Find the range of the function for the given domain. 𝑓(π‘₯) = 3π‘₯ βˆ’ 8;{βˆ’2, βˆ’1, 0, 1, 2} no mater what I do I cant find the answer.

The domain of a function consists of your choice of x values that you can use.

The range would be the set of values that are the result of using the domain values.
e.g.
if x = -2 , f(-2) = 3(-2) - 8 = -14
if x = -1 , f(-1) = 3(-1) - 8 = -11
...(x) '
if x = 2 , f(2) = 3(2) - 8 = -2

So the range would be the set of numbers from the f(x)s above.

To find the range of a function, you need to substitute the values from the given domain into the function and see what outputs you get. Let's calculate the values of the function 𝑓(π‘₯) = 3π‘₯ βˆ’ 8 for each value in the domain {βˆ’2, βˆ’1, 0, 1, 2}.

For π‘₯ = -2:
𝑓(π‘₯) = 3(βˆ’2) βˆ’ 8
= -6 - 8
= -14

For π‘₯ = -1:
𝑓(π‘₯) = 3(βˆ’1) βˆ’ 8
= -3 - 8
= -11

For π‘₯ = 0:
𝑓(π‘₯) = 3(0) βˆ’ 8
= 0 - 8
= -8

For π‘₯ = 1:
𝑓(π‘₯) = 3(1) βˆ’ 8
= 3 - 8
= -5

For π‘₯ = 2:
𝑓(π‘₯) = 3(2) βˆ’ 8
= 6 - 8
= -2

Now, look at the outputs we obtained: -14, -11, -8, -5, -2. The range of the function 𝑓(π‘₯) = 3π‘₯ βˆ’ 8 is the set of all these output values, which is {-14, -11, -8, -5, -2}. Thus, the range of the function for the given domain is {-14, -11, -8, -5, -2}.

thx!!