Exercise 6.2 · Q11: $x^2-x+1$ is neither increasing nor decreasing on $(-1,1)$

Basics you need

Solution

1. Differentiate $f$

   $f$ is a polynomial, so it is defined, continuous and differentiable at every real number, in particular at every point of $(-1, 1)$⁠. By the power rule ($\frac{d}{dx}x^n = nx^{n-1}$⁠), the sum rule, and the fact that a constant has derivative 0,

   $$f'(x) = 2x - 1 + 0 = 2x - 1.$$
   The sign of $f'$ is what tells us where $f$ rises and where it falls (Theorem 1). Continuity on every closed subinterval and differentiability inside it are exactly the conditions Theorem 1 needs, and a polynomial has both everywhere.
   Derivative of $x^n$ is $nx^{n-1}$⁠; derivative of a constant is 0 (Class XI, Limits and Derivatives).

2. Find where $f'(x) = 0$

   $$2x - 1 = 0 \iff x = \tfrac12.$$

   Since $-1 \lt \frac12 \lt 1$⁠, this point lies inside $(-1, 1)$⁠. It splits the interval into two pieces, $\left(-1, \frac12\right)$ and $\left(\frac12, 1\right)$⁠.

   If the only zero of $f'$ were outside $(-1, 1)$⁠, $f'$ would keep one sign on the whole interval and $f$ would be strictly increasing or strictly decreasing there. The whole question turns on the fact that the turning point $\frac12$ is inside.
   Points where $f' = 0$ are the only places a continuous $f'$ can change sign. Basics: factorising and solving

3. Sign of $f'$ on each piece

   If $-1 \lt x \lt \frac12$⁠, then $2x \lt 1$⁠, so $f'(x) = 2x - 1 \lt 0$⁠.

   If $\frac12 \lt x \lt 1$⁠, then $2x \gt 1$⁠, so $f'(x) = 2x - 1 \gt 0$⁠.

   To find the sign on a whole piece we use the inequality that defines the piece, not just one test value. A test value (for example $f'(0) = -1$ and $f'(0.75) = 0.5$⁠) is a useful check, but the inequality $2x \lt 1$ covers every $x$ in the piece at once.
   Multiplying $x \lt \frac12$ by the positive number 2 keeps the direction: $2x \lt 1$⁠. Basics: sign of a product

4. Where $f$ falls and where it rises

   Take any $x_1 \lt x_2$ in $\left(-1, \frac12\right]$⁠. On $[x_1, x_2]$⁠, $f$ is continuous and differentiable, and $f'(x) \lt 0$ at every point strictly between $x_1$ and $x_2$ (all such points are below $\frac12$⁠). By Theorem 1 (b), and the strict form from its proof, $f(x_1) \gt f(x_2)$⁠. So $f$ is strictly decreasing on $\left(-1, \frac12\right]$.

   In the same way, for $x_1 \lt x_2$ in $\left[\frac12, 1\right)$⁠, $f' \gt 0$ between them, so $f(x_1) \lt f(x_2)$⁠: $f$ is strictly increasing on $\left[\frac12, 1\right)$.

   Theorem 1 is stated for a closed interval $[a, b]$⁠, so to use it on the open interval $(-1, 1)$ we apply it on the closed interval $[x_1, x_2]$ between any two chosen points. The turning point $\frac12$ can be included in both pieces because Theorem 1 only asks about the sign of $f'$ strictly between the two end points, and $f$ is continuous at $\frac12$⁠.
   Theorem 1 (a), (b), Section 6.3, and the Mean Value Theorem step in its proof. Lesson: increasing and decreasing functions

5. $f$ is not increasing (so not strictly increasing)

   Choose $x_1 = -\frac12$ and $x_2 = 0$⁠. Both lie in $(-1, 1)$ and $x_1 \lt x_2$⁠. Then

   $$\begin{aligned} f\left(-\tfrac12\right) &= \tfrac14 + \tfrac12 + 1 = \tfrac74,\\ f(0) &= 0 - 0 + 1 = 1. \end{aligned}$$

   So $f(x_1) = \frac74 \gt 1 = f(x_2)$⁠. Strictly increasing needs $f(x_1) \lt f(x_2)$⁠, and even plain "increasing" needs $f(x_1) \le f(x_2)$⁠; both fail for this pair. Hence $f$ is not strictly increasing on $(-1, 1)$, and in fact not increasing either.

   We picked both points in the falling piece $\left(-1, \frac12\right]$ from Step 4, where a drop is guaranteed. One pair is enough because "increasing on $(-1, 1)$" must hold for every pair.
   $\left(-\frac12\right)^2 = \frac14$ and $-\left(-\frac12\right) = +\frac12$⁠: watch both signs. Basics: sign of a product

6. $f$ is not decreasing (so not strictly decreasing)

   Choose $x_1 = \frac12$ and $x_2 = \frac34$⁠. Both lie in $(-1, 1)$ and $x_1 \lt x_2$⁠. Then

   $$\begin{aligned} f\left(\tfrac12\right) &= \tfrac14 - \tfrac12 + 1 = \tfrac34 = \tfrac{12}{16},\\ f\left(\tfrac34\right) &= \tfrac{9}{16} - \tfrac{12}{16} + \tfrac{16}{16} = \tfrac{13}{16}. \end{aligned}$$

   So $f(x_1) = \frac{12}{16} \lt \frac{13}{16} = f(x_2)$⁠. Decreasing needs $f(x_1) \ge f(x_2)$⁠, which fails. Hence $f$ is not decreasing on $(-1, 1)$, and so not strictly decreasing either.

   Now both points are in the rising piece $\left[\frac12, 1\right)$⁠. Writing all three fractions over 16 makes the comparison $12 \lt 13$ immediate.
   To compare fractions, put them over a common denominator: $\frac34 = \frac{12}{16}$⁠.

7. Conclude

   By Step 5, $f$ is not (strictly) increasing on $(-1, 1)$⁠; by Step 6, it is not (strictly) decreasing there. So $f$ is neither strictly increasing nor decreasing on $(-1, 1)$, which is what we had to prove.

   The book's sentence can be read as "neither strictly increasing nor strictly decreasing" or as "neither strictly increasing nor decreasing". Steps 5 and 6 rule out all four properties of Definition 1, so the proof covers every reading.

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