Answer
A. The domain is $(-\infty, \infty)$
B. The y-intercept is $1$
The x-intercept is $-1$
C. The function is not an even function or an odd function.
D. $\lim\limits_{x \to -\infty} \sqrt[3] {x^3+1} = -\infty$
$\lim\limits_{x \to \infty} \sqrt[3] {x^3+1} = \infty$
The line $y=x$ is a slant asymptote.
E. The function is increasing on the interval $(-\infty, \infty)$
F. There is no local minimum or local maximum.
G. The graph is concave down on the interval $(-1, 0)$
The graph is concave up on the intervals $(-\infty,-1)\cup (0, \infty)$
The points of inflection are $(-1,0)$ and $(0,1)$
H. We can see a sketch of the curve below.
Work Step by Step
$y = \sqrt[3] {x^3+1}$
A. The function is defined for all real numbers.
The domain is $(-\infty, \infty)$
B. When $x=0$, then $y = \sqrt[3] {0^3+1} = 1$
The y-intercept is $1$
When $y = 0$:
$\sqrt[3] {x^3+1} = 0$
$x^3+1 = 0$
$x =-1$
The x-intercept is $-1$
C. The function is not an even function or an odd function.
D. $\lim\limits_{x \to -\infty} \sqrt[3] {x^3+1} = -\infty$
$\lim\limits_{x \to \infty} \sqrt[3] {x^3+1} = \infty$
There are no horizontal asymptotes.
$\lim\limits_{x \to -\infty} (\sqrt[3] {x^3+1}-x) = 0$
$\lim\limits_{x \to \infty} (\sqrt[3] {x^3+1}-x) = 0$
The line $y=x$ is a slant asymptote.
E. We can find values of $x$ such that $y' = 0$:
$y' =\frac{1}{3}(x^3+1)^{-2/3}(3x^2) = \frac{x^2}{(x^3+1)^{2/3}} = 0$
$x^2 = 0$
$x = 0$
Note that $y'$ is undefined when $x=-1$
When $x \lt -1$ or $-1 \lt x \lt 0$ or $x \gt 0$, then $y' \gt 0$
The function is increasing on the interval $(-\infty, \infty)$
F. There is no local minimum or local maximum since the function is increasing on all intervals.
G. We can try to find the values of $x$ such that $y'' = 0$:
$y'' =\frac{(2x)((x^3+1)^{2/3})- (x^2)(\frac{2x^2}{(x^3+1)^{1/3}})}{(x^3+1)^{4/3}}$
$y'' =\frac{(2x)(x^3+1)- (x^2)(2x^2)}{(x^3+1)^{5/3}}$
$y'' =\frac{2x}{(x^3+1)^{5/3}} = 0$
$2x = 0$
$x = 0$
Note that $y''$ is undefined when $x=-1$
When $-1 \lt x \lt 0$, then $y'' \lt 0$
The graph is concave down on the interval $(-1, 0)$
When $x \lt -1$ or $x \gt 0$, then $y'' \gt 0$
The graph is concave up on the intervals $(-\infty,-1)\cup (0, \infty)$
When $x=-1$, then $y = \sqrt[3] {(-1)^3+1} = 0$
When $x=0$, then $y = \sqrt[3] {0^3+1} = 1$
The points of inflection are $(-1,0)$ and $(0,1)$
H. We can see a sketch of the curve below.