Chapter 2.6, Problem 82E

Solution

To check: The functions $f=g$ are same. $f\left(x\right)=1+\frac{1}{\left(x-\frac{1}{x}\right)},g\left(x\right)=\frac{\left(x^3+x^2-x\right)}{x^3-x}$ .

Given information:

The functions are $f\left(x\right)=1+\frac{1}{\left(x-\frac{1}{x}\right)},g\left(x\right)=\frac{\left(x^3+x^2-x\right)}{x^3-x}$ .

Concept used:

Using the analyze of asymptotes, intercepts, end behavior can prove that the both given function are same.

Calculations:

Rewrite

  $\begin{array}{c}g\left(x\right)=\frac{x^3-x}{x^3-x}+\frac{x^2}{x^3-x}\\ =1+\frac{x^2}{x^2\left(x-\frac{1}{x}\right)}\\ =1+\frac{1}{\left(x-\frac{1}{x}\right)}\\ =f\left(x\right)\end{array}$

Means these functions are algebraically identical.

Now find out the domain.

The domain of the function $f\left(x\right)$ is real except $x=-1,1,0$ since the numerator is $x-\frac{1}{x}$ should not be equal to zero. The same for the function $g\left(x\right)$ since $x^3-x=0$ .Thus the two functions have the same domain.

And the vertical asymptotes are since the denominators for both function is $x\left(x^2-1\right)$ so asymptotes are at $x=1,-1,0$ and horizontal asymptote is one in both case. Thus asymptotes is also the same.

The end behavior of $\underset{x\to {\infty }^{+}}{\lim }f\left(x\right)=\underset{x\to {\infty }^{-}}{\lim }f\left(x\right)=1$ , the end behavior is same for $g\left(x\right)$ as well.

Hence both functions are equal here since domain end behavior etc are equal in both cases.