Chapter 5: Continuity and Differentiability

Ans: The given function is \[\text{f}\left( \text{x} \right)\text{=5x-3}\].

At \[\text{x=0}\], \[\text{f}\left( \text{0} \right)\text{=5 }\!\!\times\!\!\text{ 0-3=-3}\].

Taking limit as $x\to 0$ both sides of the function give

\[\underset{x\to 0}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{x\to 0}{\mathop{\lim }}\,\left( \text{5x-3} \right)\text{=5 }\!\!\times\!\!\text{ 0-3=-3}\]

$\therefore \underset{x\to 0}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{0} \right)$.

Thus, $\text{f}$ satisfies continuity at \[\text{x=0}\].

Again, at \[\text{x=-3,f}\left( \text{-3} \right)\text{=5 }\!\!\times\!\!\text{ }\left( \text{-3} \right)\text{-3=-18}\].

Now, taking limit as $x\to 3$ both sides of the function give

\[\underset{x\to 3}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{x\to 3}{\mathop{\lim }}\,\text{f}\left( \text{5x-3} \right)\text{=5 }\!\!\times\!\!\text{ }\left( \text{-3} \right)\text{-3=-18}\]

\[\therefore \underset{x\to 3}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{-3} \right)\].

Therefore, $\text{f}$ satisfies continuity at \[\text{x=-3}\].

Also, at \[\text{x=5,f}\left( \text{x} \right)\text{=f}\left( \text{5} \right)\text{=5 }\!\!\times\!\!\text{ 5-3=25-3=22}\].

Taking limit as $x\to 5$ both sides of the function give

\[\underset{x\to 5}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{x\to 5}{\mathop{\lim }}\,\left( \text{5x-3} \right)\text{=5 }\!\!\times\!\!\text{ 5-3=22}\]

\[\therefore \underset{x\to 5}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{5} \right)\].

Hence, $\text{f}$ satisfies continuity at \[\text{x=5}\].

Ans: The given function is $\text{f}\left( \text{x} \right)\text{=2}{{\text{x}}^{\text{2}}}\text{-1}$.

Now, at $\text{x=3, f}\left( \text{3} \right)\text{=2 }\!\!\times\!\!\text{ }{{\text{3}}^{\text{2}}}\text{-1=17}$.

Taking limit as $x\to 3$ both sides of the function give

$\underset{x\to 3}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{x\to 3}{\mathop{\lim }}\,\left( \text{2}{{\text{x}}^{\text{2}}}\text{-1} \right)\text{=2 }\!\!\times\!\!\text{ }{{\text{3}}^{\text{2}}}\text{-1=17}$

\[\therefore \underset{x\to 3}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f=}\left( \text{3} \right)\].

Hence, $\text{f}$ satisfies continuity at \[\text{x=3}\].

Ans: The given function is $\text{f}\left( \text{x} \right)\text{=x-5}$.

It is assured that for every real number $\text{k}$, $\text{f}$ is defined and its value at $k$ is $\text{k-5}$. Also, it can be noted that 

$\underset{x\to k}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{x\to k}{\mathop{\lim }}\,\text{f}\left( \text{x-5} \right)\text{=k=k-5=f}\left( \text{k} \right)$.

$\therefore \underset{x\to k}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{k} \right)$

Hence, $\text{f}$ satisfies continuity at every real number and so, it is a continuous function.

Ans: The given function is

$\text{f}\left( \text{x} \right)\text{=}\frac{\text{1}}{\text{x-5}}$.

Let $\text{k}\ne \text{5}$ is any real number, then taking limit as $\text{x}\to \text{k}$ both sides of the function give

$\underset{x\to k}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{x\to k}{\mathop{\lim }}\,\frac{\text{1}}{\text{x-5}}\text{=}\frac{\text{1}}{\text{k-5}}$

Also, $\text{f(k)=}\frac{\text{1}}{\text{k-5}}$ , since $\text{k}\ne \text{5}$

$\therefore \underset{x\to k}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{k} \right)$

Therefore, $f$ satisfies continuity at every point in the domain of $f$ and so, it is a continuous function.

