what would be the range of f(x)=|x-1|+|x-2| x belongs to [-1,3]

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.