Example 8: $f(x) = x^3-3x^2+4x$ is increasing on $\R$

Basics you need

Solution

1. Differentiate

   By the power, constant multiple and sum rules:

   $$f'(x) = 3x^2 - 6x + 4 .$$
   Theorem 1 converts the question "does $f$ go up?" into "is $f'$ non-negative?", which is a question about a quadratic, much easier than comparing values of a cubic.
   $\frac{d}{dx}(x^3) = 3x^2$⁠, $\frac{d}{dx}(3x^2) = 6x$⁠, $\frac{d}{dx}(4x) = 4$ (Section 5.3).

2. Show $f'(x) > 0$ for every $x$ by completing the square

   Take out 3 from the first two terms, then complete the square:

   $$\begin{aligned} f'(x) &= 3(x^2 - 2x) + 4 \\ &= 3(x^2 - 2x + 1) - 3 + 4 \\ &= 3(x-1)^2+1 . \end{aligned}$$

   Since $(x - 1)^2 \ge 0$ for every real $x$⁠, we get $f'(x) \ge 0 + 1 = 1 > 0$ for every real $x$⁠. The least value 1 occurs at $x = 1$⁠.

   Writing $f'$ as "a square times a positive number, plus a positive number" makes its sign obvious at a glance. The book writes the same chain: $3(x^2 - 2x + 1) + 1$ $= 3(x-1)^2 + 1 > 0$⁠. Equivalently, the discriminant of $3x^2 - 6x + 4$ is $36 - 48 = -12 < 0$⁠, so $f'$ has no real zero, and since its leading coefficient 3 is positive it is positive everywhere.
   Basics: factorising and the discriminant

3. Apply Theorem 1 and conclude "increasing on $\mathbf{R}$"

   Take any two real numbers $x_1 < x_2$ and the interval $[x_1, x_2]$⁠. The polynomial $f$ is continuous on $[x_1, x_2]$ and differentiable on $(x_1, x_2)$⁠, and $f'(x) > 0$⁠, so certainly $f'(x) \ge 0$⁠, on $(x_1, x_2)$⁠. By Theorem 1(a), $f$ is increasing in $[x_1, x_2]$⁠; in particular $f(x_1) \le f(x_2)$⁠.

   Since $x_1 < x_2$ were arbitrary, $f$ is increasing on $\mathbf{R}$.

   Theorem 1 is stated for a closed interval $[a, b]$⁠, while $\mathbf{R}$ is not one. The way round this is to notice that the definition of "increasing on $\mathbf{R}$" only ever compares two points at a time, and any two points lie in a closed interval where Theorem 1 applies. This is what the book means by "$> 0$⁠, in every interval of $\mathbf{R}$".
   Theorem 1(a) and Definition 1(i), Section 6.3. Lesson: increasing and decreasing functions

4. In fact $f$ is strictly increasing

   Follow the book's proof of Theorem 1: for $x_1 < x_2$ there is a $c$ between them with

   $$f(x_2) - f(x_1) = f'(c)\,(x_2 - x_1) .$$

   Here $f'(c) \ge 1 > 0$ and $x_2 - x_1 > 0$⁠, so $f(x_2) - f(x_1) > 0$⁠. Hence $f$ is strictly increasing on $\mathbf{R}$ (Definition 1(iv)), which is more than the question asks.

   Theorem 1 as printed only promises "increasing" ($\le$⁠), because it allows $f' = 0$⁠. When $f'$ is strictly positive, the same Mean Value Theorem argument gives the strict inequality. The book's printed proof in fact writes "$f'(c) > 0$ (given)" although the theorem assumes only $f' \ge 0$⁠; with $f'(c) \ge 0$ the same line gives $f(x_2) - f(x_1) \ge 0$⁠, which is exactly "increasing". The Remarks after the theorem assume $f' > 0$ on an interval without its end points, but they too conclude only "increasing".
   The Mean Value Theorem is used without proof in the book's proof of Theorem 1, which cites it as "Theorem 8 in Chapter 5"; this reprint's Chapter 5 does not contain it.

5. A check without derivatives

   For any real $a < b$⁠, factorising gives

   $$\begin{aligned} f(b) - f(a) &= (b^3 - a^3) - 3(b^2 - a^2) \\ &\quad + 4(b - a) \\ &= (b - a)\,Q, \end{aligned}$$

   where $Q = a^2 + ab + b^2 - 3(a + b) + 4$⁠. Put $p = a - 1$ and $q = b - 1$⁠; expanding shows

   $$Q = p^2 + pq + q^2 + 1 ,$$

   and $p^2 + pq + q^2 = \left(p + \tfrac{q}{2}\right)^2 + \tfrac{3}{4}q^2 \ge 0$⁠. So $Q \ge 1$⁠, and $f(b) - f(a) = (b - a)Q > 0$⁠.

   This proves "strictly increasing" straight from Definition 1, as in Example 7, with no calculus at all. It is longer, which is exactly why Theorem 1 is so useful.
   $b^3 - a^3 = (b - a)(a^2 + ab + b^2)$ and $b^2 - a^2 = (b - a)(a + b)$⁠.

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