Precalculus Examples

$f = ((2, 3), (4, 5), (7, 6))$ , $h = ((3, 2), (5, 4), (6, 7))$

Step 1

Since there is one value of $y$ for every value of $x$ in $(2, 3), (4, 5), (7, 6)$ , this relation is a function.

The relation is a function.

Step 2

The domain is the set of all the values of $x$ . The range is the set of all the values of $y$ .

Domain: ${2, 4, 7}$

Range: ${3, 5, 6}$

Step 3

Since there is one value of $y$ for every value of $x$ in $(3, 2), (5, 4), (6, 7)$ , this relation is a function.

The relation is a function.

Step 4

The domain is the set of all the values of $x$ . The range is the set of all the values of $y$ .

Domain: ${3, 5, 6}$

Range: ${2, 4, 7}$

Step 5

The domain of the first relation $f = ((2, 3), (4, 5), (7, 6))$ is equal to the range of the second relation $h = ((3, 2), (5, 4), (6, 7))$ . Also, the range of the first relation is equal to the domain of the second relation $h = ((3, 2), (5, 4), (6, 7))$ , which means that $f = ((2, 3), (4, 5), (7, 6))$ is the inverse of $h = ((3, 2), (5, 4), (6, 7))$ and vice versa.

$f = ((2, 3), (4, 5), (7, 6))$ is the inverse of $h = ((3, 2), (5, 4), (6, 7))$