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# Question Watch

1. (Original post by Zenarthra)
Yeah i understand now, just out of curiosity what would this graph look like?
Different straight lines joined together at the boundaries of each region e.g. 0 < x < 1 etc
2. (Original post by davros)
Different straight lines joined together at the boundaries of each region e.g. 0 < x < 1 etc
Just 1 question, i got x=2, x=-1, x=-2 but why is the solution only x=-2?
Thanks!
3. (Original post by Zenarthra)
Just 1 question, i got x=2, x=-1, x=-2 but why is the solution only x=-2?
Thanks!
Substitute your values back into the original equation. Only x = -2 works!
4. (Original post by davros)
Well, how else would you do it?

The point is that if you look at different regions you can replace modulus signs by + or - as appropriate e.g. if you have something like |x + 1| then in the region x < - 1 this is -(x + 1) whereas elsewhere it is just x + 1.
So say that in the region 0<X<1 that f(x)=-2x-2, is this how the original equation would look like y=-2x-2 for that particular region?
5. (Original post by Zenarthra)
So say that in the region 0<X<1 that f(x)=-2x-2, is this how the original equation would look like y=-2x-2 for that particular region?
Yes

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