Find all the points of discontinuity of f defined by

f(x) = |x| – |x + 1|.

Basic Idea:

A real function f is said to be continuous at x = c, where c is any point in the domain of f if :

where h is a very small ‘+ve’ no.

i.e. left hand limit as x → c (LHL) = right hand limit as x → c (RHL) = value of function at x = c.

This is very precise, using our fundamental idea of the limit from class 11 we can summarise it as a function is continuous at x = c if :

NOTE:

Here we have,

f(x) = |x| – |x + 1|

f(x) rewritten using idea of mod function:

…….equation 1

Clearly for x < –1 , f(x) = constant and also for x > 1 f(x) is constant

∴ in these regions f(x) is everywhere continuous.

For –1 < x < 0, plot of graph is a straight line as in this region f(x) is given by linear polynomial

∴ it is also continuous here.

∵ function is changing its expression at x = –1 and x = 0, so we need to check continuities at these points.

At x = –1 :

f(–1) = 1 [using equation 1]

LHL = [using equation 1]

RHL =

[using equation 1]

Clearly,

LHL = RHL = f(–1)

∴ it is continuous at x = –1

At x = 0 :

f(0) = –2*0–1 = –1 [using equation 1]

LHL =

[using equation 1]

RHL =

[using equation 1]

Clearly,

LHL = RHL = f(0)

∴ it is continuous at x = 0

Hence,

f(x) is continuous everywhere in its domain.

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