Answer
A. The domain is $(-\infty, \infty)$
B. The y-intercept is $2$
C. The function is not odd or even.
D. $\lim\limits_{x \to -\infty} (2-2x-x^3) = \infty$
$\lim\limits_{x \to \infty} (2-2x-x^3) = -\infty$
There are no asymptotes.
E. The function is decreasing on the interval $(-\infty,\infty)$
F. There is no local minimum or local maximum.
G. The graph is concave down on the interval $(0, \infty)$
The graph is concave up on the interval $(-\infty, 0)$
The point of inflection is $(0, 2)$
H. We can see a sketch of the curve below.
Work Step by Step
$y = 2-2x-x^3$
A. The function is defined for all real numbers. The domain is $(-\infty, \infty)$
B. When $x = 0,$ then $y = 2-2(0)-(0)^3 = 2$
The y-intercept is $2$
C. The function is not odd or even.
D. $\lim\limits_{x \to -\infty} (2-2x-x^3) = \infty$
$\lim\limits_{x \to \infty} (2-2x-x^3) = -\infty$
There are no asymptotes.
E. We can find the values of $x$ such that $y' = 0$:
$y' = -2-3x^2 = 0$
$x^2 = -\frac{2}{3}$
There are no values of $x$ such that $y' = 0$
For all $x,~~~$ $y' \lt 0$
The function is decreasing on the interval $(-\infty,\infty)$
F. There is no local minimum or local maximum.
G. We can find the values of $x$ such that $y'' = 0$:
$y'' = -6x = 0$
$x = 0$
When $x \gt 0$, then $y'' \lt 0$
The graph is concave down on the interval $(0, \infty)$
When $x \lt 0$, then $y'' \gt 0$
The graph is concave up on the interval $(-\infty, 0)$
When $x=0$, then $y = 2-2(0)-(0)^3 = 2$
The point of inflection is $(0, 2)$
H. We can see a sketch of the curve below.