Ans: The given function is

\[\text{f}\left( \text{x} \right)\text{=}\frac{{{\text{x}}^{\text{2}}}\text{-25}}{\text{x+5}}\text{, x}\ne \text{5}\]

Now let $\text{c}\ne \text{-5}$ be any real number, then taking limit as $\text{x}\to \text{c}$ on both sides of the function give

\[\underset{x\to c}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{x\to c}{\mathop{\lim }}\,\frac{{{\text{x}}^{\text{2}}}\text{-25}}{\text{x+5}}\text{=}\underset{x\to c}{\mathop{\lim }}\,\frac{\left( \text{x+5} \right)\left( \text{x-5} \right)}{\text{x+5}}\text{=}\underset{x\to c}{\mathop{\lim }}\,\left( \text{x-5} \right)\text{=}\left( \text{c-5} \right)\]

Again, \[\text{f}\left( \text{c} \right)\text{=}\frac{\left( \text{c+5} \right)\left( \text{c-5} \right)}{\text{c+5}}\text{=}\left( \text{c-5} \right)\], since $\text{c}\ne \text{5}$.

Hence, \[\text{f}\] satisfies continuity at every point in the domain of $\text{f}$ and so it is a continuous function.

Ans: The given function is $\text{f}\left( \text{x} \right)\text{=}\left| \text{x-5} \right|=\left\{ \begin{align} & \text{5-x, if x}<\text{5} \\ & \text{x-5, if x}\,\ge \text{5} \\ \end{align} \right.$

Note that, $\text{f}$ is defined at all points in the real line. So, let assume $\text{c}$ be a point on a real line. 

Then, we have \[\text{c}<\text{5}\] or \[\text{c}=\text{5}\] or \[\text{c}>\text{5}\].

Now, let's discuss these three cases one by one.

Case (i): $\text{c}<\text{5}$

Then, the function becomes $\text{f}\left( \text{c} \right)\text{=5-c}$.

Now, $\underset{x\to c}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{x\to c}{\mathop{\lim }}\,\left( \text{5-x} \right)\text{=5-c}$.

$\therefore \underset{x\to c}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right)$.

Therefore, $\text{f}$ is continuous at all real numbers which are less than $5$. 

Case (ii): $\text{c=5}$

Then, \[\text{f}\left( \text{c} \right)\text{=f}\left( \text{5} \right)\text{=}\left( \text{5-5} \right)\text{=0}\].

Now,

$\underset{x\to {{5}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)=\underset{x\to 5}{\mathop{\lim }}\,\left( \text{5-x} \right)\text{=}\left( \text{5-5} \right)\text{=0}$ and

$\underset{x\to {{5}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{x\to 5}{\mathop{\lim }}\,\left( \text{x-5} \right)\text{=0}$.

Therefore, we have

$\underset{x\to {{c}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{x\to {{c}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right)$.

Thus, $\text{f}$ satisfies continuity at \[\text{x=5}\], and so $\text{f}$ is continuous at $\text{x=5}$.

Case (iii): $\text{c}>\text{5}$

Then we have, $\text{f}\left( \text{c} \right)\text{=f}\left( \text{5} \right)\text{=c-5}$.

Now,

$\underset{x\to c}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{x\to c}{\mathop{\lim }}\,\left( \text{x-5} \right)\text{=c-5}$.

Therefore,

$\underset{x\to c}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right)$.

So, $\text{f}$ is continuous at all real numbers that are greater than $5$.

Thus, $\text{f}$ satisfies continuity at every real number and hence, it is a continuous function.

Ans: The given function is  $\text{f}\left( \text{x} \right)\text{=}{{\text{x}}^{\text{n}}}$.

We noticed that the function $\text{f}$ is defined at all positive integers $\text{n}$ and also its value at $\text{x=n}$ is ${{\text{n}}^{\text{n}}}$.

Therefore, $\underset{x\to n}{\mathop{\lim }}\,\text{f}\left( \text{n} \right)=\underset{x\to n}{\mathop{\lim }}\,\text{f}\left( {{\text{x}}^{\text{n}}} \right)\text{=}{{\text{n}}^{\text{n}}}$.

