 # The points of discontinuity of the function

Question:

The points of discontinuity of the function

$f(x)= \begin{cases}\frac{1}{5}\left(2 x^{2}+3\right), & x \leq 1 \\ 6-5 x & , \quad 1 (a) x = 1 (b) x = 3 (c) x = 1, 3 (d) none of these Solution: (b)$x=3$If$x \leq 1$, then$f(x)=\frac{1}{5}\left(2 x^{2}+3\right)$. Since$2 x^{2}+3$is a polynomial function and$\frac{1}{5}$is a constant function, both of them are continuous. So, their product will also be continuous. Thus,$f(x)$is continuous at$x \leq 1$. If$1

Since $5 x$ is a polynomial function and 6 is a constant function, both of them are continuous. So, their difference will also be continuous. Thus, $f(x)$ is continuous for every $1 If$x \geq 3$, then$f(x)=x-3$. Since$x-3$is a polynomial function, it is continuous. So,$f(x)$is continuous for every$x \geq 3$. Now, Consider the point$x=1$. Here,$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0} f(1-h)=\lim _{h \rightarrow 0}\left(\frac{1}{5}\left[2(1-h)^{2}+3\right]\right)=1\lim _{x \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0} f(1+h)=\lim _{h \rightarrow 0}(6-5(1+h))=1$Also,$f(1)=\frac{1}{5}\left(2(1)^{2}+3\right)=1$Thus,$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)=f(1)$Hence,$f(x)$is continuous at$x=1$.$\lim _{x \rightarrow 3^{-}} f(x)=\lim _{h \rightarrow 0} f(3-h)=\lim _{h \rightarrow 0}(6-5(3-h))=-9\lim _{x \rightarrow 3^{+}} f(x)=\lim _{h \rightarrow 0} f(3+h)=\lim _{h \rightarrow 0}((3+h)-3)=0$Also,$\lim _{x \rightarrow 3^{-}} f(x) \neq \lim _{x \rightarrow 3^{+}} f(x)$Hence,$f(x)$is discontinuous at$x=3$. So, the only point of discontinuity of$f(x)$is$x=3\$.