Answer
A. The domain is $(-\infty, \infty)$
B. The y-intercept is $0$
The x-intercepts are $-5.2, 0, 5.2$
C. The function is an odd function.
D. $\lim\limits_{x \to -\infty} x-3x^{1/3} = -\infty$
$\lim\limits_{x \to \infty} x-3x^{1/3} = \infty$
No asymptotes.
E. The function is decreasing on the interval $(-1,1)$
The function is increasing on the intervals $(-\infty, -1)\cup (1, \infty)$
F. The local maximum is $(-1, 2)$
The local minimum is $(1, -2)$
G. The graph is concave down on the interval $(-\infty,0)$
The graph is concave up on the interval $(0, \infty)$
The point of inflection is $(0,0)$
H. We can see a sketch of the curve below.
Work Step by Step
$y = x-3x^{1/3}$
A. The function is defined for all real numbers.
The domain is $(-\infty, \infty)$
B. When $x=0$, then $y = 0-3(0)^{1/3} = 0$
The y-intercept is $0$
When $y = 0$:
$x-3x^{1/3} = 0$
$x=3x^{1/3}$
$x^3=27x$
$x^3-27x = 0$
$x(x^2-27) = 0$
$x = -\sqrt{27}, 0, \sqrt{27}$
$x = -5.2, 0, 5.2$
The x-intercepts are $-5.2, 0, 5.2$
C. $(-x)-3(-x)^{1/3}= -(x-3x^{1/3})$
The function is an odd function.
D. $\lim\limits_{x \to -\infty} x-3x^{1/3} = -\infty$
$\lim\limits_{x \to \infty} x-3x^{1/3} = \infty$
There are no horizontal asymptotes.
E. We can find values of $x$ such that $y' = 0$:
$y' =1-x^{-2/3} = 1-\frac{1}{x^{2/3}} = 0$
$x^{2/3} = 1$
$x = -1, 1$
Note that $y'$ is undefined when $x=0$
When $-1 \lt x \lt 0$ or $0 \lt x \lt 1$, then $y' \lt 0$
The function is decreasing on the interval $(-1,1)$
When $x \lt -1$ or $x \gt 1$, then $y' \gt 0$
The function is increasing on the intervals $(-\infty, -1)\cup (1, \infty)$
F. When $x=-1$, then $y = (-1)-3(-1)^{1/3} = 2$
The local maximum is $(-1, 2)$
When $x=1$, then $y = (1)-3(1)^{1/3} = -2$
The local minimum is $(1, -2)$
G. We can find the values of $x$ such that $y'' = 0$:
$y'' =\frac{2}{3}\cdot \frac{1}{x^{5/3}} = \frac{2}{3~x^{5/3}} = 0$
There are no values $x$ such that $y'' = 0$
Note that $y''$ is undefined when $x=0$
When $x \lt 0$, then $y'' \lt 0$
The graph is concave down on the interval $(-\infty,0)$
When $x \gt 0$, then $y'' \gt 0$
The graph is concave up on the interval $(0, \infty)$
The point of inflection is $(0,0)$
H. We can see a sketch of the curve below.