Question

Find points of discontinuity of f defined by f (x) = |x + 1| - |x + 2| + |x - 1|​

Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm \: f(x) = |x + 1| - |x + 2| + |x - 1|$

Let first define the function f(x).

We know, By definition of Modulus function, we have

$\begin{gathered}\begin{gathered}\bf\: |x| = \begin{cases} &\sf{ - x \: \: if \: x < 0} \\ \\ &\sf{ \: \: x \: \: if \: x \geqslant 0} \end{cases}\end{gathered}\end{gathered}$

So, using this definition, f(x) is defined as

$\begin{gathered}\begin{gathered}\bf\: f(x) = \begin{cases} &\sf{ - (x + 1) + (x + 2) - (x - 1) \: \: if \: x \leqslant - 2} \\ &\sf{ - (x + 1) - (x + 2) - (x - 1) \: \: if \: - 2 < x \leqslant - 1}\\ &\sf{(x + 1) - (x + 2) - (x - 1) \: \: if \: - 1 < x \leqslant 1}\\ &\sf{(x + 1) - (x + 2) + (x - 1) \: \: if \: x > 1} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf\: f(x) = \begin{cases} &\sf{ 2 - x \: \: if \: x \leqslant - 2} \\ &\sf{ - (2 + 3x) \: \: if \: - 2 < x \leqslant - 1}\\ &\sf{ - x \: \: if \: - 1 < x \leqslant 1}\\ &\sf{x - 2 \: \: if \: x > 1} \end{cases}\end{gathered}\end{gathered}$

So, breaking points are -2, - 1 and 1

So, we have to check continuity at x = - 2, x = - 1, x = 1

Case :- 1 Continuity at x = - 2

$\rm \: f( - 2) = 2 - ( - 2) = 4$

Left Hand Limit at x = - 2

$\displaystyle\lim_{x \to -2^-}\rm f(x)$

$\rm \: = \: \displaystyle\lim_{x \to -2^-}\rm (2 - x)$

To evaluate this limit, we use method of Substitution,

So Substitute

$\rm \: x = - 2 - h, \: \: \: as \: x \: \to \: - 2, \: \: so \: h \: \to \: 0$

So, on substituting, we get

$\rm \: = \: \displaystyle\lim_{h \to 0}\rm [2 - ( - 2 - h)]$

$\rm \: = \: \displaystyle\lim_{h \to 0}\rm [2 + 2 + h]$

$\rm \: = \: 4$

Right Hand Limit at x = - 2

$\displaystyle\lim_{x \to -2^ + }\rm f(x)$

$\rm \: = \: \displaystyle\lim_{x \to -2^-}\rm - (2 + 3x)$

So, Substitute

$\rm \: x = - 2 + h, \: \: \: as \: x \: \to \: - 2, \: \: so \: h \: \to \: 0$

$\rm \: = \: \displaystyle\lim_{h \to 0}\rm - [2 + 3( - 2 + h)]$

$\rm \: = \: - (2 - 6)$

$\rm \: = \: 4$

$\rm\implies \:f( - 2) = \displaystyle\lim_{x \to -2^-}\rm f(x) = \displaystyle\lim_{x \to -2^ + }\rm f(x)$

$\rm\implies \:f(x) \: is \: continuous \: at \: x = - 2$

Case :- 2 Continuity at x = - 1

$\rm \: f( - 1) = - (2 - 3) = 1$

Left Hand Limit at x = - 1

$\displaystyle\lim_{x \to -1^-}\rm f(x)$

$\rm \: = \: \displaystyle\lim_{x \to -1^-}\rm - (2 + 3x)$

So, Substitute

$\rm \: x = - 1 - h, \: \: \: as \: x \: \to \: - 1, \: \: so \: h \: \to \: 0$

$\rm \: = \: \displaystyle\lim_{h \to 0}\rm - [2 + 3( - 1 - h)]$

$\rm \: = \: - (2 - 3)$

$\rm \: = \: 1$

Right Hand Limit at x = - 1

$\displaystyle\lim_{x \to -1^ + }\rm f(x)$

$\rm \: = \: \displaystyle\lim_{x \to -1^ + }\rm - x$

So, Substitute

$\rm \: x = - 1 + h, \: \: \: as \: x \: \to \: - 1, \: \: so \: h \: \to \: 0$

$\rm \: = \: \displaystyle\lim_{h \to 0}\rm - ( - 1 + h)$

$\rm \: = \: 1$

$\rm\implies \:f( - 1) = \displaystyle\lim_{x \to -1^-}\rm f(x) = \displaystyle\lim_{x \to -1^ + }\rm f(x)$

$\rm\implies \:f(x) \: is \: continuous \: at \: x = - 1$

Case :- 3 Continuity at x = 1

$\rm \: f(1) = - 1$

Left Hand Limit at x = 1

$\displaystyle\lim_{x \to 1^-}\rm f(x)$

$\rm \: = \: \displaystyle\lim_{x \to 1^-}\rm - x$

So, Substitute

$\rm \: x = 1 - h, \: \: \: as \: x \: \to \: 1, \: \: so \: h \: \to \: 0$

$\rm \: = \: \displaystyle\lim_{h \to 0}\rm - (1 - h)$

$\rm \: = \: - \: 1$

Right Hand Limit at x = 1

$\displaystyle\lim_{x \to 1^ + }\rm f(x)$

$\rm \: = \: \displaystyle\lim_{x \to 1^ + }\rm (x - 2)$

So, Substitute

$\rm \: x = 1 + h, \: \: \: as \: x \: \to \: 1, \: \: so \: h \: \to \: 0$

$\rm \: = \: \displaystyle\lim_{h \to 0}\rm (1 + h - 2)$

$\rm \: = \: - \: 1$

$\rm\implies \:f(1) = \displaystyle\lim_{x \to 1^-}\rm f(x) = \displaystyle\lim_{x \to 1^ + }\rm f(x)$

$\rm\implies \:f(x) \: is \: continuous \: at \: x = 1$

So, from this, we get

$\rm\implies \:f(x) \: is \: continuous \: at \: x = - 2, - 1, 1$

$\rm\implies \:f(x) \: is \: continuous \: everywhere$

$\rm\implies \:There \: is \: no \: point \: of \: discontinuity.$