Question

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)

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) \(0\)