Question

Find the points of discontinuity, if any, of the following functions: \[f\left( x \right) = \begin{cases}2x , & \text{ if }  & x < 0 \\ 0 , & \text{ if }  & 0 \leq x \leq 1 \\ 4x , & \text{ if }  & x > 1\end{cases}\]

Answer 1

The given function is  \[f\left( x \right) = \begin{cases}2x , & \text{ if }  & x < 0 \\ 0 , & \text{ if }  & 0 \leq x \leq 1 \\ 4x , & \text{ if }  & x > 1\end{cases}\]

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 < 0, " then "  f(c)=2c`

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

`∴lim_(N->oo)f(x)=f(c)`

Therefore, f is continuous at all points x, such that x < 0

Case II:

` " If " (0), " then "  f(c)=f(0)=0`

The left hand limit of f at x = 0 is,

`lim_(x->0)f(x)=lim_(x->0)(2x)=2xx0=0`

The right hand limit of f at x = 0 is,

`lim_(x->0)f(x)=lim_(x->0)(0)=0`

`∴lim_(x->0)(x)=f(0)`

Therefore, f is continuous at x = 0

Case III:

` " If "  0 < c< 1 " then " f(x)  " and "  lim_(x->c) f(x)=lim_(x->c)(0)=0`

`∴ lim_(x->c)f(x)=f(c)`

Therefore, f is continuous at all points of the interval (0, 1).

Case IV:

` " If "c=1 " then "  f(c)=f(1)=0`

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

`lim_(x->1)f(x)= lim_(x->1)f(1)=0`

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

`lim_(x->1)f(x)=lim_(x->1)(4x)=4xx1=4`

It is observed that the left and right hand limits of f at x = 1 do not coincide.

Therefore, f is not continuous at x = 1

Case V:

` " If " c <, " then " f(c)=4c " and " lim_(x->c)f(4x)=4c`

`∴ lim_(x->c)f(x)=f(c)`

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

Hence, f is not continuous only at x = 1