## Precalculus (6th Edition) Blitzer

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 $41$.
(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 $41$.