Question

Answer the following questions for the function
f(x)=x^3/(x^2−4)
defined on the interval [−19,16].

a.) Enter the x-coordinates of the vertical asymptotes of f(x) as a comma-separated list. That is, if there is just one value, give it; if there are more than one, enter them separated commas; and if there are none, enter NONE .

b)f(x) is concave up on the region=

Answer 1

a) Since $\lim_{x\to\pm 2}\frac{x^3}{x^2-4}=\infty$, the lines $x=-2$ and $x=2$ are the vertical asymptotes of $f(x)$.

Answer: -2, 2

Let us find fist and second derivatives of the function $f(x)=\frac{x^3}{x^2-4}$ defined on the interval $[−19,16]$ :

$f'(x)=\frac{3x^2(x^2-4)-x^32x}{(x^2-4)^2}=\frac{x^4-12x^2}{(x^2-4)^2}$

$f''(x)=\frac{(4x^3-24x)(x^2-4)^2-(x^4-12x^2)2(x^2-4)2x}{(x^2-4)^4}= \frac{(4x^3-24x)(x^2-4)-4(x^4-12x^2)x}{(x^2-4)^3}=$

$=\frac{4x^5-16x^3-24x^3+96x-4x^5+48x^3}{(x^2-4)^3}=\frac{8x^3+96x}{(x^2-4)^3}=\frac{8x(x^2+12)}{(x^2-4)^3}$

Since $f(x)$ is concave up if $f''(x)>0$, we conclude that $f(x)$ is concave up on the region $(-2,0)\cup(2,16].$

Answer: $(-2,0)\cup(2,16].$