Answer
a) The value of $\left( f\circ g \right)\left( x \right)$ is $6{{x}^{2}}-18x+17$.
b) The value of $\left( g\circ f \right)\left( x \right)$ is $36{{x}^{2}}+42x+12$.
c) The value of $\left( f\circ g \right)\left( -1 \right)$ is $41$.
Work Step by Step
(a)
Calculate $\left( f\circ g \right)\left( x \right)$ as follows:
$\begin{align}
& \left( f\circ g \right)\left( x \right)=f\left( g\left( x \right) \right) \\
& =f\left( g\left( x \right) \right) \\
& =f\left( {{x}^{2}}-3x+2 \right) \\
& =6\left( {{x}^{2}}-3x+2 \right)+5
\end{align}$
Then, solve it:
$6\left( {{x}^{2}}-3x+2 \right)+5=6{{x}^{2}}-18x+17$.
Thus, the value of $\left( f\circ g \right)\left( x \right)$ is $6{{x}^{2}}-18x+17$.
(b)
Calculate $\left( g\circ f \right)\left( x \right)$ as follows:
$\begin{align}
& \left( g\circ f \right)\left( x \right)=g\left( f\left( x \right) \right) \\
& =g\left( f\left( x \right) \right) \\
& =f\left( 6x+5 \right) \\
& ={{\left( 6x+5 \right)}^{2}}-3\left( 6x+5 \right)+2
\end{align}$
Then simplify as follows:
$\begin{align}
& \left( g\circ f \right)\left( x \right)={{\left( 6x+5 \right)}^{2}}-3\left( 6x+5 \right)+2 \\
& =36{{x}^{2}}+25+60x-18x-15+2 \\
& =36{{x}^{2}}+42x+12
\end{align}$
Thus, the value of $\left( g\circ f \right)\left( x \right)$ is $36{{x}^{2}}+42x+12$.
(c)
From part (a), $\left( f\circ g \right)\left( x \right)=6{{x}^{2}}-18x+17$
Putting $x=-1$ in the above expression, get:
$\begin{align}
& \left( f\circ g \right)\left( -1 \right)=6{{\left( -1 \right)}^{2}}-18\left( -1 \right)+17 \\
& =6+18+17 \\
& =41
\end{align}$
Thus, the value of $\left( f\circ g \right)\left( -1 \right)$ is $41$.
