Question

For all real $x$, the minimum value of the function $f(x) = \frac{1 - x + x^2}{1 + x + x^2}$ is

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

Hint:

Whenever you see a rational function of the symmetric form $\frac{x^2 - kx + 1}{x^2 + kx + 1}$, the minimum and maximum values are always reciprocals of each other! Evaluating the function at quick testing points like $x=1$ gives $f(1) = \frac{1}{3}$, which instantly highlights $\frac{1}{3}$ as a critical value bound.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks for the global minimum real value attained by the rational function $f(x) = \frac{1 - x + x^2}{1 + x + x^2}$ over its entire domain of real numbers.

Step 2: Key Formula or Approach:
We can solve this using either calculus optimization or an algebraic discriminant approach. Let's use the algebraic method by setting $f(x) = y$, rearranging it into a standard quadratic equation in terms of $x$, and applying the real root condition that the discriminant must be non-negative ($\Delta \ge 0$).

Step 3: Detailed Explanation:
Let $y = \frac{x^2 - x + 1}{x^2 + x + 1}$. Cross-multiply to clear the denominator:
$$y(x^2 + x + 1) = x^2 - x + 1$$ $$yx^2 + yx + y = x^2 - x + 1$$ Group all the terms together to create a quadratic equation in standard form ($Ax^2 + Bx + C = 0$):
$$(y - 1)x^2 + (y + 1)x + (y - 1) = 0$$ Since $x$ is a real number, this quadratic equation must possess real roots. Therefore, its mathematical discriminant $\Delta = B^2 - 4AC$ must be greater than or equal to zero:
$$\Delta = (y + 1)^2 - 4(y - 1)(y - 1) \ge 0$$ $$(y + 1)^2 - 4(y - 1)^2 \ge 0$$ Expand the polynomial terms:
$$(y^2 + 2y + 1) - 4(y^2 - 2y + 1) \ge 0$$ $$y^2 + 2y + 1 - 4y^2 + 8y - 4 \ge 0$$ $$-3y^2 + 10y - 3 \ge 0$$ Multiply the entire inequality by $-1$ and reverse the inequality sign direction:
$$3y^2 - 10y + 3 \le 0$$ Factor the quadratic expression by splitting the middle term:
$$3y^2 - 9y - y + 3 \le 0$$ $$3y(y - 3) - 1(y - 3) \le 0 \implies (3y - 1)(y - 3) \le 0$$ The roots of this expression are $y = \frac{1}{3}$ and $y = 3$. For the product to be less than or equal to zero ($\le 0$), $y$ must lie between these two values:
$$\frac{1}{3} \le y \le 3$$ This range tells us that the absolute minimum value of the function is $\frac{1}{3}$ (and its maximum value is 3). This matches option (A).

Step 4: Final Answer:
The minimum value of the function is $\frac{1}{3}$, which corresponds to option (A).