3. consider the function $f(x)=\frac{|x|(x - 3)}{9 - x^{2}}$.\na) what is the domain of $f$? express your…

Explanation:

Step1: Find the domain

The denominator of the function $f(x)=\frac{|x|(x - 3)}{9 - x^{2}}$ cannot be zero. Since $9 - x^{2}=(3 + x)(3 - x)=0$ when $x = 3$ or $x=-3$, the domain is all real - numbers except $x = 3$ and $x=-3$. In interval notation, it is $(-\infty,-3)\cup(-3,3)\cup(3,\infty)$.

Step2: Find the zeros

Set the numerator equal to zero. Since $|x|(x - 3)=0$, we have $|x| = 0$ or $x - 3=0$. But when $x = 3$, the denominator is zero. So the zero of the function is $x = 0$ (because $|0|(0 - 3)=0$).

Step3: Evaluate the limit as $x\to3$

We have $\lim_{x\to3}\frac{|x|(x - 3)}{9 - x^{2}}=\lim_{x\to3}\frac{|x|(x - 3)}{-(x^{2}-9)}=\lim_{x\to3}\frac{|x|(x - 3)}{-(x + 3)(x - 3)}=\lim_{x\to3}\frac{|x|}{-(x + 3)}$. When $x\to3^{+}$, $|x|=x$, and $\lim_{x\to3^{+}}\frac{x}{-(x + 3)}=-\frac{1}{2}$. When $x\to3^{-}$, $|x|=x$, and $\lim_{x\to3^{-}}\frac{x}{-(x + 3)}=-\frac{1}{2}$. So $\lim_{x\to3}f(x)=-\frac{1}{2}$.

Step4: Find vertical asymptotes

Vertical asymptotes occur where the denominator is zero and the numerator is non - zero. Since $9 - x^{2}=(3 + x)(3 - x)=0$ at $x=-3$ and $x = 3$, and the numerator is non - zero at these points (except when considering the limit process), the vertical asymptotes are $x=-3$ and $x = 3$.

Step5: Find horizontal asymptotes

We rewrite the function as $f(x)=\frac{|x|(x - 3)}{9 - x^{2}}$. When $x\to\pm\infty$, we consider the highest - degree terms. If $x\to\infty$, $f(x)=\frac{x(x - 3)}{-x^{2}}=\frac{x^{2}-3x}{-x^{2}}=-1+\frac{3}{x}$, and $\lim_{x\to\infty}f(x)=-1$. If $x\to-\infty$, $f(x)=\frac{-x(x - 3)}{-x^{2}}=\frac{-x^{2}+3x}{-x^{2}}=1-\frac{3}{x}$, and $\lim_{x\to-\infty}f(x)=1$. So the horizontal asymptotes are $y = 1$ and $y=-1$.

Answer:

a) $(-\infty,-3)\cup(-3,3)\cup(3,\infty)$ b) $x = 0$ c) $-\frac{1}{2}$ d) $x=-3,x = 3$ e) $y = 1,y=-1$