What is the range if f(x) = 3x - 9 and domain: -4,-3,0,1,8?

Jul 24, 2018

$y \in \left\{- 21 , - 18 , - 9 , - 6 , 15\right\}$

Explanation:

$\text{to obtain the range substitute the given values in the }$
$\text{domain into } f \left(x\right)$

$f \left(- 4\right) = - 12 - 9 = - 21$

$f \left(- 3\right) = - 9 - 9 = - 18$

$f \left(0\right) = - 9$

$f \left(1\right) = 3 - 9 = - 6$

$f \left(8\right) = 24 - 9 = 15$

$\text{range is } y \in \left\{- 21 , - 18 , - 9 , - 6 , 15\right\}$

Jul 24, 2018

Range = $\left\{- 21 , - 18 , - 9 , - 6 , + 15\right\}$

Explanation:

Here we have a lineal function $f \left(x\right) = 3 x - 9$ defined for $x = \left\{- 4 , - 3 , 0 , 1 , 8\right\}$

The slope of $f \left(x\right) = 3 \to f \left(x\right)$ is linear increasing.

Since $f \left(x\right)$ is linear increasing, its minimum and maximum values will be at the minimum and maximum values in its domain.

$\therefore {f}_{\min} = f \left(- 4\right) = - 21$

and ${f}_{\max} = f \left(8\right) = 15$

The other values of $f \left(x\right)$ are:

$f \left(- 3\right) = - 18$
$f \left(0\right) = - 9$
$f \left(1\right) = - 6$

Hence the range of $f \left(x\right)$ is $\left\{- 21 , - 18 , - 9 , - 6 , + 15\right\}$