Absolute Value and Converting to Expanded Form: Explained

[trap101]

Just a general question with absolute values:

Is it possible to have an absolute value of this form:

| f(x) - L | < -L (the minus sign is meant to be there) and if so how can I convert it into expanded form? i.e: -L < f(x) - L < -L or something of that form.

 

[LCKurtz]

Just a general question with absolute values:

Is it possible to have an absolute value of this form:

| f(x) - L | < -L (the minus sign is meant to be there) and if so how can I convert it into expanded form? i.e: -L < f(x) - L < -L or something of that form.

If L is meant to signify a positive number, then the inequality can't hold because no absolute value is negative. If L might be negative, then there is no problem. An inequality $|f(x) - a|<b$ is always equivalent to $-b < f(x)-a<b$.

 

[lol_nl]

If L is negative, then your problem reduces to (substitute K = -L, K positive) |f(x) + K| &lt; K which holds if -2K &lt; f(x) &lt; 0.

 

[trap101]

If L is negative, then your problem reduces to (substitute K = -L, K positive) |f(x) + K| &lt; K which holds if -2K &lt; f(x) &lt; 0.

So the fact that there is a 2 with the K does not affect the inequality? Wouldn't I have to get rid of the 2 to make it a standard statement?

 

[lol_nl]

So the fact that there is a 2 with the K does not affect the inequality? Wouldn't I have to get rid of the 2 to make it a standard statement?

Let me write it out more explicitly:
|f(x) + K| &lt; K means
-K &lt; f(x) + K &lt; K.
(compare with the first reply).
Now substract K from all three terms to get:
-2K &lt; f(x) &lt; 0.

Hence the 2 is simply the result of the calculation.

 

[trap101]

Ok, I was just confirming if that's all it meant. Thanks.