Answer
A. The domain is $(-\infty, \infty)$
B. The y-intercept is $0$
The x-intercepts are $-2$ and $2$
C. The function is an even function.
D. $\lim\limits_{x \to -\infty} (4-x^2)^5 = -\infty$
$\lim\limits_{x \to \infty} (4-x^2)^5 = -\infty$
There are no asymptotes.
E. The function is increasing on the intervals $(-\infty, -2)\cup (-2, 0)$
The function is decreasing on the interval $(0,2)\cup (2, \infty)$
F. $(0,1024)$ is a local maximum.
G. The graph is concave down on the intervals $(-\infty, -2)\cup (-\frac{2}{3}, \frac{2}{3})\cup (2, \infty)$
The graph is concave up on the intervals $(-2, -\frac{2}{3}) \cup (\frac{2}{3}, 2)$
The points of inflection are $(-2, 0), (-\frac{2}{3}, 568), (\frac{2}{3}, 568)$ and $(2,0)$
H. We can see a sketch of the curve below.
Work Step by Step
$y = (4-x^2)^5$
A. The function is defined for all real numbers. The domain is $(-\infty, \infty)$
B. When $x = 0,$ then $y = (4-0^2)^5 = 1024$
The y-intercept is $0$
When $y = 0$:
$(4-x^2)^5 = 0$
$4-x^2=0$
$x = -2,2$
The x-intercepts are $-2$ and $2$
C. $[4-(-x)^2]^5 = (4-x^2)^5$
The function is an even function.
D. $\lim\limits_{x \to -\infty} (4-x^2)^5 = -\infty$
$\lim\limits_{x \to \infty} (4-x^2)^5 = -\infty$
There are no asymptotes.
E. We can find the values of $x$ such that $y' = 0$:
$y' = 5(4-x^2)^4(-2x)= 0$
$x=0~~$ or $~~(4-x^2)=0$
$x=-2,0,2$
The function is increasing on the intervals $(-\infty, -2)\cup (-2, 0)$
The function is decreasing on the interval $(0,2)\cup (2, \infty)$
F. When $x = 0$, then $y = (4-(0)^2)^5= 1024$
$(0,1024)$ is a local maximum.
G. We can find the values of $x$ such that $y'' = 0$:
$y'' = (-10)(4-x^2)^4+(-10x)(4)(4-x^2)^3(-2x)$
$y'' = (-10)(4-x^2)^4+(80x^2)(4-x^2)^3 = 0$
$(-10)(4-x^2)^4+(80x^2)(4-x^2)^3 = 0$
$x = -2, 2~~$ or $~~(-10)(4-x^2)+(80x^2) = 0$
$x = -2, 2~~$ or $~~90x^2-40 = 0$
$x = -2, 2~~$ or $~~x^2 = \frac{4}{9}$
$x = -2, 2~~$ or $~~x = \pm \frac{2}{3}$
When $x \lt -2~~$ or $~~-\frac{2}{3} \lt x \lt \frac{2}{3}~~$ or $~~x \gt 2$, then $y'' \lt 0$
The graph is concave down on the intervals $(-\infty, -2)\cup (-\frac{2}{3}, \frac{2}{3})\cup (2, \infty)$
When $-2 \lt x \lt -\frac{2}{3}$ or $\frac{2}{3} \lt x \lt 2$, then $y'' \gt 0$
The graph is concave up on the intervals $(-2, -\frac{2}{3}) \cup (\frac{2}{3}, 2)$
When $x = -2$, then $y = [4-(-2)^2]^5 = 0$
When $x = -\frac{2}{3}$, then $y = [4-(-\frac{2}{3})^2]^5 = 568$
When $x = \frac{2}{3}$, then $y = [4-(\frac{2}{3})^2]^5 = 568$
When $x = 2$, then $y = [4-(2)^2]^5 = 0$
The points of inflection are $(-2, 0), (-\frac{2}{3}, 568), (\frac{2}{3}, 568)$ and $(2,0)$
H. We can see a sketch of the curve below.