Chapter 0.2, Problem 12E

(a)

To determine

To find: The domain and range of the given function.

Answer to Problem 12E

The domain is equal to $\left(-\infty ,0\right)\cup \left(0,\infty \right)$ and the range is equal to $\left(1,\infty \right)$ .

Explanation of Solution

Given information:

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

Calculation:

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

To find the domain: The denominator should not be equal to zero.

It can be written as below:

  $\begin{array}{c}{x}^2\ne 0\\ x\ne 0\end{array}$

So, the $y$ exists for all values of $x$ except $0$ .

Therefore, the domain will be

  $x\in \left(-\infty ,0\right)\cup \left(0,\infty \right)$

To find the range: Rewrite the equation of function and solve for $x$ .

  $\begin{array}{c}\frac{1}{x^2}=y-1\\ x^2=\frac{1}{y-1}\\ x=\pm \sqrt{\frac{1}{y-1}}\end{array}$

There does not exist a real square root of a negative number, so the inside of the square root, $\frac{1}{y-1}$ , must be greater than or equal to zero.

It can be simplified as below:

  $\frac{1}{y-1}\ge 0$

But the denominator should not be equal to zero, so $y-1$ must not be equal to zero.

It can be written as below:

  $\begin{array}{c}y-1\ne 0\\ y\ne 1\end{array}$

So, the $x$ exists for all values of $y$ which are greater than $1$ .

Therefore, the range will be

  $R=\left(1,\infty \right)$

Hence, the domain is $\left(-\infty ,0\right)\cup \left(0,\infty \right)$ and the range is $\left(1,\infty \right)$ .

(b)

To determine

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

Explanation of Solution

Given information:

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

Calculation:

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

The given function has a vertical asymptote at $x=0$ and a horizontal asymptote at $y=1$ .

Now, sketch the graph of the function below: