Chapter 7.1, Problem 21PPS

To determine

The value of $f\circ g\text{ and }g\circ f$ if they exist for the pair of functions $f=\left\{\left(-15,-5\right),\left(-4,12\right),\left(1,7\right),\left(3,9\right)\right\}$ and $g=\left\{\left(3,-9\right),\left(7,2\right),\left(8,-6\right),\left(12,0\right)\right\}$ .

Answer to Problem 21PPS

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

Explanation of Solution

Given information: The given functions are,

  $\begin{array}{c}f=\left\{\left(-15,-5\right),\left(-4,12\right),\left(1,7\right),\left(3,9\right)\right\}\\ g=\left\{\left(3,-9\right),\left(7,2\right),\left(8,-6\right),\left(12,0\right)\right\}\end{array}$

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,

  $\left(\begin{array}{c}g\left(3\right)=-9\\ g\left(7\right)=2\\ g\left(8\right)=-6\\ g\left(12\right)=0\end{array}\right)$

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

  $\left(\begin{array}{c}f\left[g\left(3\right)\right]=f\left(-9\right)\mathrm{....}\left(\text{not defined}\right)\\ f\left[g\left(7\right)\right]=f\left(2\right)\mathrm{....}\left(\text{not defined}\right)\\ f\left[g\left(8\right)\right]=f\left(-6\right)\mathrm{....}\left(\text{not defined}\right)\\ f\left[g\left(12\right)\right]=f\left(0\right)\mathrm{....}\left(\text{not defined}\right)\end{array}\right)$

Since $-9,2,-6\text{ and }0$ 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,

  $\left(\begin{array}{c}f\left(-15\right)=-5\\ f\left(-4\right)=12\\ f\left(1\right)=7\\ f\left(3\right)=9\end{array}\right)$

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

  $\left(\begin{array}{c}g\left[f\left(-15\right)\right]=g\left(-5\right)\mathrm{....}\left(\text{not defined}\right)\\ g\left[f\left(-4\right)\right]=g\left(12\right)\\ =0\\ g\left[f\left(1\right)\right]=g\left(7\right)\\ =2\\ g\left[f\left(3\right)\right]=g\left(9\right)\mathrm{....}\left(\text{not defined}\right)\end{array}\right)$

Since the points $-5\text{ and 9}$ do not lie in domain of $g$ therefore $g\circ f$ is not defined for the points $-5\text{ and 9}$ . But $g\left[f\left(-4\right)\right]=0\text{and }g\left[f\left(1\right)\right]=2$ .

Therefore $g\circ f=\left\{\left(-4,0\right),\left(1,2\right)\right\}$ .