##### Find all the points of discontinuity of f defined byf(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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