If f (X) = square root 3x-2 what the smallets possible value of f (X)

F (x) = √ (3x - 2), this function exists as long as the radicand (3x-2) is ≥ 0

3x - 2 ≥ 0

3x ≥ 2

x≥ 2/3

For all values of x ≥ 2/3, y exists then the smallest value of f (x) = y, is 0

or f (2/3) = √[ (3) (2/3) - 2]

f (2/3) = 0

f (x) can't be negative