How do you find the domain and range of #f(x)=sqrt(3x-2)#?

1 Answer
Nov 7, 2017

Answer:

Visualize the graph (often easier by graphing the function). Then, determine all possible values of #x# and #y#.

Explanation:

Let's list the transformations of the function before we graph it out. By graphing it out, we get a visual of the function so it's much easier to determine the domain and range.

  • Horizontal compression by a factor of #1/3#.
  • Horizontal translation requires us to isolate #x# within the radical. We get #2/3# to the right.

Now let's graph this.

The easiest way to do this is to sub in values for #x# and solve for #y#.

You get this:

graph{y=sqrt(3x-2) [-10, 10, -5, 5]}

Now we can visually see the function's domain and range.

The domain is a set of all the possible #x# values, while range is for #y#.

Because #x# can only be values that is equal to or greater than #2#, the domain is: #{x inRR | x >= 2/3}#

On the other hand, the range can only be values equal to or greater than #0#. Thus, the range is: #{y inRR | y >=0}#

Hope this helps :)