Chapter EP, Problem 7.1.8EP

To determine

To calculate: The value of $f\circ g\text{ and }g\circ f$ where

  $f=\left\{\left(10,4\right),\left(-1,2\right),\left(5,6\right),\left(-1,0\right)\right\}\text{ and }g=\left\{\left(-4,10\right),\left(2,-9\right),\left(-7,5\right),\left(-2,-1\right)\right\}$ .

Answer to Problem 7.1.8EP

The value of $f\circ g=\left\{\left(-4,4\right),\left(-7,6\right),\left(-2,0\right)\right\}\text{ and }g\circ f=\left\{\left(-1,-9\right)\right\}$ .

Explanation of Solution

Given information:

The functions $f=\left\{\left(10,4\right),\left(-1,2\right),\left(5,6\right),\left(-1,0\right)\right\}\text{ and }g=\left\{\left(-4,10\right),\left(2,-9\right),\left(-7,5\right),\left(-2,-1\right)\right\}$ .

Formula used:

The composition of two functions f and gis $\left[f\circ g\right]\left(x\right)=f\left(g\left(x\right)\right)$ , where range of g is a subset of the domain of f .

Calculation:

Consider the functions

  $f=\left\{\left(10,4\right),\left(-1,2\right),\left(5,6\right),\left(-1,0\right)\right\}\text{ and }g=\left\{\left(-4,10\right),\left(2,-9\right),\left(-7,5\right),\left(-2,-1\right)\right\}$ .

Recall that the composition of two functions f and gis $\left[f\circ g\right]\left(x\right)=f\left(g\left(x\right)\right)$ , where range of g is a subset of the domain of f .

Now, $f\circ g$ is evaluated as, first compute the value of $g\left(x\right)$ and then use the range as the domain of f to compute $f\left(x\right)$ ,

  $\begin{array}{c}f\left(g\left(-4\right)\right)=f\left(10\right)\\ =4\end{array}$

Similarly,

  $\begin{array}{c}f\left(g\left(2\right)\right)=f\left(-9\right)\\ =\text{does not exist}\end{array}$

Similarly,

  $\begin{array}{c}f\left(g\left(-7\right)\right)=f\left(5\right)\\ =6\end{array}$

Similarly,

  $\begin{array}{c}f\left(g\left(-2\right)\right)=f\left(-1\right)\\ =0\end{array}$

Since, $-9$ is not in domain of f so, $f\circ g$ is not defined at $x=-9$ .

Therefore, $f\circ g=\left\{\left(-4,4\right),\left(-7,6\right),\left(-2,0\right)\right\}$ .

Now, $g\circ f$ is evaluated as, first compute the value of $f\left(x\right)$ and then use the range as the domain of g to compute $g\left(x\right)$ ,

  $\begin{array}{c}g\left(f\left(10\right)\right)=g\left(4\right)\\ =\text{does not exist}\end{array}$

Similarly,

  $\begin{array}{c}g\left(f\left(-1\right)\right)=g\left(2\right)\\ =-9\end{array}$

Similarly,

  $\begin{array}{c}g\left(f\left(5\right)\right)=g\left(6\right)\\ =\text{does not exist}\end{array}$

Similarly,

  $\begin{array}{c}g\left(f\left(-1\right)\right)=g\left(0\right)\\ =\text{does not exist}\end{array}$

Since, $4,6\text{ and 0}$ are not in domain of f so, $f\circ g$ is not defined at $x=4,x=6\text{ and }x=0$ .

Therefore, $g\circ f=\left\{\left(-1,-9\right)\right\}$

Thus, the value of $f\circ g=\left\{\left(-4,4\right),\left(-7,6\right),\left(-2,0\right)\right\}\text{ and }g\circ f=\left\{\left(-1,-9\right)\right\}$ .