Chapter 7.1, Problem 20PPS

To determine

The value of $f\circ g\text{ and }g\circ f$ if they exist for the pair of functions $f=\left\{\left(-4,-14\right),\left(0,-6\right),\left(-6,-18\right),\left(2,-2\right)\right\}$ and $g=\left\{\left(-6,1\right),\left(-18,13\right),\left(-14,9\right),\left(-2,-3\right)\right\}$ .

Answer to Problem 20PPS

The value of $f\circ g=\varphi \text{and }g\circ f=\left\{\left(-4,9\right),\left(0,1\right),\left(-6,13\right),\left(2,-3\right)\right\}$ .

Explanation of Solution

Given information: The given functions are,

  $\begin{array}{c}f=\left\{\left(-4,-14\right),\left(0,-6\right),\left(-6,-18\right),\left(2,-2\right)\right\}\\ g=\left\{\left(-6,1\right),\left(-18,13\right),\left(-14,9\right),\left(-2,-3\right)\right\}\end{array}$

Calculation: To evaluate $f\circ g$ first evaluate $g\left(x\right)$ and then use range of $g\left(x\right)$ as domain of $f\left(x\right)$ to evaluate $f\circ g$ . Therefore $g\left(x\right)$ is evaluated as,

  $\left(\begin{array}{c}g\left(-6\right)=1\\ g\left(-18\right)=13\\ g\left(-14\right)=9\\ g\left(-2\right)=-3\end{array}\right)$

Now, evaluating $f\circ g$ using the range of $g\left(x\right)$ as domain of $f\left(x\right)$ .

  $\left(\begin{array}{c}f\left[g\left(-6\right)\right]=f\left(1\right)\mathrm{....}\left(\text{not defined}\right)\\ f\left[g\left(-18\right)\right]=f\left(13\right)\mathrm{....}\left(\text{not defined}\right)\\ f\left[g\left(-14\right)\right]=f\left(9\right)\mathrm{....}\left(\text{not defined}\right)\\ f\left[g\left(-2\right)\right]=f\left(-3\right)\mathrm{....}\left(\text{not defined}\right)\end{array}\right)$

Since $1,13,9\text{ and }-3$ do not lie in domain of $f$ , therefore $f\circ g$ is undefined for all these points. Therefore $f\circ g=\varphi$ .

Similarly to find $g\circ f$ evaluate $f\left(x\right)$ first and then use range of $f\left(x\right)$ as domain of $g\left(x\right)$ to evaluate $g\circ f$ . Therefore $f\left(x\right)$ is evaluated as,

  $\left(\begin{array}{c}f\left(-4\right)=-14\\ f\left(0\right)=-6\\ f\left(-6\right)=-18\\ f\left(2\right)=-2\end{array}\right)$

Now, evaluating $g\circ f$ using the range of $f\left(x\right)$ as domain of $g\left(x\right)$ .

  $\left(\begin{array}{c}g\left[f\left(-4\right)\right]=g\left(-14\right)\\ =9\\ g\left[f\left(0\right)\right]=g\left(-6\right)\\ =1\\ g\left[f\left(-6\right)\right]=g\left(-18\right)\\ =13\\ g\left[f\left(2\right)\right]=g\left(-2\right)\\ =-3\end{array}\right)$

Since $g\left[f\left(-4\right)\right]=9,g\left[f\left(0\right)\right]=1,g\left[f\left(-6\right)\right]=13$ and $g\left[f\left(2\right)\right]=-3$ . Therefore domain of $g\circ f$ is undefined for all these points, which means $g\circ f=\left\{\left(-4,9\right),\left(0,1\right),\left(-6,13\right),\left(2,-3\right)\right\}$ .