For all real values of x, the minimum value of (1 - x + x^2)/(1+x+x^2) is ______.

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For all real values of x, the minimum value of (1 - x + x^2)/(1+x+x^2) is underline($\frac{1}{3}$). Explanation: Let, y = (1 - x + x^2)/(1 + x = x^2) Differentiating both sides with respect to x, `dy/dx = ((-1 + 2x)(1 + x + x^2) - (1 - x + x^2)(1 + 2x))/((1 + x + x^2)^2) = ((- 1 - x + x^2) + (2x + 2x^2 + 2x^3) - [(1 - x + x^2 + 2x - 2x^2 = 2x^3)])/((1 + x + x^2)^2) = (-1 - x - x^2 + 2x + 2x^2 + 2x^3 - 1 + x - x^2 - 2x + 2x^2 - 2x^3)/((1 + x + x^2)^2) = (-2 + 2x^2)/((1 + x + x^2)^2) = (2 (x^2 - 1))/((1 + x + x^2)^2) = (2(x - 1)(x + 1))/((1 + x + x^2)^2)` For highest and lowest value, dy/dx = 0 `therefore (2(x - 1)(x + 1))/((1 + x + x^2)^2) = $\frac{0}{1}$ => (x - 1)(x + 1) = 0 therefore x = 1, -1 The sign of dy/dx` at x = 1 changes from negative to positive when the point moves through x = 1. therefore y is minimum at the point x = 1. Minimum value, f(1) `= (1 - 1 + 1)/(1 + 1 + 1) = $\frac{1}{3}$`

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