Answer
A. The domain is $(-\infty, \infty)$
B. The y-intercept is $0$
The x-intercepts are $0$ and $4$
C. The function is not an odd function or an even function.
D. $\lim\limits_{x \to -\infty} x(x-4)^3 = \infty$
$\lim\limits_{x \to \infty} x(x-4)^3 = \infty$
There are no asymptotes.
E. The function is increasing on the intervals $(1,4)\cup (4, \infty)$
The function is decreasing on the interval $(-\infty,1)$
F. $(1,-27)$ is a local minimum.
G. The graph is concave down on the interval $(2,4)$
The graph is concave up on the intervals $(-\infty, 2) \cup (4, \infty)$
The points of inflection are $(2, -16)$ and $(4,0)$
H. We can see a sketch of the curve below.
Work Step by Step
$y = x(x-4)^3$
A. The function is defined for all real numbers. The domain is $(-\infty, \infty)$
B. When $x = 0,$ then $y = (0)(0-4)^3 = 0$
The y-intercept is $0$
When $y = 0$:
$x(x-4)^3 = 0$
$x = 0,4$
The x-intercepts are $0$ and $4$
C. The function is not an odd function or an even function.
D. $\lim\limits_{x \to -\infty} x(x-4)^3 = \infty$
$\lim\limits_{x \to \infty} x(x-4)^3 = \infty$
There are no asymptotes.
E. We can find the values of $x$ such that $y' = 0$:
$y' = (x-4)^3+(x)(3)(x-4)^2= 0$
$(x-4)^3=(-3x)(x-4)^2$
$x=4$ or $(x-4)=(-3x)$
$x=4$ or $x=1$
The function is increasing on the intervals $(1,4)\cup (4, \infty)$
The function is decreasing on the interval $(-\infty,1)$
F. When $x = 1$, then $y = (1)(1-4)^3= -27$
$(1,-27)$ is a local minimum.
G. We can find the values of $x$ such that $y'' = 0$:
$y' = (x-4)^3+(3x)(x-4)^2$
$y'' = 3(x-4)^2+(3)(x-4)^2+(3x)(2)(x-4) = 0$
$x = 4~~~$ or
$3(x-4)+3(x-4)+6x = 0$
$12x-24= 0$
$x = 2$
$x = 2, 4$
When $2 \lt x \lt 4$, then $y'' \lt 0$
The graph is concave down on the interval $(2,4)$
When $x \lt 2$ or $x \gt 4$, then $y'' \gt 0$
The graph is concave up on the intervals $(-\infty, 2) \cup (4, \infty)$
When $x = 2$, then $y = 2(2-4)^3 = -16$
When $x = 4$, then $y = (4)(4-4)^3 = 0$
The points of inflection are $(2, -16)$ and $(4,0)$
H. We can see a sketch of the curve below.