Chapter 1, Problem 27SGA

To determine

To find: $\left[f\circ g\right]\left[x\right]$and $\left[g\circ f\right]\left[x\right]$ .

Answer to Problem 27SGA

  $\begin{array}{l}\left[f\circ g\right]\left[x\right]=x^2-x+7\\ \left[g\circ f\right]\left[x\right]=x^2+11x+31\end{array}$

Explanation of Solution

Given information: The functions given are as follows:

  $\begin{array}{l}f\left(x\right)=6+x\\ g\left(x\right)=x^2-x+1\end{array}$

Calculation:

Put the values of $g\left(x\right)$ in $f\left(g\left(x\right)\right)$ in the equation below and simplify as follows:

  $\begin{array}{c}\left[f\circ g\right]\left[x\right]=f\left(g\left(x\right)\right)\\ =6+g\left(x\right)\\ =6+\left(x^2-x+1\right)\left[\text{Substitute }g\left(x\right)\right]\\ =x^2-x+7\end{array}$

Thus, the value of $\left[f\circ g\right]\left[x\right]$ is $x^2-x+7$ .

Put the values of $f\left(x\right)$ in $g\left(f\left(x\right)\right)$ in the equation below and simplify as follows:

  $\begin{array}{c}\left[g\circ f\right]\left[x\right]=g\left(f\left(x\right)\right)\\ ={\left(f\left(x\right)\right)}^2-\left(f\left(x\right)\right)+1\\ ={\left(6+x\right)}^2-\left(6+x\right)+1\left[\text{Substitute }f\left(x\right)\right]\\ =x^2+12x+36-6-x+1\\ =x^2+11x+31\end{array}$

Thus, the value of $\left[g\circ f\right]\left[x\right]$ is $x^2+11x+31$ .