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. $\lim\limits_{x \to -\infty} \frac{x^3}{x^3+1} = 1$
$\lim\limits_{x \to \infty} \frac{x^3}{x^3+1} = 1$
$y = 1$ is a horizontal asymptote.
$\lim\limits_{x \to -1^-} \frac{x^3}{x^3+1} = \infty$
$\lim\limits_{x \to -1^+} \frac{x^3}{x^3+1} = -\infty$
$x = -1$ is a vertical asymptote.
E. The function is increasing on the intervals $(-\infty, -1)\cup (-1,0)\cup (0, \infty)$
F. There is no local maximum or local minimum.
G. The graph is concave down on the intervals $(-1,0)\cup (0.794,\infty)$
The graph is concave up on the intervals $(-\infty,1)\cup (0,0.794)$
The points of inflection are $(0,0)$ and $(0.794, 0.33)$
H. We can see a sketch of the curve below.
Work Step by Step
$y = \frac{x^3}{x^3+1}$
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 $y = \frac{0^3}{0^3+1}= 0$
The y-intercept is $0$
When $y = 0$:
$\frac{x^3}{x^3+1} = 0$
$x^3 = 0$
$x = 0$
The x-intercept is $0$
C. The function is not an even function or an odd function.
D. $\lim\limits_{x \to -\infty} \frac{x^3}{x^3+1} = 1$
$\lim\limits_{x \to \infty} \frac{x^3}{x^3+1} = 1$
$y = 1$ is a horizontal asymptote.
$\lim\limits_{x \to -1^-} \frac{x^3}{x^3+1} = \infty$
$\lim\limits_{x \to -1^+} \frac{x^3}{x^3+1} = -\infty$
$x = -1$ is a vertical asymptote.
E. We can find the values of $x$ such that $y' = 0$:
$y' = \frac{(3x^2)(x^3+1)-(x^3)(3x^2)}{(x^3+1)^2} = \frac{3x^2}{(x^3+1)^2} = 0$
$3x^2 = 0$
$x = 0$
When $x \lt -1$ or $-1 \lt x \lt 0$ or $x \gt 0$, then $y' \gt 0$
The function is increasing on the intervals $(-\infty, -1)\cup (-1,0)\cup (0, \infty)$
F. There is no local maximum or local minimum.
G. We can find the values of $x$ such that $y'' = 0$:
$y'' = \frac{(6x)(x^3+1)^2-(3x^2)(2)(x^3+1)(3x^2)}{(x^3+1)^4}$
$y'' = \frac{(6x)(x^3+1)-(3x^2)(2)(3x^2)}{(x^3+1)^3}$
$y'' = \frac{6x-12x^4}{(x^3+1)^3} = 0$
$6x-12x^4 = 0$
$-6x(2x^3-1) = 0$
$x = 0$ or $x = \sqrt[3] {\frac{1}{2}}$
$x = 0$ or $x = 0.794$
When $-1 \lt x \lt 0$ or $x \gt 0.794$ then $y'' \lt 0$
The graph is concave down on the intervals $(-1,0)\cup (0.794,\infty)$
When $x \lt -1$ or $0 \lt x \lt 0.794$, then $y'' \gt 0$
The graph is concave up on the intervals $(-\infty,1)\cup (0,0.794)$
When $x = 0$, then $y = \frac{0^3}{0^3+1}= 0$
When $x = 0.794$, then $y = \frac{0.794^3}{0.794^3+1}= \frac{1}{3}$
The points of inflection are $(0,0)$ and $(0.794, 0.33)$
H. We can see a sketch of the curve below.