Answer
A. The domain is $(-\infty, \infty)$
B. The y-intercept is $-1$
The x-intercepts are $-1$ and $1$
C. The function is an even function.
D. $\lim\limits_{x \to -\infty} \sqrt[3] {x^2-1} = -\infty$
$\lim\limits_{x \to \infty} \sqrt[3] {x^2-1} = \infty$
No asymptotes.
E. The function is decreasing on the interval $(-\infty,0)$
The function is increasing on the interval $(0, \infty)$
F. The local minimum is $(0, -1)$
G. The graph is concave down on the intervals $(-\infty,-1)\cup (1, \infty)$
The graph is concave up on the interval $(-1,1)$
The points of inflection are $(-1,0)$ and $(1,0)$
H. We can see a sketch of the curve below.
Work Step by Step
$y = \sqrt[3] {x^2-1}$
A. The function is defined for all real numbers.
The domain is $(-\infty, \infty)$
B. When $x=0$, then $y = \sqrt[3] {0^2-1} = -1$
The y-intercept is $-1$
When $y = 0$:
$\sqrt[3] {x^2-1} = 0$
$x^2-1 = 0$
$x =-1, 1$
The x-intercepts are $-1$ and $1$
C. $\sqrt[3] {(-x)^2-1} = \sqrt[3] {x^2-1}$
The function is an even function.
D. $\lim\limits_{x \to -\infty} \sqrt[3] {x^2-1} = -\infty$
$\lim\limits_{x \to \infty} \sqrt[3] {x^2-1} = \infty$
There are no horizontal asymptotes.
E. We can find values of $x$ such that $y' = 0$:
$y' =\frac{1}{3}(x^2-1)^{-2/3}(2x) = \frac{2x}{3(x^2-1)^{2/3}} = 0$
$2x = 0$
$x = 0$
Note that $y'$ is undefined when $x=-1$ or $x=1$
When $x \lt -1$ or $-1 \lt x \lt 0$, then $y' \lt 0$
The function is decreasing on the interval $(-\infty,0)$
When $0 \lt x \lt 1$ or $x \gt 1$, then $y' \gt 0$
The function is increasing on the interval $(0, \infty)$
F. When $x=0$, then $y = \sqrt[3] {0^2-1} = -1$
The local minimum is $(0, -1)$
G. We can try to find the values of $x$ such that $y'' = 0$:
$y'' =\frac{(2)(3(x^2-1)^{2/3}-(2x)(\frac{4x}{(x^2-1)^{1/3}})}{9(x^2-1)^{4/3}}$
$y'' =\frac{(6)(x^2-1)-(8x^2)}{9(x^2-1)^{5/3}}$
$y'' =\frac{-2x^2-6}{9(x^2-1)^{5/3}} = 0$
$-2x^2-6 = 0$
$x^2+3 = 0$
There are no values of $x$ such that $y'' = 0$
Note that $y''$ is undefined when $x=-1$ or $x=1$
When $x \lt -1$ or $x \gt 1$, then $y'' \lt 0$
The graph is concave down on the intervals $(-\infty,-1)\cup (1, \infty)$
When $-1 \lt x \lt 1$, then $y'' \gt 0$
The graph is concave up on the interval $(-1,1)$
When $x=-1$, then $y = \sqrt[3] {(-1)^2-1} = 0$
When $x=1$, then $y = \sqrt[3] {1^2-1} = 0$
The points of inflection are $(-1,0)$ and $(1,0)$
H. We can see a sketch of the curve below.