Answer
A. The domain is $(-\infty,1)\cup (1,\infty)$
B. The y-intercept is $0$
The x-intercept is $0$
C. The function is not an even function or an odd function.
D. $\lim\limits_{x \to -\infty} (\frac{x}{x^3-1}) = 0$
$\lim\limits_{x \to \infty} (\frac{x}{x^3-1}) = 0$
$y = 0$ is a horizontal asymptote.
$x = 1$ is a vertical asymptote.
E. The function is decreasing on the intervals $(-0.794, 1)\cup (1, \infty)$
The function is increasing on the interval $(-\infty, -0.794)$
F. The local maximum is $(-0.794, 0.529)$
G. The graph is concave down on the intervals $(-1.26,0)\cup (0,1)$
The graph is concave up on the intervals $(-\infty, -1.26)\cup (1, \infty)$
The point of inflection is $(-1.26, 4.2)$
H. We can see a sketch of the curve below.
Work Step by Step
$y = \frac{x}{x^3-1}$
A. The function is defined for all real numbers except $x=1$
The domain is $(-\infty,1)\cup (1,\infty)$
B. When $x=0$ then $y = \frac{(0)}{(0)^3-1} = 0$
The y-intercept is $0$
When $y = 0$:
$\frac{x}{x^3-1} = 0$
$x = 0$
The x-intercept is $0$
C. The function is not an even function or an odd function.
D. $\lim\limits_{x \to -\infty} (\frac{x}{x^3-1}) = 0$
$\lim\limits_{x \to \infty} (\frac{x}{x^3-1}) = 0$
$y = 0$ is a horizontal asymptote.
E. We can find the values of $x$ such that $y' = 0$:
$y' = \frac{(x^3-1)-(x)(3x^2)}{(x^3-1)^2}$
$y' = \frac{-2x^3-1}{(x^3-1)^2} = 0$
$-2x^3-1 = 0$
$x^3 = -\frac{1}{2}$
$x = -\sqrt[3] {\frac{1}{2}} = -0.794$
When $-0.794 \lt x \lt 1$ or $x \gt 1$ then $y' \lt 0$
The function is decreasing on the intervals $(-0.794, 1)\cup (1, \infty)$
When $x \lt -0.794$, then $y' \gt 0$
The function is increasing on the interval $(-\infty, -0.794)$
F. When $x = -0.794$ then $y = \frac{(-0.794)}{(-0.794)^3-1} = 0.529$
The local maximum is $(-0.794, 0.529)$
G. We can find the values of $x$ such that $y'' = 0$:
$y'' = \frac{(-6x^2)(x^3-1)^2-(-2x^3-1)(2)(x^3-1)(3x^2)}{(x^3-1)^4}$
$y'' = \frac{(-6x^2)(x^3-1)-(-2x^3-1)(2)(3x^2)}{(x^3-1)^3}$
$y'' = \frac{-6x^5+6x^2+12x^5+6x^2}{(x^3-1)^3}$
$y'' = \frac{6x^5+12x^2}{(x^3-1)^3} = 0$
$6x^5+12x^2 = 0$
$6x^2(x^3+2) = 0$
$x = 0$ or $x^3 = -2$
$x=0$ or $x = \sqrt[3] {-2}$
$x=0$ or $x = -1.26$
When $-1.26 \lt x \lt 0~~$ or $~~0 \lt x \lt 1~~$ then $y'' \lt 0$
The graph is concave down on the intervals $(-1.26,0)\cup (0,1)$
When $x \lt -1.26$ or $x \gt 1$, then $y'' \gt 0$
The graph is concave up on the intervals $(-\infty, -1.26)\cup (1, \infty)$
When $x = -1.26,$ then $y = \frac{(-1.26)}{(-1.26)^3-1} = 0.42$
The point of inflection is $(-1.26, 4.2)$
H. We can see a sketch of the curve below.
