Question

Find the points of discontinuity, if any, of the following functions:  \[f\left( x \right) = \begin{cases}- 2 , & \text{ if }& x \leq - 1 \\ 2x , & \text{ if } & - 1 < x < 1 \\ 2 , & \text{ if }  & x \geq 1\end{cases}\]

Answer 1

The given function f is \[f\left( x \right) = \begin{cases}- 2 , & \text{ if }& x \leq - 1 \\ 2x , & \text{ if } & - 1 < x < 1 \\ 2 , & \text{ if }  & x \geq 1\end{cases}\] 

The given function is defined at all points of the real line.

Let c be a point on the real line.

Case I:

` " If " c < -1 " then " f(c)= -2 " and " lim_(x->c)(x)=lim_(x->c)(-2)=-2`

`∴lim_(x->c)f(x)=f(c)`

Therefore, f is continuous at all points x, such that x < −1

Case II:

` " If c =1 then "  f(c)=f(-1)=-2`

The left hand limit of f at x = −1 is,

`lim_(x->-1)f(x)=lim_(x->-1)f(-2)=-2`

The right hand limit of f at x = −1 is,

`lim_(x->-1)f(x)=lim_(x->-1)f(2x)=2xx(-1)=-2`

`∴lim_(x->-1)f(x)=f(-1)`

Therefore, f is continuous at x = −1

Case III:

` " if -1 < c < 1,then " f(c)=2c`

`lim_(x->c)f(x)=lim_(x->c)f(2x)=2c`

`∴lim_(x->c)f(x)=f(c)`

Therefore, f is continuous at all points of the interval (−1, 1).

Case IV:

` " if c = 1,then "   f(c)=f(1)=2xx1=2`

The left hand limit of f at x = 1 is,

`lim_(x->1)f(x)=lim_(x->1)2=2`

The right hand limit of f at x = 1 is,

`lim_(x->1)f(x)=lim_(x->1)2=2`

`∴ lim_(x->1)f(x)=lim_(x->1)f(c)`

Therefore, f is continuous at x = 2

Case V:

` " If c > 1 , then f(c)=2 and "   lim_(x->1)f (x)=lim_(x->1)(2)=2 `

`lim_(x->c)f(x)=f(c)`

Therefore, f is continuous at all points x, such that x > 1

Thus, from the above observations, it can be concluded that f is continuous at all points of the real line.