How do I calculate x(t)+x(-t) for t<0 and t>=0?

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dervast
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Hi i have to read mathematics over 7 years and now i can't even solve a simple equation

Homework Statement


I hav3e the following
x(t)=-3t for t<0 and
x(t)=t t>=0
I want to calculate the x(t)+x(-t) for t<0 firstloy and then for t>=0

Homework Equations




The Attempt at a Solution


For t<0
x(t)+x(-t)=-3t+t (x(t)=-3t cause x is negative x(-t)=t cause the negative of the negative -t is the t which is t)
The problem is that my teachers book says that x(t)+x(-t)=-3t-t and i can't understand how he calculates that

Now for t>=0 x(t)+x(-t)=t-3t (x(t)=t cause it is positive and x(-t)=-3t cause the negative of t>=0 is the t<0 which makes x(t)=-3t..
My teacher again argues with me cause he calculates in his note
x(t)+x(-t) as =t+3t
So what i am doing wrong ?:(
 
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dervast said:
For t<0
x(t)+x(-t)=-3t+t (x(t)=-3t cause x is negative x(-t)=t cause the negative of the negative -t is the t which is t)
Since t < 0, it follows that -t > 0, hence x(-t) = -t. I can't understand your reasoning for why x(-t) = t when t < 0.
Now for t>=0 x(t)+x(-t)=t-3t (x(t)=t cause it is positive
t is not necessarily positive. It could be exactly 0.
and x(-t)=-3t cause the negative of t>=0 is the t<0 which makes x(t)=-3t..
Again, I don't know what you're doing. When t is non-negative, -t is non-positive, so either -t = 0 or -t < 0. If -t = 0, then x(-t) = 0 = 3(0) = 3t. If -t < 0, then x(-t) = -3(-t) = 3t.
 
thx for your answer one more question then
y(t)=2t+2 for -1<=t<0
y(-t)=-2t+2 for ? i first multiple with - or first i convert the > to < ?
 
If y(t)=2t+2 for -1<=t<0 is the only definition you are given, then in order to be able to find f(-t) using that definition you must have, multiplying each part by -1, -(0)< -(-t)<= -(-1). In other words, 0< t< 1. And, in that case, y(-t)= 2(-t)+ 2.
 
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