find each function value and limit. use - ∞ or ∞ where appropriate.
f(x)=\frac{3x^{3}-6x^{2}}{16x^{4}+8}
(a) f(-6)
(b) f(-12)
(c) lim_{x
ightarrow - infty} f(x)

Question

Accepted Answer

# Explanation:
## Step1: Calculate f(-6)
Substitute $x = - 6$ into $f(x)=\frac{3x^{3}-6x^{2}}{16x^{4}+8}$.
$$
\begin{align*}
f(-6)&=\frac{3(-6)^{3}-6(-6)^{2}}{16(-6)^{4}+8}\
&=\frac{3\times(-216)-6\times36}{16\times1296 + 8}\
&=\frac{-648- 216}{20736+8}\
&=\frac{-864}{20744}\
&=-\frac{108}{2593}
\end{align*}
$$
## Step2: Calculate f(-12)
Substitute $x=-12$ into $f(x)=\frac{3x^{3}-6x^{2}}{16x^{4}+8}$.
$$
\begin{align*}
f(-12)&=\frac{3(-12)^{3}-6(-12)^{2}}{16(-12)^{4}+8}\
&=\frac{3\times(-1728)-6\times144}{16\times20736+8}\
&=\frac{-5184 - 864}{331776+8}\
&=\frac{-6048}{331784}\
&=-\frac{756}{41473}
\end{align*}
$$
## Step3: Find $\lim_{x
ightarrow-\infty}f(x)$
Divide both the numerator and denominator by $x^{4}$:
$$
\begin{align*}
f(x)&=\frac{3x^{3}-6x^{2}}{16x^{4}+8}\
&=\frac{\frac{3x^{3}}{x^{4}}-\frac{6x^{2}}{x^{4}}}{\frac{16x^{4}}{x^{4}}+\frac{8}{x^{4}}}\
&=\frac{\frac{3}{x}-\frac{6}{x^{2}}}{16 + \frac{8}{x^{4}}}
\end{align*}
$$
As $x
ightarrow-\infty$, $\frac{3}{x}
ightarrow0$, $\frac{6}{x^{2}}
ightarrow0$ and $\frac{8}{x^{4}}
ightarrow0$. So $\lim_{x
ightarrow-\infty}f(x)=\frac{0 - 0}{16+0}=0$.

# Answer:
(A) $-\frac{108}{2593}$
(B) $-\frac{756}{41473}$
(C) $0$

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QUESTION IMAGE

Question

Explanation:

Step1: Calculate f(-6)

Step2: Calculate f(-12)

Step3: Find \(\lim_{x

Answer: