Chapter 12, Problem 2PS

a.

To determine

Evaluate $f(1),f(0),f\left(\frac{1}{2}\right)$ , and $f\left(-2.7\right)$ .

Answer to Problem 2PS

  $\boxed{0}$ , $\boxed{0}$ , $\boxed{-1}$ and $\boxed{-1}$

Explanation of Solution

Given information:

Sketch the graph of the function $f(x)=〚x〛+〚-x〛$ .

Evaluate $f(1),f(0),f\left(\frac{1}{2}\right)$ , and $f\left(-2.7\right)$ .

Calculation:

Consider the following limit

  $f(x)=〚x〛+〚-x〛$

To sketch the graph of the function using the graphing utility Ti-83, proceed as follows.

Press $Y$ and provide the function. The display will be

  

Now press window and choose proper scale. The display will be

  

Now press graph, the display will be

  

To evaluate the value of the function at the defined points, proceed as follows.

Use the definition of maximum integer function and evaluate.

Definition of maximum integer function:

  $\begin{array}{l}[1.05]=1\\ [-1.05]=-2\\ [1]=1\\ [-1]=-1\end{array}$

When $x=1$

  $\begin{array}{l}f(1)=[1]+[-1]\\ 1-1\\ =\boxed{0}\end{array}$

When $x=0$

  $\begin{array}{l}f\left(0\right)=[0]+[-0]\\ =0-0\\ =\boxed{0}\end{array}$

When $x=\frac{1}{2}$

  $\begin{array}{l}f\left(\frac{1}{2}\right)=\left[\frac{1}{2}\right]+\left[-\frac{1}{2}\right]\\ =0-1\\ =\boxed{-1}\end{array}$

When $x=-2.7$

  $\begin{array}{l}f\left(-2.7\right)=\left[-2.7\right]+\left[-\left(-2.7\right)\right]\\ =-3+2\\ =\boxed{-1}\end{array}$

Hence, At the integral points the value of the function is $0$ .At the decimal points the value of the function is $1$ .

b.

To determine

Evaluate the following limits.

  $\underset{x\to 1^{-}}{\lim }f(x),\underset{x\to 1^{+}}{\lim }f(x),\underset{x\to 1/2}{\lim }f(x)$ .

Answer to Problem 2PS

  $\boxed{\underset{x\to 1^{-}}{\lim }f(x)=-1}$ , $\boxed{\underset{x\to 1^{+}}{\lim }f(x)=-1}$ and $\boxed{\underset{x\to \frac{1}{2}}{\lim }f(x)=-1}$ .

Explanation of Solution

Given information:

Sketch the graph of the function $f(x)=〚x〛+〚-x〛$ .

Evaluate the following limits.

  $\underset{x\to 1^{-}}{\lim }f(x),\underset{x\to 1^{+}}{\lim }f(x),\underset{x\to 1/2}{\lim }f(x)$

Calculation:

Consider the following limit

  $f(x)=〚x〛+〚-x〛$

To sketch the graph of the function using the graphing utility Ti-83, proceed as follows.

Press $Y$ and provide the function. The display will be

  

Now press window and choose proper scale. The display will be

  

Now press graph, the display will be

  

Now consider the following limit of the function.

  $\underset{x\to 1^{-}}{\lim }f(x)$

To evaluate the limit of the function at the defined points, use the observations made in the previous section.

  $\underset{x\to 1^{-}}{\lim }f(x)=\underset{x\to 1^{-}}{\lim }\left[\left[x\right]\right]+\left[\left[-x\right]\right]$

Observe that the limiting value is less than $1$ and very close to $1$ .

Hence the limiting value is lying between $0$ and $1$ .(A decimal point)

Hence $\boxed{\underset{x\to 1^{-}}{\lim }f(x)=-1}$

Now consider the limit of the function.

  $\underset{x\to 1^{+}}{\lim }f(x)$

To evaluate the limit of the function, use the observations made in the previous section

  $\underset{x\to 1^{+}}{\lim }f(x)=\underset{x\to 1^{-}}{\lim }\left[\left[x\right]\right]+\left[\left[-x\right]\right]$

Observe that the limiting value is less than $1$ and very close to $1$ .

Hence the limiting value is lying between $0$ and $2$ .(A decimal point)

Hence $\boxed{\underset{x\to 1^{+}}{\lim }f(x)=-1}$

Now consider the following limit of the function

  $\underset{x\to 1/2}{\lim }f(x)$

To evaluate the limit of the function, use the method of direct substitution.

  $\underset{x\to \frac{1}{2}}{\lim }f(x)=\underset{x\to \frac{1}{2}}{\lim }\left[\left[x\right]\right]+\left[\left[-x\right]\right]$

  $\begin{array}{l}=\left[\frac{1}{2}\right]+\left[-\frac{1}{2}\right]\\ =0-1\end{array}$

Hence, $\boxed{\underset{x\to \frac{1}{2}}{\lim }f(x)=-1}$