So, \[\underset{\text{x}\to \text{n}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{n} \right)\].

Thus, the function $\text{f}\left( \text{x} \right)\mathbf{=}{{\text{x}}^{\text{n}}}$  is continuous at $\text{x=n}$ , where $\text{n}$ is a positive integer.

Ans: The given function is \[\text{f}\left( \text{x} \right)\text{=}\left\{ \begin{align} & \text{x,}\,\text{ if x}\le \text{1} \\ & \text{5, }\,\text{if x}\,>\text{1} \\ \end{align} \right.\]

It is obvious that the function $\text{f}$ is defined at $\text{x=}0$ and its value at $\text{x=}0$ is $0$.

Now, $\underset{x\to 0}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{x\to 0}{\mathop{\lim }}\,\text{x=0}$.

So, $\underset{x\to 0}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{0} \right)$.

Hence, the function $\text{f}$ satisfies continuity at \[\text{x=0}\].

It can be observed that $\text{f}$ is defined at $\text{x=1}$ and its value at this point is $\text{1}$.

Now, the left-hand limit of the function $\text{f}$ at \[\text{x=1}\] is

$\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,\text{x=1}$.

Also, the right-hand limit of the function $\text{f}$ at \[\text{x=1}\] is

$\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{5} \right)$

Therefore, $\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\ne \underset{x\to {{1}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)$.

Thus, $\text{f}$ is not continuous at \[\text{x=1}\]

It can be found that $\text{f}$ is defined at $\text{x=2}$ and its value at this point is $5$.

That is, \[\underset{x\to 2}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{x\to 2}{\mathop{\lim }}\,\text{f}\left( \text{5} \right)\text{=5}\].

Therefore, \[\underset{x\to 2}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{2} \right)\]

Hence, $\text{f}$ satisfies continuity at \[\text{x=2}\].

Ans: The given function is \[\text{f}\left( \text{x} \right)\text{=}\left\{ \begin{align} & \text{2x+3, if x}\le \text{2} \\ & \text{2x-3, if x}\,>\,\text{2} \\ \end{align} \right.\]

It can be observed that the function $\text{f}$ is defined at all the points in the real line.

Let consider $\text{c}$ be a point on the real line. Then, three cases may arise.

I.      $\text{c}\,<\,\text{2}$

II.     $\text{c}\,>\,\text{2}$

III.    $\text{c}=\text{2}$

Case (i): When $\text{c}\,<\text{2}$

Then, we have $\underset{x\to c}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{x\to \infty }{\mathop{\lim }}\,\left( \text{2x+3} \right)\text{=2c+3}$.

Therefore,

$\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right)$.

Hence, $\text{f}$ attains continuity at all points $\text{x}$, where \[\text{x}<\text{2}\].

Case (ii): When $\text{c}>\text{2}$

Then, we have  $\text{f}\left( \text{c} \right)\text{=2c-3}$.

So,

$\underset{\text{x}\to c}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to \infty }{\mathop{\lim }}\,\left( \text{2x-3} \right)\text{=2c-3}$.

Therefore, $\underset{\text{x}\to c}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right)$.

Hence, $\text{f}$ satisfies continuity at all points $\text{x}$ , where \[\text{x}>\text{2}\].

Case(iii): When $\text{c=2}$

Then, the left-hand limit of the function $\text{f}$ at \[\text{x=2}\] is

$\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{2}^{-}}}{\mathop{\lim }}\,\left( \text{2x+3} \right)\text{=2 }\!\!\times\!\!\text{ 2+3=7}$ and 

the right-hand limit of the function $\text{f}$ at \[\text{x=2}\] is,

$\underset{\text{x}\to {{2}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to 2+}{\mathop{\lim }}\,\left( \text{2x+3} \right)\text{=2 }\!\!\times\!\!\text{ 2-3=1}$.

Thus, at \[\text{x=2}\], $\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\ne \underset{\text{x}\to {{2}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)$.

