Answer
a) $\frac{33}{8}$
b) $\frac{31}{8}$
c) $\frac{-4{{x}^{3}}+1}{-{{x}^{3}}}$
Work Step by Step
(a)
Let us consider the function $f\left( x \right)=\frac{4{{x}^{3}}+1}{{{x}^{3}}}$.
By putting the value of $x$ as $2$ in the equation of $f\left( x \right)$ we obtain the value of $f\left( 2 \right)$ as:
$\begin{align}
& f\left( x \right)=\frac{4{{x}^{3}}+1}{{{x}^{3}}} \\
& f\left( 2 \right)=\frac{4\cdot {{2}^{3}}+1}{{{2}^{3}}} \\
& =\frac{4\cdot 8+1}{8} \\
& =\frac{33}{8}
\end{align}$
Hence, the value of $f\left( 2 \right)$ is $\frac{33}{8}$.
(b)
Let us consider the function $f\left( x \right)=\frac{4{{x}^{3}}+1}{{{x}^{3}}}$.
By putting the value of $x$ as $-2$ in the equation of $f\left( x \right)$ we obtain the value of $f\left( -2 \right)$ as:
$\begin{align}
& f\left( x \right)=\frac{4{{x}^{3}}+1}{{{x}^{3}}} \\
& f\left( -2 \right)=\frac{4\cdot {{\left( -2 \right)}^{3}}+1}{{{\left( -2 \right)}^{3}}} \\
& =\frac{-4\cdot 8+1}{-8} \\
& =\frac{-31}{-8}
\end{align}$
$=\frac{31}{8}$
Hence, the value of $f\left( -2 \right)$ is $\frac{31}{8}$.
(c)
Let us consider the function $f\left( x \right)=\frac{4{{x}^{3}}+1}{{{x}^{3}}}$.
By putting the value of $x$ as $-x$ in the equation of $f\left( x \right)$ we obtain the value of $f\left( -x \right)$ as:
$\begin{align}
& f\left( x \right)=\frac{4{{x}^{3}}+1}{{{x}^{3}}} \\
& f\left( -x \right)=\frac{4\cdot {{\left( -x \right)}^{3}}+1}{{{\left( -x \right)}^{3}}} \\
& =\frac{-4{{x}^{3}}+1}{-{{x}^{3}}}
\end{align}$
Hence, the value of $f\left( -x \right)$ is $\frac{-4{{x}^{3}}+1}{-{{x}^{3}}}$.
