Chapter 1.2, Problem 12E

(a)

To determine

To find:The domain and range of the function $y=1+\frac{1}{x^2}$ .

Answer to Problem 12E

The domain and range of the function $y=1+\frac{1}{x^2}$ is $\left(-\infty ,0\right)\cup \left(0,\infty \right)$ and $\left(1,\infty \right)$ respectively.

Explanation of Solution

Given information:The given function is $y=1+\frac{1}{x^2}$ .

Calculation:

The function $y=1+\frac{1}{x^2}$ is defined when $x^2\ne 0$ . So, the domain of the function is theinterval $\left(-\infty ,0\right)\cup \left(0,\infty \right)$ .

The function $\frac{1}{x^2}$ is always positive forevery real value of $x$ .

  $\frac{1}{x^2}>0$

Add $1$ to both the sides.

  $\begin{array}{c}1+\frac{1}{x^2}>1+0\\ y>1\end{array}$

So, the range of the function $y=1+\frac{1}{x^2}$ is the interval $\left(1,\infty \right)$ .

Therefore, the domain and range of the function $y=1+\frac{1}{x^2}$ is $\left(-\infty ,0\right)\cup \left(0,\infty \right)$ and $\left(1,\infty \right)$ respectively.

(b)

To determine

To graph:The function $y=1+\frac{1}{x^2}$ .

Explanation of Solution

Given information:The function is $y=1+\frac{1}{x^2}$ .

Graph:

The graph of the function $y=1+\frac{1}{x^2}$ is drawn below:

  

Figure(1)

Interpretation:The graph of the function $y=1+\frac{1}{x^2}$ is reciprocal function with horizontal asymptote as $y=1$ and vertical asymptote as $y$ -axis.