So, the function  $\text{f}$ does not satisfy continuity at \[\text{x=2}\].

Hence, \[\text{x=2}\] is the only point of discontinuity of the function $\text{f}\left( \text{x} \right)$.

Ans: The given function is \[\text{f}\left( \text{x} \right)\text{=}\left\{ \begin{align} & \left| \text{x} \right|\text{+3, if x}\le \text{-3} \\ & \text{-2x, if -3}<\text{x}<\text{3} \\ & \text{6x+2, if x}\ge \text{3} \\ \end{align} \right.\]

Observe that, $\text{f}$ is defined at all the points in the real line.

Now, let assume $\text{c}$ as a point on the real line.

Then five cases may arise. Either $\text{c}<\text{-3}$, or $\text{c}=\text{-3}$ or $\text{-3}<\text{c}<\text{3}$, or $\text{c}=\text{3}$, or $\text{c}>\text{3}$.

Let's discuss the five cases one by one.

Case I: When \[\text{c}<\text{-3}\]

Then, $\text{f}\left( \text{c} \right)\text{=-c+3}$ and

$\underset{\text{x}\to c}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{x\to c}{\mathop{\lim }}\,\left( \text{-x+3} \right)\text{=-c+3}$.

Therefore, $\underset{\text{x}\to c}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right)$.

Hence, $\text{f}$ satisfies continuity at all points $\text{x}$, where \[\text{x}<\text{-3}\].

Case II: When \[\text{c=-3}\]

Then, $\text{f}\left( \text{-3} \right)\text{=-}\left( \text{-3} \right)\text{+3=6}$.

Also, the left-hand limit

$\underset{\text{x}\to -{{3}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to -{{3}^{-}}}{\mathop{\lim }}\,\left( \text{-x+3} \right)\text{=-}\left( \text{-3} \right)\text{+3=6}$.

and the right-hand limit

$\underset{\text{x}\to -{{3}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)=\underset{\text{x}\to -{{3}^{+}}}{\mathop{\lim }}\,\text{f}\left( -2\text{x} \right)\text{=-2}\times \left( \text{-3} \right)\text{=6}$.

Therefore, $\underset{\text{x}\to -3}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{-3} \right)$.

Hence, $\text{f}$ satisfies continuity at $\text{x=-3}$.

Case III: When  $\text{-3}<\text{c}<\text{3}$.

Then, $f\left( c \right)=-2c$ and also

 $\underset{x\to c}{\mathop{\lim }}\,\text{f}\left( x \right)\text{=}\underset{x\to c}{\mathop{\lim }}\,\left( -2x \right)=-2c$.

Therefore, \[\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right)\].

Hence, $\text{f}$ satisfies continuity at $\text{x}$, where $\text{-3}<\text{x}<\text{3}$.

Case IV: When \[\text{c=3}\]

Then, the left-hand limit of the function $\text{f}$ at \[\text{x=3}\] is

$\underset{\text{x}\to {{3}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{3}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{-2x} \right)\text{=-2 }\!\!\times\!\!\text{ 3=-6}$ and

the right-hand limit of the function $\text{f}$ at \[\text{x=3}\] is

$\underset{\text{x}\to {{3}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{3}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{6x+2} \right)\text{=6 }\!\!\times\!\!\text{ 3+2=20}$.

Thus, at $\text{x=3}$, $\underset{\text{x}\to {{3}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\ne \underset{\text{x}\to {{3}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)$.

Hence, $\text{f}$ does not satisfy continuity at \[\text{x=3}\].

Case V: When $\text{c}>\text{3}$.

 Then $\text{f}\left( \text{c} \right)\text{=6c+2}$ and also

 $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\left( \text{6x+2} \right)\text{=6c+2}$.

Therefore, $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right)$.

So, $\text{f}$ satisfies continuity at all points $\text{x}$, when $\text{x}>\text{3}$.

Thus, \[\text{x=3}\] is the only point of discontinuity of the function $\text{f}$.

