Chapter 2.3, Problem 10E

a.

To determine

To graph:The equations $y_1=\frac{x^4+x^2-1}{x^2+1}$ and $y_2=x^2-\frac{1}{x^2+1}$ in the same viewing window.

Explanation of Solution

Given:

The given equations are $y_1=\frac{x^4+x^2-1}{x^2+1}$ and $y_2=x^2-\frac{1}{x^2+1}$

Graph:

The graph of equations $y_1=\frac{x^4+x^2-1}{x^2+1}$ and $y_2=x^2-\frac{1}{x^2+1}$ are shown in figure here.

  

Interpretation:

From the graphs of $y_1=\frac{x^4+x^2-1}{x^2+1}$ and $y_2=x^2-\frac{1}{x^2+1}$ , we observe that the shapes of graphs of equations $y_1$ and $y_2$ are identical.

b.

To determine

To verify:The expressions $y_1=\frac{x^4+x^2-1}{x^2+1}$ and $y_2=x^2-\frac{1}{x^2+1}$ are equivalent, using their graphs.

Explanation of Solution

Given:

The expressions $y_1=\frac{x^4+x^2-1}{x^2+1}$ and $y_2=x^2-\frac{1}{x^2+1}$

Concept used:

Two expressions are said to be equivalent if their graphs pass through same points and their geometry is same.

Calculation:

From graphs of theexpressions $y_1=\frac{x^4+x^2-1}{x^2+1}$ and $y_2=x^2-\frac{1}{x^2+1}$ pass through same points and their shapes are same.

  

Hence, the expressions $y_1=\frac{x^4+x^2-1}{x^2+1}$ and $y_2=x^2-\frac{1}{x^2+1}$ are equivalent

c.

To determine

To show:The expressions $y_1=\frac{x^4+x^2-1}{x^2+1}$ and $y_2=x^2-\frac{1}{x^2+1}$ are equivalent; using long division (or algebraic method).

Explanation of Solution

Given:

Theexpressions $y_1=\frac{x^4+x^2-1}{x^2+1}$ and $y_2=x^2-\frac{1}{x^2+1}$ .

Method used:

The long- division method.

Calculations:

Consider equation $y_1=\frac{x^4+x^2-1}{x^2+1}$

  $\Rightarrow y_1=x^2+1\begin{array}{c}\hfill x^2\\ \hfill \overline{)\begin{array}{l}{x}^4+x^2-1\\ \frac{\begin{array}{l}{x}^4+x^2\\ --\end{array}}{-1}\end{array}}\end{array}$

  $\Rightarrow y_1=x^2-\frac{1}{x^2+1}$.........(1)

From result (1) and given expression $y_2=x^2-\frac{1}{x^2+1}$ , it is clear that $y_1=y_2$ .

Hence, the expressions $y_1=\frac{x^4+x^2-1}{x^2+1}$ and $y_2=x^2-\frac{1}{x^2+1}$ are equivalent.