# Calculus

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Find the positive value of x where fx=x3+2x1/3 has a horizontal tangent line.

• Calculus -

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.

see
http://www.wolframalpha.com/input/?i=x%5E3+%2B+2x%5E%281%2F3%29

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