Given the function: \[ f(x) = \frac{1 - x + x^2}{1 + x + x^2} \] The objective is to simplify the function and determine its minimum value through substitution. Step 1: Substitution with \( x = \tan \theta \) Applying the substitution yields: \[ f(x) = \frac{1 - \tan \theta + \tan^2 \theta}{1 + \tan \theta + \tan^2 \theta} \] Dividing the numerator and denominator by \( \cos^2 \theta \) to utilize trigonometric identities: \[ = \frac{\cos^2 \theta - \sin \theta \cos \theta + \sin^2 \theta}{\cos^2 \theta + \sin \theta \cos \theta + \sin^2 \theta} \] Step 2: Substitution with \( x = t - 1 \) for symmetry simplification Substituting \( x = t - 1 \) into the function: \[ f(t - 1) = \frac{1 - (t - 1) + (t - 1)^2}{1 + (t - 1) + (t - 1)^2} \] Simplifying the numerator: \[ 1 - t + 1 + (t^2 - 2t + 1) = t^2 - 2t + 3 \] Simplifying the denominator: \[ 1 + t - 1 + (t^2 - 2t + 1) = t^2 - t + 1 \] The function simplifies to: \[ f(t - 1) = \frac{t^2 - 2t + 3}{t^2 - t + 1} \] Which can be represented as: \[ f(t) = \frac{t^2 - 2t + 3}{t^2 - t + 1} \] Setting \( f(t) = k \) to find the range: \[ \frac{t^2 - 2t + 3}{t^2 - t + 1} = k \Rightarrow t^2 - 2t + 3 = k(t^2 - t + 1) \] Rearranging the equation into a quadratic form in terms of \( t \): \[ t^2 - 2t + 3 = kt^2 - kt + k \Rightarrow t^2(1 - k) + t(-2 + k) + (3 - k) = 0 \] For the quadratic equation to have real roots (indicating real values for \( t \)), the discriminant must be non-negative: \[ \text{Discriminant } D \geq 0 \] Calculating the discriminant: \[ (-2 + k)^2 - 4(1 - k)(3 - k) \geq 0 \] The minimum value of \( k \) is found to be: \[ k_{\min} = \frac{1}{3} \] Therefore, the minimum value of the function is: \[ \boxed{\frac{1}{3}} \]