apparently

you see I know how they got the 5

but for the 3 all they did was say

and for some reason that lead them to 3

MAT196F Midterm Test Solutions

it's kind of hard to read but it's the first question

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- December 8th 2007, 06:39 PMakhayoonEpsilon Delta Proof

apparently

you see I know how they got the 5

but for the 3 all they did was say

and for some reason that lead them to 3

MAT196F Midterm Test Solutions

it's kind of hard to read but it's the first question - December 8th 2007, 06:55 PMJhevon
- December 8th 2007, 07:02 PMakhayoon
I know they did that

but they said that if

so this give you permission to turn the |2x-1| into a five but to get rid of the

all they said is that it's bigger than one so there

so how does that give them permission to make the that polynomial equal 1

?? - December 8th 2007, 07:17 PMJhevon
- December 8th 2007, 07:28 PMakhayoon
I'm sorry, let me try to be a little more clear, it's kind of hard with these kinds of questions though, since I don't understand whats really going on and I just kind of try to work out patterns from the solutions of all the other questions I've solved like these:)

|2x-1|<5 so that means that I could use 5 instead of this expression

I know they got this from sweet...

and then they go

but where did this come from did they use again somehow?? :eek: - December 8th 2007, 07:48 PMJhevon
- December 8th 2007, 07:56 PMakhayoon
ok thanks, that explains it, so why couldn't I just make

|2x+1|>1 make x=0 and then see that it's bigger than one and make that one as well?

why did I have to go through all that -1<x-1<1 that led me to |2x-1|<5?:( - December 8th 2007, 08:13 PMJhevon
remember, we stated out with |x - 1|< 1, which is why things played out that way. besides, if x = 0, |2x + 1| is not bigger than 1. we need to make x a range of values here, as we wanted inequalities. the actual value doesn't matter that much. we could have made |x - 1| < 1/2 or whatever. but 1 looks nice