Ans: The given function is \[\text{f}\left( \text{x} \right)\text{=}\left\{ \begin{align} & \frac{\left| \text{x} \right|}{\text{x}}\text{, if x}\ne \text{0} \\ & \text{0, if x=0} \\ \end{align} \right\}\]

Now, $\text{f}\left( \text{x} \right)$ can be rewritten as $\text{f}\left( \text{x} \right)\text{=}\left\{ \begin{align} & \frac{\left| \text{x} \right|}{\text{x}}\text{=}\frac{\text{-x}}{\text{x}}\text{=-1}\,\,\text{if}\,\,\text{x<0} \\ & \text{0, if x=0} \\ & \frac{\left| \text{x} \right|}{\text{x}}\text{=}\frac{\text{x}}{\text{x}}\text{=1 }\,\,\text{if}\,\,\text{ x>0} \\ \end{align} \right\}$

It can be noted that the function $\text{f}$ is defined at all points of the real line.

Now, let assume $\text{c}$ as a point on the real line.

Then three cases may arise, either $\text{c}<\text{0}$, or $\text{c}=\text{0}$, or $\text{c}>\text{0}$.

Let's discuss three cases one by one.

Case I: When $\text{c}<\text{0}$.

Then, $\text{f}\left( \text{c} \right)\text{=-1}$ and

$\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\left( \text{-1} \right)\text{=-1}$.

Therefore, $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right)$.

Hence, $\text{f}$ satisfies continuity at all the points $\text{x}$ where \[\text{x}<\text{0}\].

Case II: When \[\text{c=0}\].

Then, the left-hand limit of the function $\text{f}$ at \[\text{x=0}\] is

$\underset{\text{x}\to {{0}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{0}^{-}}}{\mathop{\lim }}\,\left( \text{-1} \right)\text{=-1}$ and

the right-hand limit of the function $\text{f}$ at \[\text{x=0}\] is

$\underset{\text{x}\to 0+}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{0}^{+}}}{\mathop{\lim }}\,\left( \text{1} \right)\text{=1}$.

At $\text{x=0}$, $\underset{\text{x}\to {{0}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\ne \underset{\text{x}\to {{0}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)$.

Hence, the function $\text{f}$ does not satisfy continuity at \[\text{x=0}\].

Case III: When $\text{c}>\text{0}$.

Then \[\text{f}\left( \text{c} \right)\text{=1}\] and also

\[\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\left( \text{1} \right)\text{=1}\].

Therefore, \[\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right)\].

So, the function $\text{f}$ is continuous at all the points $\text{x}$, for $\text{x}>\text{0}$.

Thus, \[\text{x=0}\] is the only point of discontinuity for the function $\text{f}$.

Ans: The given function is $\text{f}\left( \text{x} \right)\text{=}\left\{ \begin{align} & \frac{\text{x}}{\left| \text{x} \right|}\text{, if x}\,<\,\text{0} \\ & \text{-1, if x}\ge \text{0} \\ \end{align} \right.$

Now, we know that, if \[\text{x}<\text{0}\], then  \[\left| \text{x} \right|\text{=-x}\].

Therefore, the $\text{f}\left( \text{x} \right)$ can be written as

$\text{f}\left( \text{x} \right)\text{=}\left\{ \begin{align} & \frac{x}{\left| \text{x} \right|}=\frac{\text{x}}{\text{-x}}=\text{-1}\,\,\text{if}\text{x}<\text{0} \\ & \text{-1,}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{ if x}=\text{0} \\ & \text{-1, }\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{if}\,\text{x}>\text{0} \\ \end{align} \right\}$

$\Rightarrow \text{f}\left( \text{x} \right)\text{=-1}$ for all positive real numbers.

Now, let's assume $c$ as any real number.

Then three cases may arise, either $\text{c}<\text{0}$, or $\text{c}=\text{0}$, or $\text{c}>\text{0}$.

Let's discuss three cases one by one.

Case I: When \[\text{c}<\text{0}\].

Then, $\text{f}\left( \text{c} \right)\text{=-1}$ and

$\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\left( \text{-1} \right)\text{=-1}$.

Therefore, $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right)$.

