Use a graph to fine a number delta such that

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/square rootsign 4x+1/ <.5 /X-2/ < delta

2 is x and 3 is y, 3 goes to 3.5 and 2.5, i need help with the absolute value of the square root sign 4x+1
 
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Is this what you want?

[tex]\sqrt{4x+1} < \frac{x-2}{2} < \delta[/tex]
 
O wait, last post I assumed the //'s were your grouping symbols. I think theyre absolulote value signs >.<

[tex]|\sqrt{4x+1}| < \frac{|x-2|}{2} < \delta[/tex]

I hope that's what you want.
 
This is the second time you have posted this. Thanks to Gib Z for writing out the formula (you forgot the second "|" on the square root). But I still have no idea what "2 is x and 3 is y, 3 goes to 3.5 and 2.5" could mean! 3 doesn't "go" anywhere- it just sits there! And there is no "y" in your expression.
 
To Halls- Look abit closer :p its there.

To OP- I still have no idea what it is you want.
 
i should write it exactly