# Analyze and sketch the graph of f(x)=\frac{3x}{x^2-1}.

## Question:

Analyze and sketch the graph of {eq}f(x)=\frac{3x}{x^2-1}{/eq}.

## Rational Functions

When we analyze rational functions or expression over the real numbers we have the issue of the domain.

All numbers that make the denominator zero have to be excluded. The behavior of the function about these

gaps in the domain has to be investigated. The function could tend to plus or minus infinity or just have a gap,

one point missing, in the graph. The other issue is when the variable x goes to plus or minus infinity. What

is the behavior of the function? Does it also tend to plus or minus infinity or does it approach a fixed value?

There may be local extreme points, but not always and points of inflection may occur also.

## Answer and Explanation:

The very first item of the analysis is the maximal domain of {eq}f(x) {/eq} with respect to the

real numbers. That is, we have to exclude the...

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View this answerThe very first item of the analysis is the maximal domain of {eq}f(x) {/eq} with respect to the

real numbers. That is, we have to exclude the numbers that make the denominator zero.

We know that {eq}x^2 -1 = (x-1)(x+1) {/eq} and therefore the numbers {eq}x= \pm 1 {/eq} must be removed

from the reals. We have domain {eq}D_f = \mathbb{R} - \left\{ -1, 1 \right\}. {/eq}

We further investigate these gaps in the domain. We see that the numerator evaluates to

plus or minus three, at {eq}x= \pm 1. {/eq} The denominator is a symmetric function, independent of

the sign of {eq}x. {/eq} If {eq}|x|<1 {/eq} we have a negative denominator and if {eq}|x|>1 {/eq} the denominator is

positive. This is enough information to determine the behavior of the function about the

gaps or singularities {eq}x= \pm1. {/eq}

{eq}\begin{align*} \dfrac{3x}{x^2-1} \to \begin{cases} + \infty & \text{ as } x \to 1^+ \\ - \infty & \text{ as } x \to 1^- \\ + \infty & \text{ as } x \to -1^+ \\ - \infty & \text{ as } x \to -1^- \end{cases} \end{align*} {/eq}

Now, what happens as {eq}x \to \pm \infty? {/eq} We change the expression slightly and have

{eq}\begin{align*} \dfrac{3x}{x^2-1} &= \dfrac{3x}{x(x - \frac{1}{x})} \\ &= \dfrac{3}{(x - \frac{1}{x})} \to 0 \text{ as } x \to \pm \infty, \end{align*} {/eq}

observing that the function approaches the zero from below, i.e. from negative values

as {eq}x \to -\infty {/eq} and from above, i.e. from positive values as {eq}x \to + \infty. {/eq}

We have that the line {eq}y \equiv 0 {/eq} is a horizontal asymptote.

This would be enough to sketch the graph of the function. Let's however quickly look at

the derivatives

{eq}\begin{align*} \dfrac{df}{dx} &= \dfrac{3(x^2 -1)-3x \cdot 2x}{(x^2 -1)^2} \\ &= \dfrac{3x^2 -3-6x^2}{(x^2 -1)^2} \\ &= \dfrac{-3x^2 -3}{(x^2 -1)^2}, \end{align*} {/eq}

which has no roots since the numerator is always negative. Hence there are no local extreme

points. The second derivative

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Chapter 10 / Lesson 11