Chapter 6.1, Problem 26PPS

To determine

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

Answer to Problem 26PPS

The value of $f\circ g=\left\{\left(-2,3\right),\left(-4,-1\right)\right\}\text{and }g\circ f=\left\{\left(12,-1\right),\left(9,6\right),\left(8,5\right)\right\}$ .

Explanation of Solution

Given information: The given functions are,

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

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,

  $\left(\begin{array}{c}g\left(-1\right)=5\\ g\left(-2\right)=6\\ g\left(-3\right)=-1\\ g\left(-4\right)=8\end{array}\right)$

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

  $\left(\begin{array}{c}f\left[g\left(-1\right)\right]=f\left(5\right)\mathrm{....}\left(\text{not defined}\right)\\ f\left[g\left(-2\right)\right]=f\left(6\right)\\ =3\\ f\left[g\left(-3\right)\right]=f\left(-1\right)\mathrm{....}\left(\text{not defined}\right)\\ f\left[g\left(-4\right)\right]=f\left(8\right)\\ =-1\end{array}\right)$

Since the points $5\text{ and}-1$ do not lie in domain of $f$ , therefore $f\circ g$ is undefined for these points but $f\left[g\left(-2\right)\right]=3\text{ and }f\left[g\left(-4\right)\right]=-1$ . Therefore $f\circ g=\left\{\left(-2,3\right),\left(-4,-1\right)\right\}$ .

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,

  $\left(\begin{array}{c}f\left(12\right)=-3\\ f\left(9\right)=-2\\ f\left(8\right)=-1\\ f\left(6\right)=3\end{array}\right)$

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

  $\left(\begin{array}{c}g\left[f\left(12\right)\right]=g\left(-3\right)\\ =-1\\ g\left[f\left(9\right)\right]=g\left(-2\right)\\ =6\\ g\left[f\left(8\right)\right]=g\left(-1\right)\\ =5\\ g\left[f\left(6\right)\right]=g\left(3\right)\mathrm{....}\left(\text{not defined}\right)\end{array}\right)$

Since the point 3 does not lie in domain of $g$ therefore $g\circ f$ is not defined for that point but $g\left[f\left(12\right)\right]=-1,g\left[f\left(9\right)\right]=6\text{ and }g\left[f\left(8\right)\right]=5$ .

Hence $g\circ f=\left\{\left(12,-1\right),\left(9,6\right),\left(8,5\right)\right\}$ .