Hence, $\text{f}$ satisfies continuity at all the points $\text{x}$ where \[\text{x}<\text{0}\].

Case II: When \[\text{c=0}\].

Then, the left-hand limit of the function $\text{f}$ at \[\text{x=0}\] is

$\underset{\text{x}\to {{0}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{0}^{-}}}{\mathop{\lim }}\,\left( \text{-1} \right)\text{=-1}$ and

the right-hand limit of the function $\text{f}$ at \[\text{x=0}\] is

$\underset{\text{x}\to 0+}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{0}^{+}}}{\mathop{\lim }}\,\left( \text{-1} \right)\text{=-1}$.

At $\text{x=0}$, $\underset{\text{x}\to {{0}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)=\underset{\text{x}\to {{0}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)$.

Hence, the function $\text{f}$ satisfies continuity at \[\text{x=0}\].

Case III: When $\text{c}>\text{0}$.

Then \[\text{f}\left( \text{c} \right)\text{=-1}\] and also

\[\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\left( \text{-1} \right)\text{=-1}\].

Therefore, \[\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right)\].

So, the function $\text{f}$ is continuous at all the points $\text{x}$, for \[\text{x}>\text{0}\].

Then, we have $\underset{\text{x}\to c}{\mathop{\lim }}\,\text{f(x)}=\underset{\text{x}\to c}{\mathop{\lim }}\,(-1)=-1$  and \[\text{f}\left( \text{c} \right)\text{=-1}=\underset{\text{x}\to c}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\].

Therefore, the function $\text{f}\left( \text{x} \right)$ is a continuous function.

Thus, there does not exist any point of discontinuity.

Answer: The given function is$\text{f(x)=}\left\{ \begin{align} & \text{x+1, if x}\ge \text{1} \\ & {{\text{x}}^{\text{2}}}\text{+1, if x}\,<\,\text{1} \\ \end{align} \right.$

Note that, $\text{f}\left( \text{x} \right)$ is defined at all the points of the real line.

Now, let's assume $c$ as a point on the real line.

Then three cases may arise, either either $\text{c}<\text{1}$, or $\text{c}=\text{1}$, or $\text{c}>\text{1}$.

Let's discuss the three cases one by one.

Case I: When $\text{c}<\text{1}$.

 Then, \[\text{f}\left( \text{c} \right)\text{=}{{\text{c}}^{\text{2}}}\text{+1}\] and also

\[\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)=\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( {{\text{x}}^{\text{2}}}\text{+1} \right)\text{=}{{\text{c}}^{\text{2}}}\text{+1}\].

Therefore, \[\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right)\].

Hence, \[\text{f}\] satisfies continuity at all the points $\text{x}$, where $\text{x}<\text{1}$.

Case II: When \[\text{c=1}\].

Then, we have \[\text{f}\left( \text{c} \right)\text{=f}\left( \text{1} \right)\text{=1+1=2}\].

Now, the left-hand limit of $\text{f}$ at \[\text{x=1}\] is

\[\underset{\text{x}\to {{\text{1}}^{-}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{\text{1}}^{-}}}{\mathop{\lim }}\,\text{(}{{\text{x}}^{\text{2}}}\text{+1)=}{{\text{1}}^{\text{2}}}\text{+1=2}\] and the right-hand limit of $\text{f}$ at $\text{x=1}$ is, \[\underset{\text{x}\to {{1}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{1}^{+}}}{\mathop{\lim }}\,\left( \text{x+1} \right)\text{=1+1=2}\].

Therefore, \[\underset{\text{x}\to 1}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right)\].

Hence, $\text{f}$ satisfies continuity at \[\text{x=1}\].

Case III: When $\text{c}>\text{1}$.

Then, we have  $\text{f}\left( \text{c} \right)\text{=c+1}$ and 

\[\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\left( \text{x+1} \right)\text{=c+1}\].

Therefore,

\[\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right)\].

So, $\text{f}$ satisfies continuity at all the points $\text{x}$, where $\text{x}>\text{1}$.

Hence, there does not exist any discontinuity points.