WholeLottaLove wrote:
Hello, thanks for the response.
My reasoning came from this:
find-the-value-of-135023.html post, which I thought was a similar situation. In it, a-1 can be a positive or negative value, thus causing the signs to flip accordingly.
I do not understand why if -3<x<3, P=(x+3)+(x-3)=2x and I am also not understanding the relation to the graph. Sorry...absolute value problems and concepts have proven to be very difficult to understand.
That question is similar and I am gonna apply the same method here, so:
How many values can the integer \(p=|x+3|-|x-3|\) assume?
I think that we both agree that if \(x>3\), both \(|x+3|-|x-3|\) will be
positive so we can eliminate the || signs and keep \(x+3-(x-3)=6\).
Same reasoning for \(x<-3\): if that's the case then both \(|x+3|-|x-3|\) will be
negative so we must "flip" the sign according to the rule
\(|-value|=+value\). We obtain \(-(x+3)-(-)(x-3)=-6\).
I hope that this so far makes sense.
Now, what if -3<x<3? In this case the term \(|x+3|\) will be
positive (you can try with values -2+3>0 for example).
The other term will be
negative \(|x-3|\) (-2-3<0 for example). The first one will be positive=> the abs value will not affect the sign; The second one will be negative=> the abs value will turn it positive. Lets write this down:
\(p=(x+3)-(-)(x-3)=2x\) - if -3<x<3
Now we have
for x>3 we have P=(x+3)-(x-3)=6, for -3<x<3 P=(x+3)+(x-3)=2x, for x<-3 P=-(x-3)+(x-3)=-6
the graph just represents those equations. For example if x>3 the graph shows you a line P=6