Chapter 2, Problem 26RE

(a)

To determine

To find: The value of $\underset{x\to 1^{-}}{\lim }k\left(x\right)$ .

Answer to Problem 26RE

The value of $\underset{x\to 1^{-}}{\lim }k\left(x\right)$ is 1.5.

Explanation of Solution

Given information: The domain of the function is $-1\le x\le 3$ . The graph is given below:

  

Calculation:

From the given graph, it can be observed that as $x$ approaches 1 from left then the function approaches to 1.5.

  $\underset{x\to 1^{-}}{\lim }k\left(x\right)=1.5$

Therefore, the value of $\underset{x\to 1^{-}}{\lim }k\left(x\right)$ is 1.5.

(b)

To determine

To find: The value of $\underset{x\to 1^{+}}{\lim }k\left(x\right)$ .

Answer to Problem 26RE

The value of $\underset{x\to 1^{+}}{\lim }k\left(x\right)$ is 0.

Explanation of Solution

Given information: The domain of the function is $-1\le x\le 3$ . The graph is given below:

  

Calculation:

From the given graph, it can be observed that as $x$ approaches 1 from right then the function approaches to 0.

  $\underset{x\to 1^{+}}{\lim }k\left(x\right)=0$

Therefore, the value of $\underset{x\to 1^{+}}{\lim }k\left(x\right)$ is 0.

(c)

To determine

To find: The value of $k\left(1\right)$ .

Answer to Problem 26RE

The value is $k\left(1\right)$ is 0.

Explanation of Solution

Given information: The domain of the function is $-1\le x\le 3$ . The graph is given below:

  

Calculation:

From the graph, it can be observed that the value of the function $y=f\left(x\right)$ at $x=1$ is 0.

  $k\left(1\right)=0$

Therefore, the value is $k\left(1\right)$ is 0.

(d)

To determine

To find: Whether $k\left(x\right)$ is continuous at $x=1$ or not.

Answer to Problem 26RE

The function $k\left(x\right)$ is discontinuous at $x=1$ .

Explanation of Solution

Given information: The domain of the function is $-1\le x\le 3$ . The graph is given below:

  

Calculation:

From the graph, it can be observed that, at $x=1$ there is a jump on the graph of the function. The jump shows the discontinuity as the left hand limit is not equal to the right hand limit of the function.

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

Therefore, the function is discontinuous at $x=1$ .

(e)

To determine

To find: The points of discontinuity of $k\left(x\right)$ .

Answer to Problem 26RE

The point of discontinuity of the given function is $x=1$ .

Explanation of Solution

Given information: The domain of the function is $-1\le x\le 3$ . The graph is given below:

  

Calculation:

A function $y=f\left(x\right)$ is said to be continuous if $f\left(c\right)=\underset{x\to c^{+}}{\lim }f\left(x\right)=\underset{x\to c^{-}}{\lim }f\left(x\right)$ .

It is already known that at $x=1$ left hand limit is not equal to right hand limit of the function. So, function is discontinuous at $x=1$ .

Therefore, the point of discontinuity of the given function is $x=1$ .

(f)

To determine

To find: Whether any point of discontinuity can be removed, if so describe new function and if not explain.

Answer to Problem 26RE

There is no point for removable discontinuity.

Explanation of Solution

Given information: The domain of the function is $-1\le x\le 3$ . The graph is given below:

  

Calculation:

The discontinuity can be removable if left hand limit and right hand limit are equal. The right hand limit and left hand limit at $x=1$ are not equal.

The discontinuity is not removable because the function cannot be made continuous by filling in a point at $x=1$ .

Therefore, there is no point for removable discontinuity.