) is right-continuous we have no idea what f(7) equals.
Draw the graph and look at the possibilities.

Exactly what part of the question is not clear? There can be a jump discontinuity at x=3. Please post back with all the drawings you think are possible and state what questions/concerns you have about each one. This is a hard problem to help a student with if the tutoring is not face to face. However with your help it can happen but we need to see those graphs!
Click to expand...
I take back my 1st sentence. Too much drinking is my excuse.
 
If your teacher tells you that Abe, Betty and Carlos got a 93% on their exam and then David tells you he got a 93% on the exam, then is David necessarily being untruthful to you? No! Just because your teacher gave you a list of students who got a 93% does not mean that the list is complete. Just because I tell you that f(x) = 3 whenever x<7 does not mean that f(23) can't be 3. Just like it does not mean that f(7) is not 3. There is no reason at all to assume that I told you each and every x value that resulted in 3.

This may be the 1st time you really have to think carefully but then again that is what is needed to understand calculus.
 
Pardon,

I had went away for the weekend.

May the issue be that I am trying to solve this problem not simply “is there any situation where the given constraints are possible?”
 
Here is where I am at. The question asks us to examine the situation coming from the left and from the right.

f(x)=3 for all values x<7

We assume that f(x) is continuous at x=7 coming from the left. Coming from the left the value f(x)=3 rrrriiiiiiiiiiiiiiiiiight up until x=7 but because we established that x=7 is continuous from the left that disqualifies the possibility of a jump discontinuity meaning f(x)=3 at 7.

If we examine the reverse of this where we approach x=7 from the right we cannot determine the value of f(7) because we do not know the values leading up to it like we did coming from the left.
 
You totally lost me.
If you agree that there are two questions then please right down each question and answer them. I think you are confused about what the two questions are.
 
Integrate said:
Here is where I am at. The question asks us to examine the situation coming from the left and from the right.

f(x)=3 for all values x<7

We assume that f(x) is continuous at x=7 coming from the left. Coming from the left the value f(x)=3 rrrriiiiiiiiiiiiiiiiiight up until x=7 but because we established that x=7 is continuous from the left that disqualifies the possibility of a jump discontinuity meaning f(x)=3 at 7.
Click to expand...
No, it doesn't. Saying that f(x)= 3 for all x< 7 and f is "left continuous" at x= 7 means that f(7)= 3 but if x> 7 f(x) might be some other value so there could be a jump discontinuity at x= 7.

If we examine the reverse of this where we approach x=7 from the right we cannot determine the value of f(7) because we do not know the values leading up to it like we did coming from the left.
Click to expand...
Yes, if f is right continuous, since we don't know the values of f for x> 7 we cannot determine f(7).
 
You must log in or register to reply here.
Top