The function f is given by $f(x)= \begin{cases}x^3+3, & \text { if } x \neq 0 \\ ~~~4~~~, & \text { if } x=0\end{cases}$ Then number of points of discontinuity for this function is :

0 1 2 3

1

$f(x)= \begin{cases}x^3+3, & x \neq 0 \\ 4, & x=0\end{cases}$ $\lim\limits_{x→0} f(x) = 0^3 + 3 = 3$ f(0) = 4  ⇒  discontinuous at O only ⇒ Graph is continuous for all other x