Asked by Shreya
                the absolute value function 
g(x) = |5-x| can be written as a piecewise function. the piecewise function which represents g(x) is
this have to be g(x) = x-5 if x be greater to or equal to five. and the negative piece have to be 5-x. does it matter if i write -5+x instead of 5-x?
which of following is root of equation |3x-1| = -5
i get -4/3 for this.
            
        g(x) = |5-x| can be written as a piecewise function. the piecewise function which represents g(x) is
this have to be g(x) = x-5 if x be greater to or equal to five. and the negative piece have to be 5-x. does it matter if i write -5+x instead of 5-x?
which of following is root of equation |3x-1| = -5
i get -4/3 for this.
Answers
                    Answered by
            Damon
            
    from x = -oo to x = 5
g = -5 + x
from x = 5 to x = oo
g = x - 5
3x-1 = -5
3 x = -4
x = -4/3
    
g = -5 + x
from x = 5 to x = oo
g = x - 5
3x-1 = -5
3 x = -4
x = -4/3
                    Answered by
            Shreya
            
    thank you for the 3x-1 = -5 it says in back of book that there are no roots of the equation |3x-1| = -5 and i not see how that be possible.
    
                    Answered by
            Reiny
            
    By definition
| anything | has to be a positive number or at least zero
so |3x-1\ = -5 would contradict the very definition of absolute value,
So without even showing a single step, we can say that there is no solution.
if you verify your answer of -4/3, ...
LS = |3(-4/3) - 1|
= | -5|
= +5
RS = -5
LS ≠ RS , so .... no solution
    
| anything | has to be a positive number or at least zero
so |3x-1\ = -5 would contradict the very definition of absolute value,
So without even showing a single step, we can say that there is no solution.
if you verify your answer of -4/3, ...
LS = |3(-4/3) - 1|
= | -5|
= +5
RS = -5
LS ≠ RS , so .... no solution
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