Single Variable Calculus, Chapter 4, 4.5, Section 4.5, Problem 14

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

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 our case, there is no real solution that would make the function undefined. Therefore, the domain is $(-\infty,\infty)$

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


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

D. Asymptotes.
For vertical asymptotes, we equate the denominator to 0. That is $x^2 + 9 =0$ but there is no real solution, therefore we have no vertical asymptotes.
For horizontal asymptotes, we divide the coefficient of the highest degree in both numerator and denominator. We have, $y = \frac{1}{1} = 1$.

E. Intervals of Increase or Decrease.
If we take the derivative of $f(x)$, by using Quotient Rule,
$\displaystyle f'(x) = \frac{(x^2 + 9)(2x) - (x^2)(2x)}{(x^2+9)^2} = \frac{18x}{(x^2+9)^2}$
When $f'(x) = 0$,
$\displaystyle 0 = \frac{18x}{(x^2+9)^2}$
We have, $x = 0$ as a critical number
If we divide the interval, we can determine the intervals of increase or decrease.


$\begin{array}{|c|c|c|} \hline\\ \text{Interval} & f'(x) & f \\ \hline\\ x < 0 & - & \text{decreasing on } ( - \infty, 0)\\ \hline\\ x > 0 & + & \text{increasing on } ( 0, \infty)\\ \hline \end{array}$


F. Local Maximum and Minimum Values.
since $f'(x)$ changes from negative to positive at $x = 0$ , $f(0) = 0$ is a local minimum.


G. Concavity and Points of Inflection.

$\begin{equation} \begin{aligned} \text{if } f'(x) &= \frac{18x}{(x^2+9)^2}, \text{ then}\\ \\ f''(x) &= \frac{(x^2 + 9)^2(18) - 18x \left( 2(x^2+9)(2x)\right)}{\left[(x^2+9)^2\right]^2} \end{aligned} \end{equation}$

Which can be simplified as...
$\displaystyle f''(x) = \frac{-54x^2+162}{(x^2+9)^2}$
When $f''(x) = 0$,

$\begin{equation} \begin{aligned} 0 &= -54x^2 + 162\\ \\ 54x^2 &= 162\\ \\ y^2 &= \frac{162}{54} \end{aligned} \end{equation}$


The inflection points are, $x = \pm \sqrt{3} = \pm 1.7321$
If we divide the interval we can now determine the concavity...

$\begin{array}{|c|c|c|} \hline\\ \text{Interval} & f'(x) & f\\ \hline\\ x < -1.7321 & - & \text{Downward}\\ \hline\\ x > 1.7321 & + & \text{Upward}\\ \hline\\ x > 1.7321 & - & \text{Downward}\\ \hline \end{array}$


H. Sketch the Graph.