Question

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

Answer 1

f(x) can be redefined as,

$f\left(x\right)=\{\begin{array}{ccccccc}1& & & & & & x\le -1\\ -2x-1& & & & & & x\in (-1,0)\\ -1& & & & & & x\ge 0\end{array}$

From redefined f(x) we can easily claim that f(x) is continuous everywhere except $-1and0$ ( To be checked ) .

Let's first check continuity of f(x) at $x=-1$,

LHL of f(x) at $x=-1\to {lim}_{h\to 0}f(-1-h)=1$

RHL of f(x) at $x=-1\to {lim}_{h\to 0}f(-1+h)={lim}_{h\to 0}\{-2(-1+h)-1\}=1$

Since LHL=RHL, f(x) is continuous at $x=-1$ .

Now let's check continuity of f(x) at $x=0$,

LHL of f(x) at $x=0\to {lim}_{h\to 0}f(0-h)={lim}_{h\to 0}\{-2(0-h)-1\}=-1$

RHL of f(x) at $x=0\to {lim}_{h\to 0}f(0+h)=-1$

Since LHL=RHL, f(x) is also continuous at $x=0$

Thus , f(x) is continuous for all Real numbers .