Single Variable Calculus, Chapter 4, 4.5, Section 4.5, Problem 18

Use the guidelines of curve sketching to sketch the curve. $\displaystyle y = \frac{x}{x^3 -1 }$

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, $x = 1$. Therefore,the domain is $(-\infty, 1) \bigcup (1, \infty)$


B. Intercepts.
Solving for $y$-intercept, when $x = 0$
$\displaystyle y = \frac{0}{0^3 -1 } = 0 $
Solving for $x$-intercept, when $y=0$
$\displaystyle 0 = \frac{x}{x^3 -1}$
We have, $x = 0$

C. Symmetry.
The function is not symmetric to either $y$-axis or origin by using symmetry test.

D. Asymptotes.
For the vertical asymptotes, we set the denominator equal to 0.
$x^3 -1 = 0\\
x=1$


For the horizontal asymptotes, since the degree of denominator is greater than the degree of numerator, we have $y=0$ as horizontal asymptote.

E. Intervals of Increase or Decrease.
If we take the derivative of $f(x)$, by using Quotient Rule..
$\displaystyle y' = \frac{(x^3-1)(1)-(x)(3x^2)}{(x^3-1)^2} = \frac{-2x^3-1}{(x^3-1)^2}$
When $y' =0$,

$
\begin{equation}
\begin{aligned}
0 & = -2x^3 -1\\
\\
2x^3 &= -1\\
\\
x^3 &= \frac{-1}{2} = \sqrt[3]{\frac{-1}{2}}
\end{aligned}
\end{equation}
$

Therefore, the critical number is $ x = -0.7937$

Hence, the intervals of increase or decrease are

$
\begin{array}{|c|c|c|}
\hline\\
\text{Interval} & f'(x) & f\\
\hline\\
x < -0.7937 & + & \text{increasing on } (-\infty, -0.7937)\\
\hline\\
-0.7937 < x < 1 & - & \text{decreasing on } (-0.7937, 1)\\
\hline\\
x > 1 & - & \text{decreasing on } (1,\infty)\\
\hline
\end{array}
$



F. Local Maximum and Minimum Values.
Since $f'(x)$ changes from positive to negative at $ x= -0.7937$, then $f(-0.7937) = 0.5291$ is a local minimum.

G. Concavity and Points of Inflection.

$
\begin{equation}
\begin{aligned}
\text{if } f'(x) &= \frac{-2x^3 - 1 }{(x^3-1)^2} \quad \text{, then by using Quotient Rule and Chain Rule,}\\
\\
f''(x) &= \frac{(x^3-1)^2 (-6x^2)-(-2x^3-1)\left(2(x^3-1)(3x^2)\right)}{\left[ (x^3-1)^2\right]^2}
\end{aligned}
\end{equation}
$

Which can be simplified as, $\displaystyle f''(x) = \frac{6x^2(x^3+2)}{(x^3-1)^3}$
When $f''(x) =0$
$0 = 6x^2(x^3+2)$
The inflection points is $\displaystyle x = \sqrt[3]{-2} = -1.2599$
Thus, the concavity can be determined by dividing the interval to...

$
\begin{array}{|c|c|c|}
\hline\\
\text{Interval} & f''(x) & \text{Concavity}\\
\hline\\
x < -1.2599 & + & \text{Upward}\\
\hline\\
-1.2599 < x < -1 & - & \text{Downard}\\
\hline\\
-1 < x < 1 & - & \text{Downward}\\
\hline\\
x > 1 & + & \text{Upward}\\
\hline
\end{array}
$


H. Sketch the Graph.