Exercise 6.2 · Q8: Where is $[x(x-2)]^2$ increasing?

Basics you need

Solution

1. Differentiate with the chain rule

   Let $u = x(x - 2) = x^2 - 2x$⁠, so $y = u^2$ and $\frac{du}{dx} = 2x - 2$⁠. Then

   $$\begin{aligned} \frac{dy}{dx} &= 2u\cdot\frac{du}{dx} = 2(x^2 - 2x)(2x - 2)\\ &= 2\cdot x(x - 2)\cdot 2(x - 1)\\ &= 4x(x - 1)(x - 2). \end{aligned}$$
   Squaring is the outer operation and $x(x - 2)$ is the inner one. Keeping the inner function in factorised form, $x(x - 2)$⁠, and writing $2x - 2 = 2(x - 1)$⁠, gives a fully factorised derivative straight away, which is exactly what a sign chart needs.
   Outer $u^2$ (derivative $2u$⁠), inner $u = x^2 - 2x$ (derivative $2x - 2$⁠). Basics: the chain rule

2. Zeros of the derivative

   $$\frac{dy}{dx} = 0 \iff x = 0,\ 1 \text{ or } 2.$$

   These points divide the line into $(-\infty, 0)$⁠, $(0, 1)$⁠, $(1, 2)$ and $(2, \infty)$⁠.

   Each factor $x$⁠, $x - 1$⁠, $x - 2$ changes sign only at its own zero, so the product $4x(x - 1)(x - 2)$ can change sign only at $0$⁠, $1$ or $2$⁠. On each of the four intervals every factor, and hence the product, keeps one sign.

3. Sign chart

   • $x < 0$⁠: all three factors $x, x - 1, x - 2$ negative (odd number), so $\frac{dy}{dx} < 0$⁠. (Test: at $x = -1$⁠, $4(-1)(-2)(-3) = -24$⁠.)
   • $0 < x < 1$⁠: two negative factors, so $\frac{dy}{dx} > 0$⁠. (Test: at $x = \frac12$⁠, $4\cdot\frac12\cdot(-\frac12)(-\frac32) = \frac32$⁠.)
   • $1 < x < 2$⁠: one negative factor, so $\frac{dy}{dx} < 0$⁠. (Test: at $x = \frac32$⁠, the value is $-\frac32$⁠.)
   • $x > 2$⁠: no negative factor, so $\frac{dy}{dx} > 0$⁠. (Test: at $x = 3$⁠, $4\cdot 3\cdot 2\cdot 1 = 24$⁠.)
   With three simple factors, each crossing of a zero flips exactly one factor, so the sign alternates. Counting negative factors interval by interval makes this visible.
   Odd number of negative factors: negative; even: positive. Basics: sign of a product

4. Conclude on intervals

   By Theorem 1 (applied on $[x_1, x_2]$ for pairs inside each interval):

   • $y$ is (strictly) increasing on $(0, 1)$ and on $(2, \infty)$⁠;
   • $y$ is (strictly) decreasing on $(-\infty, 0)$ and on $(1, 2)$⁠.
   The picture agrees: $y$ is a square, so $y \ge 0$ with $y = 0$ exactly at $x = 0$ and $x = 2$ (the bottoms of the W). Between them $y$ must go up and come down; $x = 1$⁠, halfway, is where $|x(x - 2)|$ is largest on $[0, 2]$⁠.

5. The points 0, 1 and 2

   The question asks for the values of $x$ where $y$ is increasing, which is Definition 2: $y$ is increasing at $x_0$ if it is increasing on some open interval containing $x_0$⁠.

   • Every $x_0$ in $(0, 1)$ or $(2, \infty)$ qualifies: that interval itself is an open interval containing $x_0$ on which $y$ is increasing.
   • $x_0 = 0$ does not: every open interval around 0 contains some $-\delta < 0$⁠, and $y(-\delta) = [\delta(\delta + 2)]^2 > 0 = y(0)$⁠, so a smaller input gives a larger output.
   • $x_0 = 1$ does not: for small $\delta > 0$⁠, $y(1 + \delta) = (1 - \delta^2)^2 < 1 = y(1)$⁠.
   • $x_0 = 2$ does not: $y(2 - \delta) = [\delta(2 - \delta)]^2 > 0 = y(2)$ for small $\delta > 0$⁠.

   So the values of $x$ are exactly $0 < x < 1$ and $x > 2$.

   This matches the answer key's open intervals exactly. Note the two different questions: "on which intervals is $y$ increasing?" would also allow the closed intervals $[0, 1]$ and $[2, \infty)$ (by continuity, Theorem 1), but "at which points?" excludes 0, 1 and 2, because near each of them $y$ also decreases on one side.
   Definition 2, Section 6.3: increasing at a point means increasing on some open interval around it. Lesson: increasing and decreasing functions

Every fact used, and where it comes from