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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