Answer
See the explanation below.
Work Step by Step
(a)
Consider the function,
$\begin{align}
& \left( f\circ g \right)\left( x \right)=f\left( g\left( x \right) \right) \\
& =f\left( 4x-1 \right) \\
& ={{\left( 4x-1 \right)}^{2}}+3
\end{align}$
On simplification, we get,
$\begin{align}
& \left( f\circ g \right)\left( x \right)={{\left( 4x-1 \right)}^{2}}+3 \\
& =16{{x}^{2}}-8x+1+3 \\
& =16{{x}^{2}}-8x+4
\end{align}$
Therefore, the value of $\left( f\circ g \right)\left( x \right)$ for the functions $f\left( x \right)={{x}^{2}}\text{+3 and }g\left( x \right)=4x-1$ is $16{{x}^{2}}-8x+4$.
(b)
Consider the function
$\begin{align}
& \left( g\circ f \right)\left( x \right)=g\left( f\left( x \right) \right) \\
& =g\left( {{x}^{2}}+3 \right) \\
& =4\left( {{x}^{2}}+3 \right)-1 \\
& =4{{x}^{2}}+11
\end{align}$
Therefore, the value of $\left( g\circ f \right)\left( x \right)$ for the functions $f\left( x \right)={{x}^{2}}\text{+3 and }g\left( x \right)=4x-1$ is $4{{x}^{2}}+11$.
(c)
From part (a), $\left( f\circ g \right)\left( x \right)={{\left( 4x-1 \right)}^{2}}+3$.
Therefore,
$\begin{align}
& \left( f\circ g \right)\left( 3 \right)={{\left( 4\times 3-1 \right)}^{2}}+3 \\
& ={{\left( 11 \right)}^{2}}+3 \\
& =121+3 \\
& =124
\end{align}$
Therefore, the value of $\left( f\circ g \right)\left( 3 \right)$ for the functions $f\left( x \right)={{x}^{2}}\text{+3 and }g\left( x \right)=4x-1$ is $124$.
