Chapter 2.3, Problem 14E

To determine

Tofind:The values of $x$ for which the function $f$ is continuous.

Answer to Problem 14E

The function $f\left(x\right)$ is continuous for all the values of in the domain of $x\in \left[-1,3\right)$ except $x=0,1,2$ .

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:

  

Calculation:

As observed from the graph, the value of function $f\left(x\right)$ is $0$ for the end point $x=-1$ and also the value of right hand limit of $f\left(x\right)$ at $x=-1$ is $0$ .

So, the function is continuous at $x=-1$ .

As observed from the graph that, the left hand limit of $f\left(x\right)$ at $x=0$ is $-1$ and the value of right hand limit at $x=0$ and $f\left(0\right)$ is $0$ .The left hand limit of $f\left(x\right)$ is not equal to the right hand limit of $f\left(x\right)$ at $x=0$ .

So, the function is discontinuous at $x=0$ .

As observed from the graph, the left hand and right hand limits of function $f\left(x\right)$ at $x=1$ is equal to $2$ and the value of $f\left(1\right)$ is equal to $1$ .

It can be concluded that $\underset{x\to 1}{\lim }f\left(x\right)\ne f\left(1\right)$ .So, the function $f\left(x\right)$ is discontinuous at $x=1$ .

As observed from the graph, the left hand and right hand limits of function $f\left(x\right)$ at $x=2$ is equal to $0$ , but the value of $f\left(2\right)$ does not exist.

So, the function is discontinuous at $x=2$ .

As observed from the graph, the left hand limit of function $f\left(x\right)$ at $x=3$ is equal to $0$ , but the value of $f\left(3\right)$ does not exist.

So, the function is discontinuous at $x=3$ .

Therefore, the function $f\left(x\right)$ is continuous for all the values of in the domain of $x\in \left[-1,3\right)$ except $x=0,1,2$ .