Single Variable Calculus, Chapter 4, 4.5, Section 4.5, Problem 12

Use the guidelines of curve sketching to sketch the curve. $\displaystyle y = \frac{x}{x^2-9}$

The guidelines of Curve Sketching
A. Domain.
We know that $f(x)$ is a rational function that is defined everywhere except for the value of $x$ that would make its denominator equal to 0. In this case, $x = 3$ and $x = -3$ therefore, the domain is $(-\infty, -3)\bigcup(3,\infty)$

B. Intercepts.
Solving for $y$-intercept, when $x = 0$
$\displaystyle y = \frac{0}{0^2 - 9 } = 0$
Solving for $x$-intercept, when $y = 0$
$\displaystyle 0 = \frac{x}{x^2 - 9}$
We have $x = 0$


C. Symmetry.
Since, $f(-x) = -f(x)$, the function is symmetric to origin.

D. Asymptotes.
For vertical asymptotes, we equate the denominator to 0, that is $x^2 - 9 = 0$. We have, $x = 3$ and $x = -3$
For horizontal asymptotes, since the degree is greater in the denominator than in the numerator, we have $y = 0$ as horizontal asymptotes.


E. Intervals of Increase or Decrease.
If we take the derivative of $f(x)$, by using Quotient Rule,
$\displaystyle f'(x) = \frac{(x^2-9)(1)-(x)(2x)}{(x^2-9)^2} = \frac{-x^2 - 9}{(x^2-9)^2}$
When $f'(x) = 0$
$\displaystyle 0 = \frac{-x^2-9}{(x^2-9)^2}$
$f'(x) = 0$ does not exist, therefore we have no critical numbers
If we divide the interval by its domain, we can determine the intervals of increase or decrease.

$\begin{array}{|c|c|c|} \hline\\ \text{Interval} & f'(x) & f\\ \hline\\ x < - 3 & - & \text{decreasing on } (-\infty,-3)\\ \hline\\ -3 < x < 3& - & \text{decreasing on } (-3,3)\\ \hline\\ x > 3 & - & \text{decreasing on } (3, \infty)\\ \hline \end{array}$



F. Local Maximum and Minimum Values.
Since $f'(x)$ doesn't change sign, we can say that the function has no local maximum and minimum.


G. Concavity and Points of Inflection.

$\begin{equation} \begin{aligned} \text{if } f'(x) &= \frac{-x^2 - 9 }{(x^2 - 9)^2} , \text{ then}\\ \\ f''(x) &= \frac{(x^2 - 9)^2 (-2x) - (-x^2 - 9) \left( 2(x^2 -9)(2x)\right)}{\left[ (x^2 - 9)^2 \right]^2} \end{aligned} \end{equation}$

Which can be simplified as,

$\begin{equation} \begin{aligned} f''(x) &= \frac{2x^3 + 54x}{(x^2-9)^3}\\ \\ \text{when } f''(x) &= 0,\\ \\ 0 & = 2x^3 +54x\\ \\ 0 &= 2x(x^2 + 27) \end{aligned} \end{equation}$

We have a real solution of, $x = 0$ (Inflection point)
If we divide the interval within the domain, we can determine the concavity as...

$\begin{array}{|c|c|c|} \hline\\ \text{Interval} & f''(x) & \text{Concavity}\\ \hline\\ x < -3 & - & \text{Downward}\\ \hline\\ -3 < x < 0 & + & \text{Upward}\\ \hline\\ 0 < x < 3 & - & \text{Downward}\\ \hline\\ x > 3 & + & \text{Upward}\\ \hline \end{array}$



H. Sketch the Graph.