The domain and range of the relation R given | vedclass

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(D) Given the relation $R = \{(x, y) : y = x + \frac{6}{x}, x, y \in N, x < 6\}$.We test each value of $x \in \{1, 2, 3, 4, 5\}$ to see if $y$ is a na

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Domain $= \{1, 2, 3\}$,Range $= \{5, 7\}$.

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(D) Given the relation $R = \{(x, y) : y = x + \frac{6}{x}, x, y \in N, x We test each value of $x \in \{1, 2, 3, 4, 5\}$ to see if $y$ is a natural number $(N)$:For $x = 1, y = 1 + \frac{6}{1} = 7 \in N$.For $x = 2, y = 2 + \frac{6}{2} = 2 + 3 = 5 \in N$.For $x = 3, y = 3 + \frac{6}{3} = 3 + 2 = 5 \in N$.For $x = 4, y = 4 + \frac{6}{4} = 4 + 1.5 = 5.5 
otin N$.For $x = 5, y = 5 + \frac{6}{5} = 5 + 1.2 = 6.2 
otin N$.Thus,the relation $R = \{(1, 7), (2, 5), (3, 5)\}$.The domain is the set of first elements: $\{1, 2, 3\}$.The range is the set of second elements: $\{5, 7\}$.

Function	Range
$A. f(x) = \|x\|$	$I. [0, \infty)$
$B. f(x) = x^2$	$II. \mathbb{R}$
$C. f(x) = x^3$	$III. [0, \infty)$
$D. f(x) = \text{sgn}(x)$	$IV. \{-1, 0, 1\}$

The domain and range of the relation $R$ given by $R = \{(x, y) : y = x + \frac{6}{x}, x, y \in N \text{ and } x < 6\}$ are

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