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:

For reciprocal-style symmetric functions like $\frac{x^2-x+1}{x^2+x+1}$, the critical points almost always occur at $x = 1$ and $x = -1$.
Doing a rapid mental check:
At $x = 1 \rightarrow f(1) = \frac{1-1+1}{1+1+1} = \frac{1}{3}$
At $x = -1 \rightarrow f(-1) = \frac{1-(-1)+1}{1-1+1} = \frac{3}{1} = 3$
This lets you find both the minimum ($\frac{1}{3}$) and maximum ($3$) values in seconds without working through the discriminant!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question requires finding the absolute minimum value of the rational function $f(x) = \frac{1 - x + x^2}{1 + x + x^2}$ over all real values of $x$.

Step 2: Key Formula or Approach:
We can solve this problem using an algebraic range method. Set $f(x) = y$, rearrange the terms into a quadratic equation in $x$, and apply the real-roots condition. Since $x$ is real, the discriminant of this quadratic equation must be greater than or equal to zero ($\Delta \geq 0$).

Step 3: Detailed Explanation:
Set the expression equal to $y$:
$$y = \frac{x^2 - x + 1}{x^2 + x + 1}$$ Cross-multiply to clear the fraction (noting that $x^2 + x + 1 > 0$ for all real $x$):
$$y(x^2 + x + 1) = x^2 - x + 1$$ $$yx^2 + yx + y = x^2 - x + 1$$ Group the terms by powers of $x$ to form a standard quadratic equation $Ax^2 + Bx + C = 0$:
$$(y - 1)x^2 + (y + 1)x + (y - 1) = 0$$ Since $x \in \mathbb{R}$, this quadratic equation must possess real roots, meaning its discriminant $\Delta = B^2 - 4AC \geq 0$:
$$(y + 1)^2 - 4(y - 1)(y - 1) \geq 0$$ $$(y + 1)^2 - 4(y - 1)^2 \geq 0$$ Expand the algebraic expressions:
$$(y^2 + 2y + 1) - 4(y^2 - 2y + 1) \geq 0$$ $$y^2 + 2y + 1 - 4y^2 + 8y - 4 \geq 0$$ $$-3y^2 + 10y - 3 \geq 0$$ Multiply by $-1$ and flip the inequality sign:
$$3y^2 - 10y + 3 \leq 0$$ Factor the quadratic inequality:
$$3y^2 - 9y - y + 3 \leq 0$$ $$3y(y - 3) - 1(y - 3) \leq 0$$ $$(3y - 1)(y - 3) \leq 0$$ Solving this inequality gives the range for $y$:
$$\frac{1}{3} \leq y \leq 3$$ The minimum value that the function can achieve is therefore $\frac{1}{3}$.

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