Discuss the continuity of the function f, where f is defined by

Question:

Discuss the continuity of the function f, where f is defined by

$f(x)= \begin{cases}2 x, & \text { if } x<0 \\ 0, & \text { if } 0 \leq x \leq 1 \\ 4 x, & \text { if } x>1\end{cases}$

Solution:

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

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 $f(c)=2 c$

$\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(2 x)=2 c$

$\therefore \lim _{x \rightarrow c} f(x)=f(c)$

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

Case II:

If $c=0$, then $f(c)=f(0)=0$

The left hand limit of at x = 0 is,

$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}(2 x)=2 \times 0=0$

$\therefore \lim _{x \rightarrow 0} f(x)=f(0)$

Thereforef is continuous at x = 0

Case III:

If $0$\therefore \lim _{x \rightarrow c} f(x)=f(c)$Thereforef is continuous at all points of the interval (0, 1). Case IV: If$c=1$, then$f(c)=f(1)=0$The left hand limit of at x = 1 is,$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow \perp^{-}}(0)=0$The right hand limit of f at = 1 is,$\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}(4 x)=4 \times 1=4$It is observed that the left and right hand limits of f at x = 1 do not coincide. Thereforef is not continuous at x = 1 Case V: If$c<1$, then$f(c)=4 c$and$\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(4 x)=4 c\therefore \lim _{x \rightarrow c} f(x)=f(c)$Therefore,$f$is continuous at all points$x$, such that$x>1\$

Hence, is not continuous only at = 1