Chapter 2.4, Problem 8E

To determine

To sketch: The graph of a function f is continuous except neither left nor right continuous at −2, continuous only from the left at 2.

Explanation of Solution

Definition used:

1 A function f is continuous from the right at a number a if $\underset{x\to a^{+}}{\lim }f\left(x\right)=f\left(a\right)$.

2. A function f is continuous from the left at a number a if $\underset{x\to a^{-}}{\lim }f\left(x\right)=f\left(a\right)$.

Note 1: The function f is said to be discontinuous at $x=a$ if anyone of the following conditions does not satisfied.

• $f\left(a\right)$ is defined
• The limit of the function at the number a exists.
• $\underset{x\to a}{\lim }f\left(x\right)=f\left(a\right)$

Jump discontinuity: A function $f\left(x\right)$ jumps one value to another value at $x=a$. That is, the limit does not exists at $x=a$ because $\underset{x\to a^{+}}{\lim }f\left(x\right)\ne \underset{x\to a^{-}}{\lim }f\left(x\right)$. Also the function f may have any constant value at $x=a$.

Removable discontinuity: The limit exists at $x=a$ because $\underset{x\to a^{+}}{\lim }f\left(x\right)=\underset{x\to a^{-}}{\lim }f\left(x\right)$ and

the actual value of $f\left(x\right)$ at $x=a$ is defined. But $\underset{x\to a}{\lim }f\left(x\right)\ne f\left(a\right)$.

Calculation:

A function f is discontinuous from the left at a number −2 if $\underset{x\to -2^{-}}{\lim }f\left(x\right)\ne f\left(-2\right)$ and a function f is discontinuous from the right at a number −2 if $\underset{x\to -2^{+}}{\lim }f\left(x\right)\ne f\left(-2\right)$. Thus the actual value of $f\left(x\right)$ at $x=-2$ is defined in $ℝ$ except L(the limit value of $f\left(x\right)$ as x approaches −2 or the left and right limit value of $f\left(x\right)$ as x approaches −2).

By definition 2, a function f is continuous only from the left at a number 2means that the function f is discontinuous from the right at a number 2. That is, $\underset{x\to 2^{-}}{\lim }f\left(x\right)=f\left(2\right)$ and $\underset{x\to 2^{+}}{\lim }f\left(x\right)\ne f\left(2\right)$.

Thus, the graph of the function is a piecewise function.

So define $f\left(x\right)=\{\begin{array}{l}{f}_1\left(x\right)\text{if }x<-2\\ k\text{if }x=-2\\ f_2\left(x\right)\text{if }-2<x\le 2\\ f_3\left(x\right)\text{if }x>2\end{array}$ where $f_1\left(x\right)$, $f_2\left(x\right)$ and $f_3\left(x\right)$ has any three functions and $k$ is a constant in $ℝ$ except L(the limit value of $f\left(x\right)$ as x approaches 3).

Draw the graph of the function f as shown below in Figure 1.

From Figure 1, it is observed that the function $f\left(x\right)$ has a hole at $x=-2$ and $f\left(-2\right)=1$ is defined. Thus, the function f is neither continuous except left nor right continuous at −2.

From Figure 1, it is observed that the function $f\left(x\right)$ has jumps −2 to 1 at $x=2$. That is, $f\left(2\right)=-2$ is defined and the function $f\left(x\right)$ approaches −2 as x approaches left side of 2 but the function $f\left(x\right)$ approaches 1 as x approaches right side of 2.

Thus, the function continuous only from the left at 2.