Chapter 1, Problem 29SGA

To determine

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

Answer to Problem 29SGA

  $\begin{array}{l}\left[f\circ g\right]\left[x\right]=-2x^2-7\\ \left[g\circ f\right]\left[x\right]=2x^2-12x+28\end{array}$

Explanation of Solution

Given information: The functions given are as follows:

  $\begin{array}{l}f\left(x\right)=3-x\\ g\left(x\right)=2x^2+10\end{array}$

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:

  $\begin{array}{c}\left[f\circ g\right]\left[x\right]=f\left(g\left(x\right)\right)\\ =3-\left(g\left(x\right)\right)\\ =3-\left(2x^2+10\right)\left[\text{Substitute }g\left(x\right)\right]\\ =-2x^2-7\end{array}$

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

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

  $\begin{array}{c}\left[g\circ f\right]\left[x\right]=g\left(f\left(x\right)\right)\\ =2{\left(f\left(x\right)\right)}^2+10\\ =2{\left(3-x\right)}^2+1\text{0 }\left[\text{Substitute }f\left(x\right)\right]\\ =2\left(9+x^2-6x\right)+1\text{0}\\ =18+2x^2-12x+1\text{0}\\ =2x^2-12x+28\end{array}$

Thus, the value of $\left[g\circ f\right]\left[x\right]$ is $2x^2-12x+28$ .