Finding Values of Delta that correspond to epsilon

[realism877]

Homework Statement

#8

(4x+1)/(3x-4)=4.5

Homework Equations

The Attempt at a Solution

4<(4x+1)/(3x-4)<5

What do I have to do next?

 

[Mark44]

You need to figure out how big the interval centered around x = 2 needs to be so that every value of x in this interval satisfies this inequality, 4<(4x+1)/(3x-4)<5. IOW, you need to find a number δ so that if x is in (2 - δ, 2 + δ), then 4.5 - ε<(4x+1)/(3x-4)< 4.5 + ε, where ε = .5

With a graphing calculator, this should be fairly easy to do.

BTW, this is the same problem you posted in this thread - https://www.physicsforums.com/showthread.php?t=499392. Why are you starting a new thread for the same problem?

 

[realism877]

You need to figure out how big the interval centered around x = 2 needs to be so that every value of x in this interval satisfies this inequality, 4<(4x+1)/(3x-4)<5. IOW, you need to find a number δ so that if x is in (2 - δ, 2 + δ), then 4.5 - ε<(4x+1)/(3x-4)< 4.5 + ε, where ε = .5

With a graphing calculator, this should be fairly easy to do.

BTW, this is the same problem you posted in this thread - https://www.physicsforums.com/showthread.php?t=499392. Why are you starting a new thread for the same problem?

I want the probl;em to appear more clearly.

I used a graphing calculator, but there are two curves for f(x)=(4x+1)/(3x-4). Which intersection do I use?

 

[Mark44]

You're interested in what happens around x = 2, which is on the right branch of the graph. There is a vertical asymptote at x = 4/3.

 

[realism877]

Where did you get the 2 from?

 

[Mark44]

The limit is as x approaches 2. Didn't you notice this?

 

[Mark44]

It might be helpful for you to post Definition 2 here.

 

[realism877]

Ohh, okay.

Is the intersection at x=2.125?

 

[Mark44]

The intersection of what? Try to be more clear in what you're saying.

I actually understand what you're asking, but I had to do some work that I shouldn't have had to do, to understand it, and I want to get you to phrase your questions to make more sense.

 

[Mark44]

One more thing - the image you uploaded is vary large, 1200 x 1600 pixels. Can you shrink the image to about 600 x 800 and upload a new copy in post 1? You'll need to edit your first post, which you can do up to a day or so after you post it.

 

[realism877]

Is the solution delta=.9?

 

[vela]

No, it isn't.

 

[Mark44]

Definitely not. You were on the right track, and then deleted your post.

For the sake of brevity, let's say that f(x) = (4x + 1)/(3x - 4). You found that f^-1(4) = 2.125. That's within .125 of 2. Did you check the other side? IOW, what is f^-1(5)? If f^-1(5) is closer to 2 than .125, then take that difference (|f^-1(5) - 2| to be your δ. Otherwise, take .125 for your δ.

For the other part of this problem, with the smaller value for ε, do pretty much the same that you did for this part. You'll get a smaller value for δ though.

 

[realism877]

Okay, I'm going to put this in layman's term. I'm a math student. I'm not a mathmatician or math wiz. So my logic might be off to your standards.

When y=(4x+1)/(3x-4) and y=5 intersect, x=1.909, which I round up to 1.91.

When y=(4x+1)/(3x-4) and y=4 intersect, x=2.125, which I leave it as 2.12.

I put this in an absolute value:(2-2.1) and got .12

I did the same with (2-1.91) and got .09

As I was trained, you're supposed to take the smallest of the two, which in this case is .09.

This is the best I can do for now. If you don't mind, please show me where I screwed up or what I have to do.

 

[Mark44]

.09 works for me. Good job!

 

[realism877]

So, did I do it right?

 

[Mark44]

Yes. That's what I meant by this:

.09 works for me. Good job!