Exercise 6.3 · Q29 (MCQ): Greatest value of $[x(x-1)+1]^{1/3}$ on $[0, 1]$

Basics you need

Solution

1. Simplify the inside

   $$x(x - 1) + 1 = x^2 - x + 1 = \left(x - \tfrac{1}{2}\right)^2 + \tfrac{3}{4} \ \ge\ \tfrac{3}{4}.$$

   So $f(x) = \left(x^2 - x + 1\right)^{1/3}$⁠, and the quantity inside the cube root is always positive.

   Expanding gives a familiar quadratic; completing the square shows at a glance that it never reaches zero, which matters in the next step because the cube root is not differentiable at 0.
   Completing the square (Basic 1). Basics: factorising

2. Differentiate with the chain rule

   Inner function $g(x) = x^2 - x + 1$⁠, with $g'(x) = 2x - 1$⁠; outer function $u^{1/3}$⁠. Since $g(x) \ge \frac{3}{4} > 0$⁠:

   $$f'(x) = \frac{1}{3}\,g(x)^{-2/3}\,(2x - 1) = \frac{2x - 1}{3\left(x^2 - x + 1\right)^{2/3}}.$$

   This holds at every $x$ in $[0, 1]$⁠, so $f$ is continuous on $[0, 1]$ and differentiable on $(0, 1)$⁠.

   The derivative of $u^{1/3}$ is $\frac{1}{3}u^{-2/3}$⁠, which blows up at $u = 0$⁠; because $g$ never gets below $\frac{3}{4}$⁠, that never happens here, so there are no points where $f$ fails to be differentiable.
   Chain rule (Basic 2): Theorem 4 (Chain Rule), Section 5.3.1. Basics: the chain rule

3. Critical points in the interval

   The denominator $3\left(x^2 - x + 1\right)^{2/3}$ is positive, so $f'(x) = 0$ exactly when $2x - 1 = 0$⁠:

   $$x = \tfrac{1}{2}.$$

   This is the only critical point in $(0, 1)$⁠. Moreover $f'(x) < 0$ for $0 < x < \frac{1}{2}$ and $f'(x) > 0$ for $\frac{1}{2} < x < 1$⁠, so $x = \frac{1}{2}$ is a point of local minima.

   Step 1 of the working rule asks for every point where $f' = 0$ or $f$ is not differentiable; Step 2 showed there are none of the second kind. The sign of $f'$ is the sign of $2x - 1$⁠.
   Critical point: the Note after Theorem 2, Section 6.4. First Derivative Test: Theorem 3, Section 6.4. Lesson: first and second derivative tests

4. Working rule: values at the critical point and the ends

   $x$	$x^2 - x + 1$	$f(x)$
   $0$ (end)	$1$	$1^{1/3} = 1$
   $\frac{1}{2}$ (critical)	$\frac{1}{4} - \frac{1}{2} + 1 = \frac{3}{4}$	$\left(\frac{3}{4}\right)^{1/3} \approx 0.9086$
   $1$ (end)	$1$	$1$

   The largest of these is 1. So the absolute maximum value of $f$ on $[0, 1]$ is 1, attained at both $x = 0$ and $x = 1$⁠; the absolute minimum value is $\left(\frac{3}{4}\right)^{1/3}$⁠, at $x = \frac{1}{2}$⁠.

   On a closed interval the greatest value can be at an end point, where $f'$ need not be zero. That is exactly what happens here: the only critical point is a minimum, and the maximum sits at the two ends.
   Working rule of Section 6.4.1, Steps 1 to 4; Theorem 5 guarantees the maximum exists. Lesson: absolute extrema

5. A direct check without calculus

   For $0 \le x \le 1$ we have $x \ge 0$ and $x - 1 \le 0$⁠, so $x(x - 1) \le 0$ and

   $$x(x - 1) + 1 \le 1, \qquad\text{hence}\qquad f(x) \le 1^{1/3} = 1,$$

   with equality exactly when $x(x - 1) = 0$⁠, that is at $x = 0$ and $x = 1$⁠.

   The printed form $x(x - 1) + 1$ is a hint: on $[0, 1]$ the product $x(x - 1)$ is never positive, so the inside never exceeds 1. This confirms the working rule's answer by a completely independent route.
   The cube root keeps order (Basic 4); a non-negative number times a non-positive number is non-positive. Basics: sign of a product

6. Examine every option

   On $[0, 1]$ the values of $f$ fill the range from $\left(\frac{3}{4}\right)^{1/3} \approx 0.9086$ up to 1.

   Option	Value	Verdict
   (A) $\left(\frac{1}{3}\right)^{1/3}$	$\approx 0.6934$	wrong: below the least value $0.9086$⁠, so $f$ never even takes this value
   (B) $\frac{1}{2}$	$0.5$	wrong: also below the least value; not a value of $f$ on $[0, 1]$
   (C) 1	$1$	correct: $f(0) = f(1) = 1$ and $f(x) \le 1$ throughout
   (D) 0	$0$	wrong: $x^2 - x + 1 \ge \frac{3}{4}$⁠, so $f(x) \ge 0.9086 > 0$

   (A) looks plausible because it has the same "cube root" shape as the answer, but the number inside is wrong: the inside quadratic is never as small as $\frac{1}{3}$ (its least value is $\frac{3}{4}$⁠).
   $\left(\frac{1}{3}\right)^{1/3} < \left(\frac{3}{4}\right)^{1/3}$ because $\frac{1}{3} < \frac{3}{4}$ (Basic 4).

Every fact used, and where it comes from