I have a function:
f(x) = (x-2) / ( |x| - 2)
why is it only at x=2 is removable but not x= - 2?
2007-02-19 12:06:53 · 3 answers · asked by Brian S 1 in Education & Reference ➔ Homework Help
So when you plug in 2 this happens:
(2-2)/(2-2) = 0/0
when you plug in -2 this happens:
(-2-2)/2-2) = -4/0
The first is removable because the same number is on the top and bottom...you can think of it as any number divided by itself is 1.
This second is an asymptote. The top is fine...the bottom is what give you the problem.
2007-02-19 12:19:16 · answer #1 · answered by theFo0t 3 · 0⤊ 0⤋
at x = 2, there is a hole in the graph, and so you can remove the discontinuity by redefining the equation as f(x) when x doesn't =2, and 1 when x =2. however, at x = -2, you have f(x) going to different infinites from the left and right. this provides a verticle asymptote, which cannot be removed. no matter how you redefine this equation, you can't remove the discontinuity at x = -2.
does that help a little?
2007-02-19 20:13:27 · answer #2 · answered by Anonymous · 1⤊ 0⤋
-2-2 = -4.
-4/0 is not a rational number.
2007-02-19 20:12:26 · answer #3 · answered by iketronic 2 · 0⤊ 0⤋