Chapter 2.3, Problem 18E

To determine

Tocheck:Whether it is possible to extend $f$ to be continuous at $x=3$ , if so check the value assigned to the extended function and if not why not.

Answer to Problem 18E

Yes, it is possible to extend $f$ to be continuous at $x=3$ by setting the value of $f\left(3\right)$ equal to $0$ in the extended function.

Explanation of Solution

Given information:

The given function is $f\left(x\right)=\{\begin{array}{l}{x}^2-1,-1\le x<0\\ 2x,0<x<1\\ 1,x=1\\ -2x+4,1<x<2\\ 0,2<x<3\end{array}$ and the graph of the function is given below:

  

As observed from the graph, the left hand limit of $f\left(x\right)$ at $x=3$ is $0$ .

The point $x=3$ is an endpoint. For the continuity of $f\left(x\right)$ at $x=3$ the condition is $\underset{x\to 3^{-}}{\lim }f\left(x\right)=f\left(3\right)$ .

For the continuity of $f\left(x\right)$ at $x=3$ , the value of $f\left(3\right)$ must also be equal to $0$ in the extended function.

Therefore,it is possible to extend $f$ to be continuous at $x=3$ by setting the value of $f\left(3\right)$ equal to $0$ in the extended function.