domain and range of the square root of 3x-1

[evan399]

the square root of 3x-1

is the domain x is greater than or equal to -1/3
and is the range [1,infinite symbol)

 

[royhaas]

No, you must have \(\displaystyle 3x-1 \ge 0\).

 

[stapel]

the square root of 3x-1

Your formatting is ambigous. Do you mean either of the following?

. . . . .f(x) = sqrt[3x] - 1

. . . . .f(x) = sqrt[3x - 1]

. . . . .f(x) = sqrt[3]x - 1

Or something else?

What was your reasoning for finding your answers?

Please be complete. Thank you!

Eliz.

 

[evan399]

sqrt(3x-1)

 

[stapel]

sqrt(3x-1)

Then the first reply you received gives you the set-up for finding the domain. (You can find the range by looking at the graph, by the way.)

Eliz.

 

[evan399]

so the answer is domain is x is greater than or equal to -1/3
and the range is [1, infinity symbol)
right?

 

[skeeter]

the range is \(\displaystyle \L [0, \infty)\)

 

[Mrspi]

so the answer is domain is x is greater than or equal to -1/3
and the range is [1, infinity symbol)
right?

you have

f(x) = sqrt(3x - 1)

It has been explained that you can't take the square root of a negative number (in the real number system)

and it has been explained that this means

3x - 1 must be greater than or equal to 0.

OR, that

3x - 1 >= 0

Please show us HOW you got x >= -1/3 out of this......

 

[evan399]

What I'm saying is to solve the equation, you take
3x-1 greater than or equal to 0 then
3x is greater than or equal to 1 then
x is greater than or equal to 1/3
right?

 

[evan399]

Forgive me, the equation is 3x+1 is greater than or equal to 0
so you have 3x is greater than or equal to -1 then
x is greater than or equal to -1/3
right?