Question

Given f(x)=√(3x-12)+2, what is the range in set notation? {yly € R, y ≥

Answer 1

Answer:

{y€R | x$\geq$12}

Step-by-step explanation:

First by looking at the function, f(x)=sqrt(3x-12)+2, we have a square root so the domain would be restricted.

Note that, $\sqrt{ 3x-12}$ cannot be less than 0 since the domain of square root functions are all positive numbers within its domain including 0.

So $x\geq 12$ for the domain of this function.

Therefore, the range would be {y€R | x$\geq$12}