posted by TG .
Find the positive value of x where fx=x3+2x1/3 has a horizontal tangent line.
I will assume you meant:
f(x) = x^3 + 2x^(1/3)
then f'(x) = 3x^2 + (2/3)x^(-2/3)
for a horizontal tangent, f'(x) = 0
3x^2 + (2/3)/(x^(2/3)) = 0
multiply by x(2/3)
3x^(8/3) = -2/3
x^(8/3) = -2/9
x = (-2/9)^(3/8) , which is undefined (can't take an even root of a negative)
Unless you typed something different from what I assumed, there is no horizontal tangent.