How Do You Solve Piecewise Calculus Functions?

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Homework Help Overview

The discussion revolves around understanding piecewise functions in calculus, specifically focusing on evaluating limits as x approaches a certain value. Participants are examining the behavior of a function defined in two segments based on the value of x.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss how to evaluate the limit of a piecewise function at a specific point, questioning the correct approach to demonstrate the limit from both sides. There is also uncertainty about the clarity of the original question and the relevance of certain statements made by the original poster.

Discussion Status

Some participants have provided guidance on how to express the limit properly, emphasizing the need to show limits from both sides. There is an ongoing exploration of how to articulate the evaluation process clearly, with no explicit consensus reached on the final answer.

Contextual Notes

There is a mention of the original poster's difficulty in formatting piecewise functions and a general assumption that the problem is from an early calculus course, which may not require rigorous proofs.

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Homework Statement



si9vr73ny4o1.jpg


Note: The functions above are in piece wise form, I just didn't know how to put them in piece wise function here.

Homework Equations


The Attempt at a Solution



http://www.mathway.com/math_image.aspx?p=SMB02LSMB03x:-1,-xSMB02ESMB032SMB02eSMB03+2SMB02lSMB03?p=87?p=38 = 1

http://www.mathway.com/math_image.aspx?p=SMB02LSMB03x:-1,4x+5SMB02lSMB03?p=83?p=38 = 1

1/1 = 1

Answer: limit as x approaches f(x) = 1
 
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I'm not sure what your question is..
Your answer is correct
 
genericusrnme said:
I'm not sure what your question is..
Your answer is correct

Edited the question! Hopefully you will be able to see it and understand it and help me out.
 
choboplayer said:
Edited the question! Hopefully you will be able to see it and understand it and help me out.

I understand the question in the image but I don't understand your question
I'll try and explain piecewise functions but I may be going in the wrong direction

your f(x) can be thought of as two separate functions in the two separate areas, when x is greater than or equal to -1 we treat the function as 4x+5
when x is less than -1 we treat the function as -x^2 + 2

In trying to find the limit, in this simple case you can just evaluate the function at x=-1. So we would use the region where x is greater than or equal to -1 and get -4+5 = 1.
I'm assuming this is all that you're expected to be doing since it seems like this is an early calculus class that isn't going to be all rigorous.

I attatched a graph of your function if it will help
 

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genericusrnme said:
I understand the question in the image but I don't understand your question
I'll try and explain piecewise functions but I may be going in the wrong direction

your f(x) can be thought of as two separate functions in the two separate areas, when x is greater than or equal to -1 we treat the function as 4x+5
when x is less than -1 we treat the function as -x^2 + 2

In trying to find the limit, in this simple case you can just evaluate the function at x=-1. So we would use the region where x is greater than or equal to -1 and get -4+5 = 1.
I'm assuming this is all that you're expected to be doing since it seems like this is an early calculus class that isn't going to be all rigorous.

I attatched a graph of your function if it will help

So the answer to my question is just limit as x approaches -1 f(x) = 1 ? And will it be acceptable for me to just show the two cases that I have in my original answer, where I have plugged in a -1 every where I have an x?
 
choboplayer said:

Homework Statement



si9vr73ny4o1.jpg


Note: The functions above are in piece wise form, I just didn't know how to put them in piece wise function here.

Homework Equations



The Attempt at a Solution



http://www.mathway.com/math_image.aspx?p=SMB02LSMB03x:-1,-xSMB02ESMB032SMB02eSMB03+2SMB02lSMB03?p=87?p=38 = 1

http://www.mathway.com/math_image.aspx?p=SMB02LSMB03x:-1,4x+5SMB02lSMB03?p=83?p=38 = 1

1/1 = 1

Answer: limit as x approaches f(x) = 1
While it is true that [itex]\displaystyle \lim_{x\to-1}\,f(x)=1\,,[/itex] what you wrote isn't quite what you need to say.

Also, what does 1/1 = 1 have to do with this question ?

You need to show that [itex]\displaystyle \lim_{x\to-1^-}\ f(x)=\lim_{x\to-1^+}\ f(x)\,.[/itex] If that is true then that common result is the limit you're looking for.

While it is true that [itex]\displaystyle \lim_{x\to-1^-}\ f(x)= \lim_{x\to-1}-x^2+2\,,[/itex] and [itex]\displaystyle \lim_{x\to-1^+}\ f(x)= \lim_{x\to-1}4x+5\,,[/itex] you really should state that this is what you're doing.
 
Last edited by a moderator:
SammyS said:
While it is true that [itex]\displaystyle \lim_{x\to-1}\,f(x)=1\,,[/itex] what you wrote isn't quite what you need to say.

Also, what does 1/1 = 1 have to do with this question ?

You need to show that [itex]\displaystyle \lim_{x\to-1^-}\ f(x)=\lim_{x\to-1^+}\ f(x)\,.[/itex] If that is true then that common result is the limit you're looking for.

While it is true that [itex]\displaystyle \lim_{x\to-1^-}\ f(x)= \lim_{x\to-1}-x^2+2\,,[/itex] and [itex]\displaystyle \lim_{x\to-1^+}\ f(x)= \lim_{x\to-1}4x+5\,,[/itex] you really should state that this is what you're doing.

So can I just rewrite your last sentence and add something like:

The above statements hold true therefore limit as x approaches -1 f(x) = 1. I mean would that be an acceptable answer?
 
As an instructor I would want you to indicate that you're taking a limit as x approaches -1 from the left, and taking a limit as x approaches -1 from the right.

That may be picky on my part, and many of my students would agree.
 

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