# Examine the continuity of this absolute value function

1. Aug 19, 2014

### needingtoknow

1. The problem statement, all variables and given/known data

y = 1-abs(x) / abs(1-x)

3. The attempt at a solution

For x < 0, abs(x) = -x

y = (1+x) / -(1-x)
= -(1+x)/(1-x)

I stopped here because this is the part I got wrong. For x < 0, my solutions manual got (1+x) / (1-x).
What did I do wrong?

2. Aug 19, 2014

### LCKurtz

For $x<0$, $-x>0$ so $1-x>0$ so $|1-x|=1-x$. No minus sign needed in the denominator.

3. Aug 19, 2014

### Ray Vickson

$$y = 1 - \frac{|x|}{|1-x|}$$
you have $y = 1 + x/(1-x) = 1/(1-x)$ for $x < 0$.

On the other hand, if you really meant something different from what you wrote, namely,
$$y = \frac{1-|x|}{|1-x|}$$
then you would have $y = (1+x)/(1-x)$ for $x < 0$. Which function do you mean? Us parentheses for clarity, like this: y = (1-|x|)/|1-x|, or y = (1-abs(x))/ abs(1-x) if your keyboard does not have an "|" key.

4. Aug 20, 2014

### needingtoknow

It was the second one: y = (1-|x|)/|1-x|. So how did you get y=(1+x)/(1−x) for x<0?

5. Aug 20, 2014

### LCKurtz

6. Aug 20, 2014

### needingtoknow

Since -x > 0 can be written as x < 0, therefore -(1-x) > 0 can also be written as 1-x < 0. If this is true then can't this apply to any time we need to consider the negative case for an absolute value function. For the numerator, why can't this same logic apply?

7. Aug 20, 2014

### Ray Vickson

Have you forgotten that -(-x) = x and that for x < 0 we have |x| = -x?

8. Aug 20, 2014

### LCKurtz

I can't make any sense out of that reply. Why don't you just follow the steps I gave you? You start with $x<0$. Multiply both sides by $-1$ so $-x>0$. Since $-x$ is positive, adding $1$ to it is still positive, so $1-x > 0$. The absolute value of a positive number is the number itself so $|1-x| = 1-x$. You don't get $-(1-x)$.

9. Aug 20, 2014

### needingtoknow

My mind must have been somewhere else when I was reading your first reply. It makes crystal clear sense. Thank you for your time.