The derivative of "f(x) = x + x^2" is "f'(x) = 1 + 2 x". Equation "f'(x) = 0" has solution "x = -\\frac{1}{2}", and the signs of the derivative change from negative to positive in the intervals "(-\\infty, -\\frac{1}{2})" , "(-\\frac{1}{2}, \\infty)" , hence "x = -1\/2" is a local minimum. The function is not bounded from above, hence its maximum value is "+\\infty".
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