Chapter 1.2, Problem 21E

To determine

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

Answer to Problem 21E

  $\begin{array}{l}\left[f\circ g\right]\left[x\right]=\frac{x}{x-1}\left(\text{where }x\ne 1\right)\\ \left[g\circ f\right]\left[x\right]=\frac{1}{x}\left(\text{where }x\ne 0\right)\end{array}$

Explanation of Solution

Given information: The functions given are as follows:

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

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.

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

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

  $\begin{array}{c}f\left(\frac{1}{x-1}\right)=\left(\frac{1}{x-1}\right)+1\\ f\left(\frac{1}{x-1}\right)=\frac{1+x-1}{x-1}\\ f\left(\frac{1}{x-1}\right)=\frac{x}{x-1}\\ \left[f\circ g\right]\left[x\right]=\frac{x}{x-1}\text{ [Rewrite the function in}\left(f\circ g\right)\left(x\right)\text{]}\end{array}$

Therefore, $\left[f\circ g\right]\left[x\right]$ is $\frac{x}{x-1}$ , where $x\ne 1$ .

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.

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

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

  $\begin{array}{c}g\left(x+1\right)=\frac{1}{\left(x+1\right)-1}\\ g\left(x+1\right)=\frac{1}{x}\\ \left[g\circ f\right]\left[x\right]=\frac{1}{x}\text{ [Rewrite the function in}\left(g\circ f\right)\left(x\right)\text{]}\end{array}$

Therefore, $\left[g\circ f\right]\left[x\right]$ is $\frac{1}{x}$ , where $x\ne 0$ .