Example 18: $2x^3-6x^2+6x+5$⁠: a critical point that is not an extremum

Basics you need

Solution

1. Differentiate and factorise

   $$\begin{aligned} f'(x) &= 6x^2 - 12x + 6 \\ &= 6(x^2 - 2x + 1) = 6(x-1)^2. \end{aligned}$$
   After taking out 6, the quadratic is a perfect square (Basic 3): its discriminant is $(-2)^2 - 4 = 0$⁠, so it has one repeated root. That repeated root is the whole story of this example.
   Basics: factorising and the discriminant

2. The only critical point is $x = 1$

   $f$ is a polynomial, so differentiable everywhere. $f'(x) = 6(x-1)^2 = 0 \iff x = 1$⁠. So $x = 1$ is the only critical point, and the only candidate for a local extremum (Theorem 2).

3. The sign of $f'$ does not change at $x = 1$

   For every $x \ne 1$⁠, $(x-1)^2 > 0$⁠, so $f'(x) = 6(x-1)^2 > 0$⁠. Thus $f'(x) \ge 0$ for all $x \in \R$⁠, and in particular $f'(x) > 0$ at points close to 1 on the left and on the right.

   For example $f'(0.9) = 6(0.01) = 0.06$ and $f'(1.1) = 0.06$⁠, both positive. By Theorem 3(iii), $x = 1$ is neither a point of local maxima nor a point of local minima; it is a point of inflexion.

   To reach a hilltop the graph must stop rising and start falling; to reach a valley it must stop falling and start rising. Here it rises before 1 and rises after 1, so neither happens.
   Theorem 3(iii), Section 6.4. Lesson: first and second derivative tests

4. Confirm directly: $f$ is a shifted cube

   Expand $2(x-1)^3 + 7$⁠:

   $$\begin{aligned} 2(x-1)^3 + 7 &= 2(x^3 - 3x^2 + 3x - 1) + 7 \\ &= 2x^3 - 6x^2 + 6x + 5. \end{aligned}$$

   So $f(x) = 2(x-1)^3 + 7$⁠, and $f(1) = 7$⁠. By Basic 4, for $x < 1$ we get $f(x) < 7$⁠, and for $x > 1$ we get $f(x) > 7$⁠. Every interval $(1-h, 1+h)$ contains points with values below 7 and points with values above 7, so neither inequality of Definition 4 can hold: no local maximum and no local minimum at 1.

   This is the Remark after Theorem 2 in disguise: $x^3$ at 0 has derivative 0 and no extremum; our $f$ is the same curve stretched by 2 and moved so that the flat point sits at $(1, 7)$⁠.

5. The book's Remark: no turning points at all

   Since $f'(x) \ge 0$ on $\R$ and never changes sign, the graph of $f$ has no turning points, hence no local maxima or minima anywhere. (The book's Remark refers to "Example 30", a numbering slip: Example 30 in this reprint is about a car; the Remark is about the present function.) In fact $f$ is strictly increasing on $\R$⁠: if $x_1 < x_2$ then $(x_1-1)^3 < (x_2-1)^3$⁠, so $f(x_1) < f(x_2)$⁠.

   A turning point needs a change in direction, which needs $f'$ to change sign. A derivative that is a perfect square times a positive number can touch 0 but can never go negative.
   The second derivative $f''(x) = 12(x-1)$ gives $f''(1) = 0$⁠, so the Second Derivative Test fails here (Theorem 4(iii)); the book revisits this in Example 21 and goes back to the first derivative test, as we did.

Every fact used, and where it comes from