Exercise 6.3 · Q28 (MCQ): Least value of $\frac{1-x+x^2}{1+x+x^2}$ is $\frac{1}{3}$

Basics you need

Solution

1. The function is defined for every real $x$

   Both the numerator and the denominator are quadratics with discriminant $1 - 4 = -3 < 0$ and leading coefficient 1:

   $$\begin{aligned} 1 + x + x^2 &= \left(x + \tfrac{1}{2}\right)^2 + \tfrac{3}{4} > 0, \\ 1 - x + x^2 &= \left(x - \tfrac{1}{2}\right)^2 + \tfrac{3}{4} > 0. \end{aligned}$$

   So $f(x)$ is defined, positive and differentiable for every real $x$⁠.

   The question says "for all real values of $x$", which only makes sense if the denominator never vanishes; and the quotient rule needs $v \ne 0$⁠. Positivity of the numerator will also settle option (A).
   Completing the square: $x^2 + x = \left(x + \frac{1}{2}\right)^2 - \frac{1}{4}$ (Basic 1). Basics: the discriminant

2. Differentiate with the quotient rule

   Take $u = 1 - x + x^2$⁠, $u' = 2x - 1$⁠, and $v = 1 + x + x^2$⁠, $v' = 2x + 1$⁠. The numerator of $f'$ is

   $$\begin{aligned} u'v - uv' &= (2x - 1)(x^2 + x + 1) \\ &\quad - (x^2 - x + 1)(2x + 1) \\ &= (2x^3 + x^2 + x - 1) \\ &\quad - (2x^3 - x^2 + x + 1) \\ &= 2x^2 - 2. \end{aligned}$$ $$f'(x) = \frac{2(x^2 - 1)}{(1 + x + x^2)^2} = \frac{2(x - 1)(x + 1)}{(1 + x + x^2)^2}.$$
   Expanding each product separately before subtracting is the safest way to avoid sign slips: most terms cancel and only $2x^2 - 2$ survives.
   Quotient rule (Basic 2). Expanding, $(2x - 1)(x^2 + x + 1)$ gives $2x^3 + x^2 + x - 1$⁠, and $(x^2 - x + 1)(2x + 1)$ gives $2x^3 - x^2 + x + 1$⁠.

3. Critical points and the sign of $f'$

   The denominator $(1 + x + x^2)^2$ is positive, so $f'(x) = 0$ exactly at $x = 1$ and $x = -1$⁠, and the sign of $f'$ is the sign of $(x - 1)(x + 1)$⁠:

   Interval	$x + 1$	$x - 1$	$f'(x)$	$f$
   $x < -1$	−	−	+	increasing
   $-1 < x < 1$	+	−	−	decreasing
   $x > 1$	+	+	+	increasing

   Each factor changes sign only at its own zero, so testing the factors on each interval gives the sign of $f'$ everywhere (Basic 3). The table tells us where $f$ rises and falls.
   Theorem 1, Section 6.3 (positive derivative: increasing; negative: decreasing). Basics: sign of a product

4. First derivative test

   At $x = 1$⁠, $f'$ changes from negative to positive: $x = 1$ is a point of local minima, with value

   $$f(1) = \frac{1 - 1 + 1}{1 + 1 + 1} = \frac{1}{3}.$$

   At $x = -1$⁠, $f'$ changes from positive to negative: a point of local maxima, with value $f(-1) = \frac{1 + 1 + 1}{1 - 1 + 1} = 3$⁠.

   The sign chart is already done, so the first derivative test is the natural choice and needs no second derivative.
   Theorem 3 (First Derivative Test), Section 6.4. Lesson: first and second derivative tests

5. $\frac{1}{3}$ is the least value for every real $x$

   A local minimum on the whole real line need not be the least value, so we prove it directly:

   $$\begin{aligned} f(x) - \frac{1}{3} &= \frac{3(1 - x + x^2) - (1 + x + x^2)}{3(1 + x + x^2)} \\ &= \frac{2x^2 - 4x + 2}{3(1 + x + x^2)} \\ &= \frac{2(x - 1)^2}{3(1 + x + x^2)} \ \ge\ 0. \end{aligned}$$

   The numerator is a square times 2 and the denominator is positive (Step 1), so $f(x) \ge \frac{1}{3}$ for every real $x$⁠, with equality only at $x = 1$⁠. Hence the minimum value is $\frac{1}{3}$.

   In the same way, $3 - f(x) = \frac{2(x + 1)^2}{1 + x + x^2} \ge 0$⁠, so 3 is the maximum value. Far out, $f(x) \to 1$ as $x \to \pm\infty$ (divide top and bottom by $x^2$⁠).

   The sign chart shows $f$ increasing on $x < -1$⁠, so values there could in principle be smaller than $\frac{1}{3}$⁠; the algebra rules this out for every $x$ at once. This is the least value over all real $x$⁠, as the question asks.
   A square is never negative; dividing by a positive number keeps the sign. Lesson: absolute extrema

6. Examine every option

   Option	Verdict	Reason
   (A) 0	wrong	Numerator and denominator are both positive (Step 1), so $f(x) > 0$ for every $x$⁠: the value 0 is never reached. It is a lower bound, but not the least value.
   (B) 1	wrong	$f(0) = 1$ and $f(x) \to 1$ far out, but $f(1) = \frac{1}{3} < 1$⁠, so 1 is not the minimum.
   (C) 3	wrong	3 is the maximum value, at $x = -1$ (Steps 4 and 5).
   (D) $\frac{1}{3}$	correct	$f(x) \ge \frac{1}{3}$ for all real $x$⁠, with equality at $x = 1$ (Step 5).

   Each wrong option corresponds to a natural slip: confusing a lower bound with a minimum (A), reading off $f(0)$ or the limit (B), or picking the other critical value (C).
   A minimum value must be attained: it is $f(c)$ for some $c$⁠, and no value of $f$ is smaller.

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