# How do you find the range of y=3x+4 for the domain {2,9, 11.5}?

Jul 31, 2016

Range$= \left\{10 , 31 , 38.5\right\}$

#### Explanation:

Let $y = f \left(x\right) = 3 x + 4 , w h e r e , x \in \left\{2 , 9 , 11.5\right\} = {D}_{f} , s a y$

The Range of $y = f$ is denoted by ${R}_{f}$, and, is defined by,

${R}_{f} = \left\{f \left(x\right) : x \in {D}_{f}\right\}$

$= \left\{f \left(2\right) , f \left(9\right) , f \left(11.5\right)\right\}$

Here, $f \left(x\right) = 3 x + 4 \Rightarrow f \left(2\right) = 3 \times 2 + 4 = 10$

Similarly, $f \left(9\right) = 31 , \mathmr{and} , f \left(11.5\right) = 38.5$

Hence, ${R}_{f} = \left\{10 , 31 , 38.5\right\}$