Chapter 1.2, Problem 20E

To determine

To find: $\left[f\circ g\right]\left[x\right]$and $\left[g\circ f\right]\left[x\right]$ for $f\left(x\right)=1+x\text{ and }g\left(x\right)=x^2+5x+6$ .

Answer to Problem 20E

  $\begin{array}{l}\left[f\circ g\right]\left[x\right]=x^2+5x+7\\ \left[g\circ f\right]\left[x\right]=x^2+7x+12\end{array}$

Explanation of Solution

Given information: The functions given are as follows:

  $\begin{array}{l}f\left(x\right)=1+x\\ g\left(x\right)=x^2+5x+6\end{array}$

Calculation:

  $\left[f\circ g\right]\left[x\right]=f\left(g\left(x\right)\right)$

Substitute $g\left(x\right)$ into the function $f\left(x\right)$ as shown below.

  $\begin{array}{c}f\left(x\right)=1+x\\ f\left(g\left(x\right)\right)=1+\left(g\left(x\right)\right)\end{array}$

Now substitute the value of $g\left(x\right)$ in the above equation and simplify as follows:

  $\begin{array}{c}f\left(x^2+5x+6\right)=1+\left(x^2+5x+6\right)\\ f\left(x^2+5x+6\right)=x^2+5x+7\\ \left[f\circ g\right]\left[x\right]=x^2+5x+7\text{ [Rewrite the function in}\left(f\circ g\right)\left(x\right)\text{ ]}\end{array}$

Therefore, $\left[f\circ g\right]\left[x\right]$ is $x^2+5x+7$ .

Similarly, $\left[g\circ f\right]\left[x\right]=g\left(f\left(x\right)\right)$

Substitute $g\left(x\right)$ into the function $f\left(x\right)$ as shown below.

  $\begin{array}{c}g\left(x\right)=x^2+5x+6\\ g\left(f\left(x\right)\right)={\left(f\left(x\right)\right)}^2+5\left(f\left(x\right)\right)+6\end{array}$

Now substitute the value of $g\left(x\right)$ in the above equation and simplify as follows:

  $\begin{array}{c}g\left(x^2+5x+6\right)={\left(x+1\right)}^2+5\left(x+1\right)+6\\ g\left(x^2+5x+6\right)=x^2+2x+1+5x+5+6\\ g\left(x^2+5x+6\right)=x^2+7x+12\\ \left[g\circ f\right]\left[x\right]=x^2+7x+12\text{ [Rewrite the function in}\left(g\circ f\right)\left(x\right)\text{]}\end{array}$

Therefore, $\left[g\circ f\right]\left[x\right]$ is $x^2+7x+12$ .