How Do You Solve Piecewise Calculus Functions?

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The discussion focuses on solving piecewise calculus functions, specifically finding limits as x approaches -1. Participants clarify that the function can be evaluated separately in two regions: for x ≥ -1, it is 4x + 5, and for x < -1, it is -x² + 2. To find the limit, one must show that the left-hand limit and right-hand limit at x = -1 are equal, both yielding a result of 1. There is an emphasis on clearly stating the evaluation process for limits from both sides. The conclusion is that the limit as x approaches -1 for f(x) is indeed 1, provided the proper notation is used.
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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 \displaystyle \lim_{x\to-1}\,f(x)=1\,, 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 \displaystyle \lim_{x\to-1^-}\ f(x)=\lim_{x\to-1^+}\ f(x)\,. If that is true then that common result is the limit you're looking for.

While it is true that \displaystyle \lim_{x\to-1^-}\ f(x)= \lim_{x\to-1}-x^2+2\,, and \displaystyle \lim_{x\to-1^+}\ f(x)= \lim_{x\to-1}4x+5\,, you really should state that this is what you're doing.
 
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SammyS said:
While it is true that \displaystyle \lim_{x\to-1}\,f(x)=1\,, 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 \displaystyle \lim_{x\to-1^-}\ f(x)=\lim_{x\to-1^+}\ f(x)\,. If that is true then that common result is the limit you're looking for.

While it is true that \displaystyle \lim_{x\to-1^-}\ f(x)= \lim_{x\to-1}-x^2+2\,, and \displaystyle \lim_{x\to-1^+}\ f(x)= \lim_{x\to-1}4x+5\,, 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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