Answer
a) $\frac{15}{4}$
b) $\frac{15}{4}$
c) $\frac{4{{x}^{2}}-1}{{{x}^{2}}}$
Work Step by Step
(a)
Let us consider the function $f\left( x \right)=\frac{4{{x}^{2}}-1}{{{x}^{2}}}$.
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
$\begin{align}
& f\left( x \right)=\frac{4{{x}^{2}}-1}{{{x}^{2}}} \\
& f\left( 2 \right)=\frac{4\cdot \left( {{2}^{2}} \right)-1}{{{2}^{2}}} \\
& =\frac{16-1}{4} \\
& =\frac{15}{4}
\end{align}$
Therefore, the value of $f\left( 2 \right)$ is $\frac{15}{4}$.
(b)
Let us consider the function $f\left( x \right)=\frac{4{{x}^{2}}-1}{{{x}^{2}}}$.
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
$\begin{align}
& f\left( x \right)=\frac{4{{x}^{2}}-1}{{{x}^{2}}} \\
& f\left( -2 \right)=\frac{4\cdot {{\left( -2 \right)}^{2}}-1}{{{\left( -2 \right)}^{2}}} \\
& =\frac{16-1}{4} \\
& =\frac{15}{4}
\end{align}$
Hence, the value of $f\left( -2 \right)$ is $\frac{15}{4}$.
(c)
Let us consider the function $f\left( x \right)=\frac{4{{x}^{2}}-1}{{{x}^{2}}}$.
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
$\begin{align}
& f\left( x \right)=\frac{4{{x}^{2}}-1}{{{x}^{2}}} \\
& f\left( -x \right)=\frac{4\cdot {{\left( -x \right)}^{2}}-1}{{{\left( -x \right)}^{2}}} \\
& =\frac{4{{x}^{2}}-1}{{{x}^{2}}}
\end{align}$
Hence, the value of $f\left( -x \right)$ is $\frac{4{{x}^{2}}-1}{{{x}^{2}}}$.
