SOLUTION: Let y = cube root of (x^2 - x +1). For which value of x is y a minimum?

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Question 974804: Let y = cube root of (x^2 - x +1). For which value of x is y a minimum?
Answer by josgarithmetic(39616) About Me  (Show Source):
You can put this solution on YOUR website!
%28d%2Fdx%29%28x%5E2-x%2B1%29%5E%281%2F3%29

%281%2F3%29%28x%5E2-x%2B1%29%5E%28-2%2F3%29%282x-1%29

%282x-1%29%2F%283%28x%5E2-x%2B1%29%5E%282%2F3%29%29

For any extremes value,
%282x-1%29%2F%283%28x%5E2-x%2B1%29%5E%282%2F3%29%29=0

The numerator now is what counts.
2x-1=0
x=1%2F2------------This is the best bet for y as a minimum value. LOCAL minimum.