Exercise 6.2 · Q15: $x+\frac1x$ is increasing on any interval disjoint from $[-1,1]$

Basics you need

Solution

1. $f$ is defined and differentiable on I

   $f(x) = x + \frac{1}{x}$ is defined for every $x \ne 0$⁠. Since $0 \in [-1, 1]$ and I has no point in common with $[-1, 1]$⁠, $0 \notin I$⁠. So $f$ is defined at every point of I, and it is continuous and differentiable there (a sum of $x$ and $\frac1x$⁠, each differentiable for $x \ne 0$⁠).

   Before talking about increasing on I, we must know $f$ is defined on all of I. The only trouble point, $x = 0$⁠, is inside $[-1, 1]$⁠, so the hypothesis keeps it out.
   $\frac1x$ is continuous and differentiable wherever $x \ne 0$ (Section 5.2, 5.3).

2. I lies wholly on one side of $[-1, 1]$

   Every point $x$ of I is outside $[-1, 1]$⁠, so $x \lt -1$ or $x \gt 1$⁠. Suppose I had a point $p \lt -1$ and a point $q \gt 1$⁠. Being an interval, I would contain every number between $p$ and $q$⁠, in particular $0$⁠, which is in $[-1, 1]$⁠. That contradicts disjointness. So

   $$\text{either } I \subseteq (-\infty, -1) \text{ or } I \subseteq (1, \infty).$$

   Either way, every $x \in I$ satisfies $|x| \gt 1$, that is, $x^2 \gt 1$⁠.

   This is where the word "interval" is used. A set like $\{-2, 2\}$ avoids $[-1, 1]$ and has points on both sides, but it is not an interval. An interval cannot skip over a gap.
   An interval contains everything between any two of its points. Basics: interval notation

3. Differentiate and factorise

   Write $\frac{1}{x} = x^{-1}$⁠; the power rule gives $\frac{d}{dx}x^{-1} = -x^{-2}$⁠. So

   $$\begin{aligned} f'(x) &= 1 - \frac{1}{x^2} = \frac{x^2 - 1}{x^2}\\ &= \frac{(x - 1)(x + 1)}{x^2}, \quad x \ne 0. \end{aligned}$$
   The factorised form turns "what is the sign of $f'$⁠?" into "what are the signs of $x - 1$⁠, $x + 1$ and $x^2$⁠?", each easy to read.
   $x^2 - 1 = (x - 1)(x + 1)$ (difference of squares). Basics: factorising and the discriminant

4. $f'(x) \gt 0$ at every point of I

   • If $I \subseteq (1, \infty)$⁠: for $x \in I$⁠, $x - 1 \gt 0$ and $x + 1 \gt 0$⁠, so the top is positive.
   • If $I \subseteq (-\infty, -1)$⁠: for $x \in I$⁠, $x - 1 \lt 0$ and $x + 1 \lt 0$⁠, so the top (negative times negative) is positive.

   The bottom $x^2$ is positive for $x \ne 0$⁠. So in both cases $f'(x) \gt 0$ for every $x \in I$.

   Equivalently, $f'(x) = \frac{x^2 - 1}{x^2}$ and $x^2 \gt 1$ on I by Step 2. The sign chart shows the whole picture, including why the interval $[-1, 1]$ has to be excluded.
   Negative times negative is positive. Basics: sign of a product

5. Apply Theorem 1: $f$ is increasing on I

   Take any $x_1 \lt x_2$ in I. Because I is an interval, $[x_1, x_2] \subseteq I$⁠. On $[x_1, x_2]$⁠, $f$ is continuous, it is differentiable on $(x_1, x_2)$⁠, and $f' \gt 0$ there (Step 4). By Theorem 1 (a), $f$ is increasing on $[x_1, x_2]$⁠, so $f(x_1) \le f(x_2)$⁠; the Mean Value Theorem step in the proof even gives

   $$f(x_2) - f(x_1) = f'(c)(x_2 - x_1) \gt 0$$

   for some $c$ between $x_1$ and $x_2$⁠. As $x_1 \lt x_2$ in I were arbitrary, $f$ is increasing (indeed strictly increasing) on I.

   This works whatever kind of interval I is: open, closed or half-open, bounded like $[2, 5]$ or unbounded like $(-\infty, -3)$⁠. That is why the proof works with an arbitrary pair $x_1 \lt x_2$ and not with end points of I.
   Theorem 1 (a), Section 6.3, and its proof. Lesson: increasing and decreasing functions

6. Why $[-1, 1]$ has to be avoided

   On $(0, 1)$⁠, $f' \lt 0$ (sign chart), and indeed $f\left(\frac12\right) = \frac12 + 2 = 2.5$ while $f(0.8) = 0.8 + 1.25 = 2.05$⁠: $f$ falls. So the statement is false for an interval such as $\left[\frac12, 0.8\right]$⁠, which meets $[-1, 1]$⁠.

   The hypothesis "disjoint from $[-1, 1]$" is sufficient, not necessary: the same proof works on $[1, \infty)$⁠, since Theorem 1 only needs $f' \gt 0$ inside the interval and $f$ is continuous at $1$⁠. The book's condition simply guarantees $|x| \gt 1$ everywhere on I.
   Points where $f' = 0$ or $f'$ is undefined ($-1$⁠, $0$⁠, $1$⁠) split the real line into the pieces of the sign chart. Basics: sign of a product

Every fact used, and where it comes from