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: Understand the goal.
We want the smallest value of $f(x) = \dfrac{1 - x + x^2}{1 + x + x^2}$ over all real $x$.
Step 2: Set the value equal to $y$.
Let $y = \dfrac{x^2 - x + 1}{x^2 + x + 1}$. Note the bottom $x^2 + x + 1$ is always positive, so we can multiply across safely.
Step 3: Form a quadratic in $x$.
\[ y(x^2 + x + 1) = x^2 - x + 1 \] Bring all to one side: \[ (y-1)x^2 + (y+1)x + (y-1) = 0 \]
Step 4: Use the real-root idea.
Since $x$ is real, this quadratic must have real roots, so its discriminant is $\ge 0$: \[ (y+1)^2 - 4(y-1)^2 \ge 0 \]
Step 5: Simplify the inequality.
\[ (y^2 + 2y + 1) - 4(y^2 - 2y + 1) \ge 0 \;\Rightarrow\; -3y^2 + 10y - 3 \ge 0 \] Multiply by $-1$ (flip sign): $3y^2 - 10y + 3 \le 0$, which factors as $(3y - 1)(y - 3) \le 0$.
Step 6: Read the range.
This holds for $\dfrac{1}{3} \le y \le 3$. So the smallest value is $\dfrac{1}{3}$. \[ \boxed{\dfrac{1}{3}} \]