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in Continuity and Differentiability by (27.4k points)
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Find all the points of discontinuity of f defined by 

f(x) = |x| – |x + 1|.

1 Answer

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Best answer

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, 

We can summarise it as a function is continuous at x = c 

if :

NOTE :

Definition of mod function : 

|x| = \( \begin{cases} -x,x <{0}\\ x,x≥{0} \end{cases} \) 

Here we have, 

f(x) = |x| – |x + 1| 

f(x) rewritten using idea of mod function :

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]

[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]

[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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