Chapter 11.3, Problem 10QR

Solution

To find: The points of continuity and points of discontinuity of $f\left(x\right)=\left\{\begin{array}{cc}3x+1& ifx\le 1\\ 4-x^2& ifx>1\end{array}\right.$ .

Points of continuity of $f\left(x\right)=\left\{\begin{array}{cc}3x+1& ifx\le 1\\ 4-x^2& ifx>1\end{array}\right.$ are $\left(-\infty ,1\right)\cup \left(1,\infty \right)$ .

Points of discontinuity of $f\left(x\right)=\left\{\begin{array}{cc}3x+1& ifx\le 1\\ 4-x^2& ifx>1\end{array}\right.$ is $x=1$ .

Given information:

The given function is $f\left(x\right)=\left\{\begin{array}{cc}3x+1& ifx\le 1\\ 4-x^2& ifx>1\end{array}\right.$ .

Formula used:

Continuity of function at a point: The function $f\left(x\right)$ is said to be continuous at point $x=c$ if $\underset{x\to c}{\lim }f\left(x\right)=f\left(c\right)$ or $\underset{x\to c^{-}}{\lim }f\left(x\right)=f\left(c\right)=\underset{x\to c^{+}}{\lim }f\left(x\right)$ .

Points of discontinuity: The points of domain where the function is not continuous are called points of discontinuity of that function.

Calculation:

Consider the function $f\left(x\right)=\left\{\begin{array}{cc}3x+1& ifx\le 1\\ 4-x^2& ifx>1\end{array}\right.$

When $x<1$ then

  $\begin{array}{c}\underset{x\to c}{\lim }f\left(x\right)=\underset{x\to c}{\lim }\left(3x+1\right)\\ \underset{x\to c}{\lim }f\left(x\right)=3c+1\\ \underset{x\to c}{\lim }f\left(x\right)=f\left(c\right)\end{array}$

And when $x>1$

  $\begin{array}{c}\underset{x\to c}{\lim }f\left(x\right)=\underset{x\to c}{\lim }\left(4-x^2\right)\\ \underset{x\to c}{\lim }f\left(x\right)=4-c^2\\ \underset{x\to c}{\lim }f\left(x\right)=f\left(c\right)\end{array}$

Clearly $f\left(x\right)=\left\{\begin{array}{cc}3x+1& ifx\le 1\\ 4-x^2& ifx>1\end{array}\right.$ is continuous when $x<1$ and $x>1$ .

Now at $x=1$

  $f\left(1\right)=4$

  $\begin{array}{c}\underset{x\to 1^{+}}{\lim }f\left(x\right)=\underset{x\to 1}{\lim }\left(4-x^2\right)\\ \underset{x\to 1^{+}}{\lim }f\left(x\right)=4-1\\ \underset{x\to 1^{+}}{\lim }f\left(x\right)=3\end{array}$

And

  $\begin{array}{c}\underset{x\to 1^{-}}{\lim }f\left(x\right)=\underset{x\to 1}{\lim }\left(3x+1\right)\\ \underset{x\to 1^{-}}{\lim }f\left(x\right)=3+1\\ \underset{x\to 1^{-}}{\lim }f\left(x\right)=4\end{array}$

Since $\underset{x\to 1^{-}}{\lim }f\left(x\right)\ne \underset{x\to 1^{+}}{\lim }f\left(x\right)$ .

Thus given function is not continuous at $x=1$ .

Hence $f\left(x\right)=\left\{\begin{array}{cc}3x+1& ifx\le 1\\ 4-x^2& ifx>1\end{array}\right.$ is continuous at $\left(-\infty ,1\right)\cup \left(1,\infty \right)$ and discontinuous at $x=1$ .