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].