Question

For all real x, the minimum value of the function f(x) = (1 - x + x^2) / (1 + x + x^2) is:

• A. \( \frac{1}{3} \)
• B. \( 0 \)
• C. \( 3 \)
• D. \( 1 \)

Hint:

Instead of using derivatives, you can also use algebraic tricks like substitution or symmetry to simplify and find extreme values.

Answer 1

The Correct Option is A

We are given: \[ f(x) = \frac{1 - x + x^2}{1 + x + x^2} \] Let’s try to simplify and find the minimum value by using a substitution. Step 1: Let \( x = \tan \theta \) Then, \[ f(x) = \frac{1 - \tan \theta + \tan^2 \theta}{1 + \tan \theta + \tan^2 \theta} \] Now divide numerator and denominator by \( \cos^2 \theta \) (to convert into 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: Let \( x = t - 1 \) (a substitution to simplify symmetry) Then, \[ f(t - 1) = \frac{1 - (t - 1) + (t - 1)^2}{1 + (t - 1) + (t - 1)^2} \] Simplify numerator: \[ 1 - t + 1 + (t^2 - 2t + 1) = t^2 - 2t + 3 \] Simplify denominator: \[ 1 + t - 1 + (t^2 - 2t + 1) = t^2 - 2t + t + 1 = t^2 - t + 1 \] So, \[ f(t - 1) = \frac{t^2 - 2t + 3}{t^2 - t + 1} \] \[ f(t) = \frac{t^2 - 2t + 3}{t^2 - t + 1} \] Let’s assume \( f(t) = k \), then: \[ \frac{t^2 - 2t + 3}{t^2 - t + 1} = k \Rightarrow t^2 - 2t + 3 = k(t^2 - t + 1) \] \[ t^2 - 2t + 3 = kt^2 - kt + k \Rightarrow t^2(1 - k) + t(-2 + k) + (3 - k) = 0 \] For minimum or maximum value, the quadratic must have real roots, so: \[ \text{Discriminant } D \geq 0 \] Discriminant: \[ (-2 + k)^2 - 4(1 - k)(3 - k) \geq 0 \] \[ k_{\min} = \frac{1}{3} \] Hence, the minimum value of the function is: \[ \boxed{\frac{1}{3}} \]