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July 25, 2014

July 25, 2014

Posted by **TG** on Friday, February 10, 2012 at 11:12pm.

Answer =

- Calculus -
**Reiny**, Friday, February 10, 2012 at 11:54pmI 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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