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Mathematics 10 Online
OpenStudy (anonymous):

Give the definition of the modulus function R -> R, often denoted by x -> |x|. (This is also called the absolute value function.) Draw the graph of the function R -> R defined by x -> ||x - 1| -2|. Express {x e R ||x-2| < 3g as} an open interval (a; b) for suitable a; b e R.

OpenStudy (anonymous):

try this graph http://www.wolframalpha.com/input/?i= ||x-1|-2|

OpenStudy (anonymous):

First question : (PNG image) http://tinyurl.com/3sql2qk

OpenStudy (anonymous):

Or more correctly according to the question : http://tinyurl.com/3g98cew

OpenStudy (anonymous):

cheers guys, any ideas how to do the last part?

OpenStudy (anonymous):

is the graph clear? because i can send it step by step

OpenStudy (anonymous):

\[|x-1|\]

OpenStudy (anonymous):

\[|x-1|-2\]

OpenStudy (anonymous):

thanks for that, i get it.

OpenStudy (anonymous):

any ideas for last part?

OpenStudy (anonymous):

\[||x-1|-2|\]

OpenStudy (anonymous):

i don't really understand what the last one says. are you supposed to write this as a piecewise function?

OpenStudy (anonymous):

it says "express \[|x-2|<3g\] as an open interval?

OpenStudy (anonymous):

i am confused by the question. if that is what it is asking, then i think the answer is easy. if g = 0 you get x = 2 if g < 0 no answer if g > 0 you solve \[|x-2|<3g\] \[-3g<x-2<3g\] \[2-3g<x<2+3g\] so open interval is \[(2-3g,2+3g)\]

OpenStudy (anonymous):

yes, sorry

OpenStudy (anonymous):

ok then that is the answer above.

OpenStudy (anonymous):

thanks very much, i get it now. appreciate it

OpenStudy (anonymous):

yw

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