Miscellaneous · Q4: Where $x^3+\frac{1}{x^3}$ increases and decreases

Basics you need

Solution

1. Domain: two separate intervals

   $\frac{1}{x^3}$ needs $x \ne 0$⁠, so the domain is

   $$(-\infty, 0) \cup (0, \infty).$$

   On each of these two intervals $f$ is a sum of differentiable functions, so it is continuous and differentiable there.

   Definition 1 talks about an interval I. The domain here is not one interval but two, with a gap at 0. Every interval we use must lie inside one of the two pieces.

2. Differentiate

   Write $f(x) = x^3 + x^{-3}$⁠. By the power rule,

   $$\begin{aligned} f'(x) &= 3x^2 - 3x^{-4} \\ &= \frac{3x^6 - 3}{x^4} = \frac{3(x^6 - 1)}{x^4}. \end{aligned}$$
   Writing $\frac1{x^3}$ as $x^{-3}$ lets us use the power rule directly. Then multiplying $3x^2$ by $\frac{x^4}{x^4}$ puts both terms over the common denominator $x^4$⁠, so the sign can be read from factors.
   Power rule for negative exponents (Basic 3), Table 5.3.

3. Factorise the numerator

   $$\begin{aligned} x^6 - 1 &= (x^2 - 1)(x^4 + x^2 + 1) \\ &= (x - 1)(x + 1)(x^4 + x^2 + 1), \end{aligned}$$

   so

   $$f'(x) = \frac{3(x - 1)(x + 1)(x^4 + x^2 + 1)}{x^4}.$$

   Here $x^4 + x^2 + 1 \ge 1 > 0$ for every $x$⁠, and $x^4 > 0$ for every $x \ne 0$⁠. So $f'(x) = 0$ only at $x = \pm1$⁠, and the sign of $f'$ is the sign of $(x - 1)(x + 1)$⁠.

   Check the first line by expanding: $(x^2 - 1)(x^4 + x^2 + 1)$ is $x^6 + x^4 + x^2 - x^4 - x^2 - 1$⁠, which is $x^6 - 1$⁠. The factor $x^4 + x^2 + 1$ is a sum of even powers plus 1, so it can never be zero or negative. Discarding the positive factors leaves a quadratic sign question.
   Factorising (Basic 4). Basics: factorising and the discriminant

4. Sign chart, with the break at 0

   The points $-1$⁠, $0$⁠, $1$ split the line into four pieces. $x = 0$ is not a zero of $f'$ but a point where $f$ and $f'$ are not defined.

   Test points: $f'(-2) = \frac{3 \cdot 63}{16} = \frac{189}{16} > 0$⁠, $f'\!\left(-\frac12\right) = \frac{3\left(\frac1{64} - 1\right)}{\frac1{16}}$⁠, which is $-\frac{189}{4} < 0$⁠, $f'\!\left(\frac12\right) = -\frac{189}4 < 0$⁠, $f'(2) = \frac{189}{16} > 0$⁠.

   $(x - 1)(x + 1)$ is negative exactly when one factor is negative and the other positive, that is for $-1 < x < 1$⁠; at $x = 0$⁠, inside that range, $f'$ does not exist, so the negative stretch is really two pieces.
   Sign of a product (Basic 5). Basics: sign of a product

5. Conclude on each interval with Theorem 1

   Take any two points $x_1 < x_2$ in one of the pieces below. $f$ is continuous on $[x_1, x_2]$ (this closed interval avoids 0), and $f'$ has one sign on $(x_1, x_2)$⁠. Theorem 1 and its proof give:

   • on $(-\infty, -1]$⁠: $f' > 0$ inside, so $f$ is increasing;
   • on $[-1, 0)$⁠: $f' < 0$ inside, so $f$ is decreasing;
   • on $(0, 1]$⁠: $f' < 0$ inside, so $f$ is decreasing;
   • on $[1, \infty)$⁠: $f' > 0$ inside, so $f$ is increasing.

   Each is even strict.

   Theorem 1 is stated for a closed interval $[a, b]$ on which $f$ is continuous. For an interval such as $[-1, 0)$ or $(-\infty, -1]$ we apply it to each closed sub-interval $[x_1, x_2]$⁠, which is enough to compare any two points.
   Theorem 1, Section 6.3 (Basic 2). Lesson: increasing and decreasing functions

6. What happens across $x = 0$⁠?

   Decreasing pieces. $f$ is not decreasing on $(-1, 0) \cup (0, 1)$ taken together: $-\frac12 < \frac12$⁠, yet

   $$f\!\left(-\tfrac12\right) = -\tfrac18 - 8 = -\tfrac{65}{8} < \tfrac{65}{8} = f\!\left(\tfrac12\right).$$

   Increasing pieces. On $(-\infty, -1]$⁠, $f$ increases up to $f(-1) = -2$⁠, so $f(x) \le -2$ there; on $[1, \infty)$⁠, $f$ increases from $f(1) = 2$⁠, so $f(x) \ge 2$ there. Hence any point of the left piece has a smaller value than any point of the right piece, and $x_1 < x_2 \Rightarrow f(x_1) < f(x_2)$ holds even across the two pieces.

   This is why answers must name intervals, not unions. Near 0 the function jumps from very large negative values (left of 0) to very large positive values (right of 0), so the two decreasing pieces do not join into one decreasing stretch. Strictly speaking, Definition 1 is only about intervals, and neither union is an interval; the computations above show what would happen if we tried.

7. Compare with the answer key

   The key gives (i) $x < -1$ and $x > 1$⁠; (ii) $-1 < x < 1$⁠. Read with the condition $x \ne 0$ from the question, (ii) means the two intervals $(-1, 0)$ and $(0, 1)$⁠, on each of which $f$ is decreasing (Step 5), not on both together (Step 6). Our intervals add the endpoints $\pm1$⁠, which Theorem 1 allows because $f$ is continuous there; the key's open intervals are also correct.

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