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 :
Answer & explanation
Correct answer: option 2
$f(x)= \begin{cases}x^3+3, & x \neq 0 \\ 4, & x=0\end{cases}$
$\lim\limits_{x→0} f(x) = 0^3 + 3 = 3$
f(0) = 4
⇒ discontinuous at O only
⇒ Graph is continuous for all other x