Examine the continuity of this absolute value function

[needingtoknow]

Homework Statement

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

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?

 

[LCKurtz]

Homework Statement

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

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?

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

 

[Ray Vickson]

Homework Statement

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

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?

For your written function
$$y = 1 - \frac{|x|}{|1-x|}$$
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.

 

[needingtoknow]

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

 

[LCKurtz]

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

Didn't you read post #2?

 

[needingtoknow]

Sorry I should have addressed your reply as well. So LCKurtz let me start off by saying what I understand from your reply.

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

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?

 

[Ray Vickson]

Sorry I should have addressed your reply as well. So LCKurtz let me start off by saying what I understand from your reply.



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?

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

 

[LCKurtz]

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

Sorry I should have addressed your reply as well. So LCKurtz let me start off by saying what I understand from your reply.



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?

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 $-(1-x)$.

 

[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.