Example 21: $2x^3-6x^2+6x+5$⁠: when the second derivative test fails

Basics you need

Solution

1. Differentiate twice

   $$\begin{aligned} f'(x) &= 6x^2 - 12x + 6 \\ &= 6(x^2 - 2x + 1) = 6(x-1)^2, \\ f''(x) &= 12x - 12 = 12(x-1). \end{aligned}$$
   Both tests need derivatives: the first test needs $f'$⁠, the second needs $f''$ as well. Factorising $f'$ now pays off twice: it shows at once where $f' = 0$ and what sign $f'$ has elsewhere.
   Power rule, term by term: $\frac{d}{dx}x^n = nx^{n-1}$⁠, constants differentiate to 0. $x^2 - 2x + 1 = (x-1)^2$ is a perfect square (Basic 5).

2. Find the critical points

   $$f'(x) = 0 \iff 6(x-1)^2 = 0 \iff x = 1.$$

   $f$ is a polynomial, so it is differentiable at every real $x$⁠; there are no critical points of the "not differentiable" kind. So $x = 1$ is the only critical point, and $f(1) = 2 - 6 + 6 + 5 = 7$⁠.

   By Theorem 2, a local maximum or minimum of a function on an open interval can only happen where $f' = 0$ or $f$ is not differentiable. So $x = 1$ is the only candidate, and the whole question reduces to: what happens at $x = 1$⁠?
   Critical point: $f'(c) = 0$ or $f$ not differentiable at $c$ (Basic 2). Lesson: maxima and minima

3. Try the second derivative test: it fails

   $$f''(1) = 12(1 - 1) = 0.$$

   So $f'(1) = 0$ and $f''(1) = 0$⁠. This is case (iii) of Theorem 4: the test fails.

   "The test fails" means the test gives no information. It does not mean "neither maximum nor minimum". Functions with $f'(c) = f''(c) = 0$ can have a minimum at $c$ ($x^4$ at 0), a maximum ($-x^4$ at 0) or neither ($x^3$ at 0). So we need a different tool.
   Theorem 4 (Second Derivative Test), part (iii): the test fails if $f'(c) = 0$ and $f''(c) = 0$⁠; "in this case, we go back to the first derivative test". Lesson: first and second derivative tests

4. Apply the first derivative test at $x = 1$

   Read the sign of $f'(x) = 6(x-1)^2$ on each side of 1:

   $x$	$(x-1)^2$	$f'(x) = 6(x-1)^2$	$f$ is
   $x < 1$ (e.g. 0.9)	$> 0$	$+$ (0.06)	increasing
   $x = 1$	$0$	$0$	flat for an instant
   $x > 1$ (e.g. 1.1)	$> 0$	$+$ (0.06)	increasing

   $f'$ is positive just to the left of 1 and positive just to the right: it does not change sign as $x$ increases through 1. 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.

   A local maximum needs the graph to rise and then fall ($f'$ from $+$ to $-$⁠); a local minimum needs fall then rise ($-$ to $+$⁠). Here the graph rises, pauses with a horizontal tangent at $(1, 7)$⁠, and rises again. The book reaches the same conclusion in Example 18, which is this very function done with the first derivative test from the start.
   Theorem 3 (First Derivative Test), part (iii): if $f'(x)$ does not change sign as $x$ increases through $c$⁠, then $c$ is neither a point of local maxima nor of local minima. Basics: sign of a product

5. Confirm from the definition

   Complete the cube: $2(x-1)^3 = 2x^3 - 6x^2 + 6x - 2$⁠, so

   $$f(x) = 2(x-1)^3 + 7.$$

   If $x < 1$ then $(x-1)^3 < 0$ and $f(x) < 7 = f(1)$⁠. If $x > 1$ then $(x-1)^3 > 0$ and $f(x) > 7 = f(1)$⁠. So every interval $(1-h, 1+h)$⁠, however small, contains points where $f$ is below $f(1)$ and points where it is above. Neither "$f(1) \ge f(x)$ for all nearby $x$" nor "$f(1) \le f(x)$ for all nearby $x$" can hold, so by Definition 4, $x = 1$ is neither kind of point.

   The tests are shortcuts; the definition is the final word. Writing $f$ around the point $x = 1$ shows the conclusion with no theorem at all, and it also shows that $f$ is strictly increasing on all of $\R$ (a cube is strictly increasing). A strictly increasing function never has a turning point.
   $(x-1)^3 = x^3 - 3x^2 + 3x - 1$ (expansion of $(a-b)^3$⁠). Definition 4 is Basic 1.

6. Why the book calls $x = 1$ a point of inflexion

   $f''(x) = 12(x-1)$ is negative for $x < 1$ and positive for $x > 1$⁠. So the graph bends downward (its slope is decreasing) to the left of 1 and bends upward to the right of 1: the bending changes direction exactly at $(1, 7)$⁠, where the tangent is horizontal.

   Theorem 3(iii) uses the name "point of inflection" for a critical point where $f'$ keeps its sign. The picture shows why the word fits: the curve crosses its own tangent line there, switching from bending down to bending up.
   The sign of $f''$ tells whether the slope $f'$ is decreasing ($f'' < 0$⁠) or increasing ($f'' > 0$⁠). Lesson: first and second derivative tests

Every fact used, and where it comes from