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Its all about understanding what work ‘modulus’ functions perform.
So |x-1| has its critical point at x=1 (the point where it’s graph’s behaviour change) and |x-2|has its critical point at x=2.
So we have to solve separately in three regions that are at x>=2, 1
x-1+x-2=2x-3 whise range is[1,3) since x>=2.
Both |x-1| and |x-2| are positive so mdulus function does nothing,
At x>=2,
-(x-1)+(x-2)=-1 so its range=nothing since x is always between 1 and 2.
|x-1| is negative while |x-2| is positive so mdulus function adds minus sign to |x-1| and leaves |x-2| as it is,
At 1
-(x-1)-(x-2)=-2x+3 whose range is (1,5)
|x-1| and |x-2| both are negative so modulus functionjust add a negative sign to it,i.e.
At x,
x.
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