Let f : [--1,3] [ ] R be defined as [f(x) = pmatrix | x | + [ x ] & , & - 1 x

Question

Let f : [--1,3] $$ \to $$ R be defined as

$$f(x) = \left\{ {\begin{pmatrix} {\left| x \right| + \left[ x \right]} & , & { - 1 \le x < 1}  \
   {x + \left| x \right|} & , & {1 \le x < 2}  \
   {x + \left[ x \right]} & , & {2 \le x \le 3}  \ \end{pmatrix} } \right.$$

where [t] denotes the greatest integer less than
or equal to t. Then, f is discontinuous at:

Accepted Answer

$$f(x) = \left\{ {\begin{pmatrix} {\left| x \right| + \left[ x \right]} & , & { - 1 \le x < 1}  \
   {x + \left| x \right|} & , & {1 \le x < 2}  \
   {x + \left[ x \right]} & , & {2 \le x \le 3}  \

\end{pmatrix} } \right.$$

= $$ = \left\{ {\begin{pmatrix} { - x - 1,} & { - 1 \le x < 0}  \
   {x,} & {0 \le x < 1}  \
   {2x,} & {1 \le x < 2}  \
   {x + 2,} & {2 \le x < 3}  \
   {6,} & {x = 3}  \

\end{pmatrix} } \right.$$

f(-1) = $$\mathop {\lim }\limits_{x \to  - 1} \left( { - x - 1} \right)$$ =  -( -1) - 1 = 0

f(-1+) = $$\mathop {\lim }\limits_{x \to  - {1^ + }} \left( { - x - 1} \right)$$ = 0

$$ \therefore $$ f(x) is continuous at x = - 1

f(0-) =$$\mathop {\lim }\limits_{x \to {0^ - }} \left( { - x - 1} \right)$$

=  -(0) - 1 = - 1

f(0) = $$\mathop {\lim }\limits_{x \to 0} \left( x \right)$$ = 0

f(0+) = $$\mathop {\lim }\limits_{x \to {0^ + }} \left( x \right)$$ = 0

$$ \therefore $$ f(x) is discontinuous at x = 0

f(1-) = $$\mathop {\lim }\limits_{x \to {1^ - }} \left( x \right)$$ = 1

f(1) = $$\mathop {\lim }\limits_{x \to 1} \left( {2x} \right)$$ = 2

f(1+) = $$\mathop {\lim }\limits_{x \to {1^ + }} \left( {2x} \right)$$ = 2

$$ \therefore $$ f(x) is discontinuous at x = 1

f(3-) = $$\mathop {\lim }\limits_{x \to {3^ - }} \left( {x + 2} \right)$$ = 3 + 2 = 5

f(3) = 6

$$ \therefore $$ f(x) is discontinuous at x = 3

So, f(x) is discontinuous at x = {0, 1, 3}.

Solutions · 1