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August 23, 2014

August 23, 2014

Posted by **Anon** on Friday, July 12, 2013 at 1:05pm.

|x+1|<=|x-3|

I promise I'm not posting my whole homework assignment. These are questions that I skipped and have no clue how to do out of a huge packet.

- AP Calc -
**Steve**, Friday, July 12, 2013 at 2:52pmThe key to absolute value problems is to remember the definition of |n|.

|n| = n if n >= 0

|n| = -n if n < 0

So, here we have

|x+1|<=|x-3|

if (x-3) >= 0, |x-3| = x-3

Also, since that means x > 3, x+1 >= 0, so |x+1| = x+1 and we have

x+1 <= x-3

1 < -3

not possible

If (x-3) < 0, (that is, x<3) we have

|x+1| <= -(x-3)

Now, if x >= -1, x+1 >= 0, and we have

x+1 <= -x+3

2x <= 2

x <= 1

But, we required above that x >= -1, so we get a solution set -1 <= x <= 1

If (x-3) < 0 and (x+1) < 0 we have

-(x+1) <= -(x-3)

-x-1 <= -x+3

-1 < 3

So x < -1 is also a solution.

Combining the solution sets, we see that

x <= 1 is the complete solution set

To see how this works graphically, visit

http://rechneronline.de/function-graphs/

and enter

abs(x+1)

abs(x-3)

as your two functions, and set the domain for x from -5 to 5, and the range y from 0 to 5

This is an excellent place to try out viewing function graphs to confirm your algebra.

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