Answer :

Given function f(x) = |x| + |x - 1|


A function f(x) is said to be continuous on a closed interval [a, b] if and only if,


(i) f is continuous on the open interval (a, b)


(ii)


(iii)


Let’s check continuity on the open interval (-1, 2)


As -1 < x < 2


Left hand limit:



=|-1-0| + |(-1-0) – 1|


=1 + 2


= 3


Right hand limit:



=|2| + |2 – 1|


=2+1


= 3


Left hand limit = Right hand limit


Here a = -1 and b = 2


Therefore,



= |-1 + 0| + |(-1 + 0) - 1|


= |- 1| + |-1 – 1|


= 1 + 2 = 3


Also f(-1) = |-1| + |-1 - 1| = 1 + 2 = 3


Now,



= |2 - 0| + |(2 - 0) - 1|


= | 2| + |2 – 1|


= 2 + 1 = 3


Also f(2) = |2| + |2 - 1| = 2 + 1 = 3


Therefore,


f(x) is continuous on the closed interval [-1, 2].


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