Exercise 6.2 · Q18: $x^3-3x^2+3x-100$ is increasing on $\R$

Basics you need

Solution

1. Differentiate $f$

   $$f'(x) = 3x^2 - 6x + 3.$$

   $f$ is a polynomial, so it is continuous and differentiable at every real number.

   Power rule term by term: $\frac{d}{dx}x^3 = 3x^2$⁠, $\frac{d}{dx}(-3x^2) = -6x$⁠, $\frac{d}{dx}(3x) = 3$⁠, and the constant $-100$ gives $0$⁠.
   Derivative of $x^n$ is $nx^{n-1}$ (Class XI, Limits and Derivatives).

2. Write $f'$ as a square

   $$\begin{aligned} f'(x) &= 3(x^2 - 2x + 1)\\ &= 3(x - 1)^2. \end{aligned}$$
   Taking out the common factor 3 leaves $x^2 - 2x + 1$⁠, which is the perfect square $(x - 1)^2$⁠. (Its discriminant is $(-2)^2 - 4\cdot1\cdot1 = 0$⁠, the sign of a repeated root.) Compare Example 8 of the book, where $3x^2 - 6x + 4 = 3(x - 1)^2 + 1$⁠.
   $(a - b)^2 = a^2 - 2ab + b^2$ with $a = x$⁠, $b = 1$⁠. Basics: factorising and the discriminant

3. $f'(x) \ge 0$ for every real $x$

   $(x - 1)^2 \ge 0$ for every real $x$⁠, and multiplying by the positive number 3 keeps this: $f'(x) = 3(x - 1)^2 \ge 0$⁠. Equality holds only when $x - 1 = 0$⁠, that is, at $x = 1$⁠.

   This is the condition of Theorem 1 (a). The derivative touches $0$ at one point but never goes below it.
   A square is never negative. Basics: sign of a product

4. Theorem 1 gives "increasing on $\R$"

   Take any real numbers $x_1 \lt x_2$⁠. On $[x_1, x_2]$⁠, $f$ is continuous, it is differentiable on $(x_1, x_2)$⁠, and $f'(x) \ge 0$ for each $x \in (x_1, x_2)$ (Step 3). By Theorem 1 (a), $f$ is increasing on $[x_1, x_2]$⁠; in particular $f(x_1) \le f(x_2)$⁠. Since $x_1 \lt x_2$ were arbitrary, Definition 1 (i) gives: $f$ is increasing on $\R$.

   Theorem 1 is about a closed interval $[a, b]$⁠, but $\R$ has no end points. So we apply it to the closed interval between any two chosen points; that is enough, because Definition 1 only ever compares two points at a time.
   Theorem 1 (a), Section 6.3. Lesson: increasing and decreasing functions

5. Going further: $f$ is even strictly increasing

   Since $(x - 1)^3 = x^3 - 3x^2 + 3x - 1$⁠,

   $$f(x) = (x - 1)^3 - 99.$$

   Let $x_1 \lt x_2$ and put $u_1 = x_1 - 1$⁠, $u_2 = x_2 - 1$⁠, so $u_1 \lt u_2$⁠. Then

   $$\begin{aligned} f(x_2) - f(x_1) &= u_2^3 - u_1^3\\ &= (u_2 - u_1)(u_1^2 + u_1u_2 + u_2^2). \end{aligned}$$

   The first factor is positive. The second equals $\left(u_1 + \frac{u_2}{2}\right)^2 + \frac34u_2^2$⁠, a sum of two squares, which is $0$ only if $u_2 = 0$ and then $u_1 = 0$⁠, impossible as $u_1 \lt u_2$⁠. So it is positive, and $f(x_2) - f(x_1) \gt 0$⁠: $f$ is strictly increasing on $\R$.

   The question asks only for "increasing", which Step 4 settles. This step shows the zero of $f'$ at $x = 1$ is harmless: the graph levels off for an instant but keeps rising. The same holds whenever $f' \ge 0$ and $f'$ is zero only at isolated points: if $f(x_1) = f(x_2)$ for some $x_1 \lt x_2$⁠, then $f$⁠, being increasing, would be constant on $[x_1, x_2]$⁠, forcing $f' = 0$ on a whole interval.
   $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$⁠, and completing the square. Basics: factorising and the discriminant

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