anonymous
  • anonymous
How do you graph this? g (x) defined piecewise: 2 for x <= -2 -x for -2 < x < 1 2 for x = 1 -1/x for 1 < x < 2 3 for x >= 2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Yes I do!
anonymous
  • anonymous
Can you draw it for me really fast?
anonymous
  • anonymous
Wait so I figured out how to graph it but now I am having troubles trying to figure out if it is continuous at x=-2,1,2

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anonymous
  • anonymous
& if not where are the discontinuities at
anonymous
  • anonymous
whats the definition of continuous?
anonymous
  • anonymous
|dw:1378592577375:dw|
anonymous
  • anonymous
I believe it looks something like this
anonymous
  • anonymous
forms an unbroken whole; without interruption.
phi
  • phi
I would not put an open circle at (-2,2). The function is defined there
anonymous
  • anonymous
^ because of this 2 for x <= -2
anonymous
  • anonymous
oh so it is a value... oh ok
anonymous
  • anonymous
same with (1,-1) ?
anonymous
  • anonymous
So x=-2 is continuous and x=1 is cont and x=2 is jump disc
phi
  • phi
the function has a value at (1,2) so you put an open circle at (1,-1) the < operator means this -1/x for 1 < x < 2 does not define g(1) so it is an open circle at the "edges" the filled circle goes at (1,2)
anonymous
  • anonymous
So x=-2 is continuous and x=1 is cont and x=2 is jump disc
anonymous
  • anonymous
I don't think x=1 is continuous....
anonymous
  • anonymous
Hole!
phi
  • phi
x=-2 is continuous but at x just a little less than 1 g(x)= -1 and at x=1 g(1)= 2 that is a jump
anonymous
  • anonymous
So x=-2 is continuous and x=1 Hole Disc and x=2 is jump disc
phi
  • phi
yes that sounds good
anonymous
  • anonymous
I think it looks right....
anonymous
  • anonymous
Oh cool thank y'all~!
anonymous
  • anonymous
These are fun! If you have more post them here!
anonymous
  • anonymous
okay sounds good!

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