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Find all points of discontinuity of f : where f is defined by: f(x)= \begin{cases}x^{3}-3, & \text { if } x \leq 2 \\ x^{2}+1, & \text { if } x>2\end{cases}
CBSE | Class 12 | Continuity and Differentiability | Exercise 5.1 | Maths | NCERT
Find all points of discontinuity of f : where f is defined by: f(x)= \begin{cases}x^{3}-3, & \text { if } x \leq 2 \\ x^{2}+1, & \text { if } x>2\end{cases}

Solution: The provided function is

f(x)=\left\{\begin{array}{lll}x^{3}-3, & \text { if } & x \leq 2 \\ x^{2}+1, & \text { if } & x>2\end{array}\right.

At x=2,
Left Hand Limit = \lim _{x \rightarrow 2^{-}}\left(x^{3}-3\right)=8-3=5

Right Hand Limit =\lim _{x \rightarrow 2^{-}}\left(x^{2}+1\right)=4+1=5

f(2)=2^{3}-3=8-3=5

Since, L.H.L. = R.H.L. =f(2)

As a result, f(x) is a continuous at x=2

For
x=c<0 \lim _{x \rightarrow C}\left(x^{3}-3\right)=c^{3}-3=f(c)
and \lim _{x \rightarrow c}\left(x^{2}+1\right)=c^{2}+1=f(c)

Hence, \lim _{x \rightarrow c^{-}}=f(x)

Therefore it implies that, the f(x) is a continuous for all x \in \mathbb{R}.

As a result, the function has no point of discontinuity.

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