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# Find all the points of discontinuity of f defined by f (x) = | x| - | x + 1|.

34. Find all the points of discontinuity of f defined by $f (x) = | x| - | x + 1|.$

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Given function is
$f (x) = | x| - | x + 1|$
Let g(x) = |x|  and h(x)  = |x+1|
Now,
$g(x)\begin{cases} -x & \text{ if } x<0 \\ 0 & \text{ if } x= 0\\ x& \text{ if } x>0 \end{cases}$
g(x) is defined for all real numbers k
case(i)  k < 0
$g(k) = -k\\ \lim_{x\rightarrow k}g(x) = -k\\ \lim_{x\rightarrow k}g(x) = g(k)$
Hence, g(x) is continuous when k < 0

case (ii) k > 0
$g(k) = k\\ \lim_{x\rightarrow k}g(x) = k\\ \lim_{x\rightarrow k}g(x) = g(k)$
Hence, g(x) is continuous when k > 0

case (iii) k = 0
$g(0) = 0\\ \lim_{x\rightarrow 0^-}g(x) = -x = 0\\ \lim_{x\rightarrow 0^+}g(x ) = x = 0\\ \lim_{x\rightarrow 0^-}g(x) = g(0) = \lim_{x\rightarrow 0^+}g(x )$
Hence, g(x) is continuous when k = 0
Therefore, g(x) = |x| is continuous for all real values of x

Now,
$h(x)\begin{cases} -(x+1) & \text{ if } x<-1 \\ 0 & \text{ if } x= -1\\ (x+1)& \text{ if } x>-1 \end{cases}$
g(x) is defined for all real numbers k
case(i)  k < -1
$h(k) = -(k+1)\\ \lim_{x\rightarrow k}h(x) = -(k+1)\\ \lim_{x\rightarrow k}h(x) = h(k)$
Hence, h(x) is continuous when k < -1

case (ii) k > -1
$h(k) = k+1\\ \lim_{x\rightarrow k}h(x) = k+1\\ \lim_{x\rightarrow k}h(x) = h(k)$
Hence, h(x) is continuous when k > -1

case (iii) k = -1
$h(-1) = 0\\ \lim_{x\rightarrow -1^-}h(x) = -(x-1) = 0\\ \lim_{x\rightarrow -1^+}h(x ) = x+1 = 0\\ \lim_{x\rightarrow -1^-}h(x) = h(0) = \lim_{x\rightarrow -1^+}h(x )$
Hence, h(x) is continuous when k = -1
Therefore, h(x) = |x+1| is continuous for all real values of x
g(x) is continuous and h(x) is continuous
Therefore, f(x) = g(x) - h(x) = |x| - |x+1| is also continuous

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