Question

The points of discontinuity of the function
$f\left(x\right)=\left\{\begin{array}{cc}\frac{1}{5}\left(2x^2+3\right),& x\le 1\\ 6-5x,& 1<x<3\\ x-3,& x\ge 3\end{array}\right.\mathrm{is}\left(\mathrm{are}\right)$
(a) x = 1
(b) x = 3
(c) x = 1, 3
(d) none of these

Answer 1

(b) x = 3

If $x\le 1$, then $f\left(x\right)=\frac{1}{5}\left(2x^2+3\right)$.
Since $2x^2+3$ 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, $f\left(x\right)$ is continuous at $x\le 1$.

If $1<x<3$, then $f\left(x\right)=6-5x$.

Since $5x$ is a polynomial function and $6$ is a constant function, both of them are continuous. So, their difference will also be continuous.
Thus, $f\left(x\right)$ is continuous for every $1<x<3$.

If $x\ge 3$, then $f\left(x\right)=x-3$.
Since $x-3$ is a polynomial function, it is continuous. So, $f\left(x\right)$ is continuous for every $x\ge 3$.

Now,
Consider the point $x=1$. Here,

$\underset{\mathrm{x}\to 1^{-}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(1-h\right)=\underset{h\to 0}{\lim }\left(\frac{1}{5}\left[2{\left(1-h\right)}^2+3\right]\right)=1$

$\underset{\mathrm{x}\to 1^{+}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(1+h\right)=\underset{h\to 0}{\lim }\left(6-5\left(1+h\right)\right)=1$

Also,
$f\left(1\right)=\frac{1}{5}\left(2{\left(1\right)}^2+3\right)=1$

Thus,
$\underset{\mathrm{x}\to 1^{-}}{\lim }f\left(x\right)=\underset{\mathrm{x}\to 1^{+}}{\lim }f\left(x\right)=f\left(1\right)$

Hence, $f\left(x\right)$ is continuous at $x=1$.

Now,
Consider the point $x=3$. Here,

$\underset{\mathrm{x}\to 3^{-}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(3-h\right)=\underset{h\to 0}{\lim }\left(6-5\left(3-h\right)\right)=-9$

$\underset{\mathrm{x}\to 3^{+}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(3+h\right)=\underset{h\to 0}{\lim }\left(\left(3+h\right)-3\right)=0$

Also,
$f\left(1\right)=\frac{1}{5}\left(2{\left(1\right)}^2+3\right)=1$

Thus,
$\underset{\mathrm{x}\to 3^{-}}{\lim }f\left(x\right)\ne \underset{\mathrm{x}\to 3^{+}}{\lim }f\left(x\right)$

Hence, $f\left(x\right)$ is discontinuous at $x=3$.

So, the only point of discontinuity of $f\left(x\right)$ is $x=3$.