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 $x=3$.

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