Question

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

Answer 1

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