find each function value and limit. use -∞ or ∞ where appropriate.\nf(x) = \\frac{9x^{4}-18x^{2}}{18x^{5}+9}\…

Explanation:

Step1: Calculate f(-6)

Substitute $x = - 6$ into $f(x)=\frac{9x^{4}-18x^{2}}{18x^{5}+9}$. $f(-6)=\frac{9\times(-6)^{4}-18\times(-6)^{2}}{18\times(-6)^{5}+9}=\frac{9\times1296 - 18\times36}{18\times(-7776)+9}=\frac{11664-648}{-139968 + 9}=\frac{11016}{-139959}=-\frac{11016}{139959}$

Step2: Calculate f(-12)

Substitute $x=-12$ into $f(x)$. $f(-12)=\frac{9\times(-12)^{4}-18\times(-12)^{2}}{18\times(-12)^{5}+9}=\frac{9\times20736-18\times144}{18\times(-248832)+9}=\frac{186624 - 2592}{-4478976+9}=\frac{184032}{-4478967}=-\frac{184032}{4478967}$

Step3: Calculate $\lim_{x\rightarrow-\infty}f(x)$

For $\lim_{x\rightarrow-\infty}\frac{9x^{4}-18x^{2}}{18x^{5}+9}$, divide both numerator and denominator by $x^{5}$ (the highest - power of $x$ in the denominator). We get $\lim_{x\rightarrow-\infty}\frac{\frac{9}{x}-\frac{18}{x^{3}}}{18+\frac{9}{x^{5}}}$. As $x\rightarrow-\infty$, $\frac{9}{x}\rightarrow0$, $\frac{18}{x^{3}}\rightarrow0$ and $\frac{9}{x^{5}}\rightarrow0$. So $\lim_{x\rightarrow-\infty}\frac{\frac{9}{x}-\frac{18}{x^{3}}}{18+\frac{9}{x^{5}}}=0$

Answer:

(A) $-\frac{11016}{139959}$ (B) $-\frac{184032}{4478967}$ (C) $0$