Miscellaneous · Q10: Local maxima, minima and inflexion of $f(x)=(x-2)^4(x+1)^3$

Basics you need

Solution

1. Differentiate and factor

   Product rule with $u = (x-2)^4$⁠, $v = (x+1)^3$⁠:

   $$\begin{aligned} f'(x) &= 4(x-2)^3(x+1)^3 + 3(x-2)^4(x+1)^2 \\ &= (x-2)^3(x+1)^2\,\big[4(x+1) + 3(x-2)\big] \\ &= (x-2)^3(x+1)^2\,(7x-2). \end{aligned}$$
   Never expand a degree-7 polynomial here. Taking out the common factor $(x-2)^3(x+1)^2$ leaves a linear bracket, $4x + 4 + 3x - 6 = 7x - 2$⁠, and the fully factored form shows at once where $f'$ is zero and what sign it has.
   $u' = 4(x-2)^3\cdot 1$ and $v' = 3(x+1)^2\cdot 1$ by the chain rule (Theorem 4, Section 5.3.1). Basics: the chain rule

2. Critical points

   $f$ is a polynomial, so it is differentiable at every real $x$⁠. The critical points are therefore exactly the zeros of $f'$⁠:

   $$f'(x) = 0 \iff x = 2,\ \ x = -1,\ \ x = \frac{2}{7}.$$
   A product is zero exactly when one of its factors is zero: $(x-2)^3 = 0$ at 2, $(x+1)^2 = 0$ at $-1$⁠, $7x - 2 = 0$ at $\frac{2}{7}$⁠. By Theorem 2 (Section 6.4) local maxima and minima can only occur at these points, so they are the only candidates.
   Critical point: $f'(c) = 0$ or $f$ not differentiable at $c$ (Note after Theorem 2, Section 6.4). Lesson: maxima and minima

3. Sign chart for $f'$

   The three critical points split the line into four intervals. On each, every factor keeps one sign:

   • $(x-2)^3$⁠: odd power, so it has the sign of $x - 2$⁠: negative for $x < 2$⁠, positive for $x > 2$⁠.
   • $(x+1)^2$⁠: even power, positive except 0 at $x = -1$⁠.
   • $7x - 2$⁠: negative for $x < \frac{2}{7}$⁠, positive for $x > \frac{2}{7}$⁠.
   For instance on $\left(\frac{2}{7}, 2\right)$⁠: $(x-2)^3 < 0$⁠, $(x+1)^2 > 0$⁠, $7x - 2 > 0$⁠, so exactly one factor is negative and $f' < 0$⁠. On $(-\infty, -1)$ and $\left(-1, \frac{2}{7}\right)$⁠, two factors are negative ($(x-2)^3$ and $7x-2$⁠), so $f' > 0$⁠. Test values confirm: $f'(0) = (-8)(1)(-2) = 16 > 0$⁠, $f'(1) = (-1)(4)(5) = -20 < 0$⁠.
   An even power of a non-zero number is positive, so $(x+1)^2$ never changes the sign of the product. Basics: sign of a product

4. (i) Local maxima: $x = \frac{2}{7}$

   As $x$ increases through $\frac{2}{7}$⁠, $f'$ changes from positive to negative. $f$ is continuous there (a polynomial), so by the First Derivative Test, Theorem 3 (i), $x = \frac{2}{7}$ is a point of local maxima. The local maximum value is

   $$\begin{aligned} f\left(\tfrac{2}{7}\right) &= \left(-\tfrac{12}{7}\right)^4\left(\tfrac{9}{7}\right)^3 = \frac{12^4\cdot 9^3}{7^7} \\ &= \frac{15116544}{823543} \approx 18.36. \end{aligned}$$
   Rising just before $\frac{2}{7}$ and falling just after means the graph has a peak there. The value uses $\frac{2}{7} - 2 = -\frac{12}{7}$ and $\frac{2}{7} + 1 = \frac{9}{7}$⁠; the even power removes the minus sign. The Second Derivative Test agrees here, since $f''\left(\frac{2}{7}\right) < 0$⁠, but the sign chart already settles it.
   Theorem 3 (i), Section 6.4: $f' > 0$ just to the left and $f' < 0$ just to the right of $c$ gives local maxima. Lesson: derivative tests

5. (ii) Local minima: $x = 2$

   As $x$ increases through 2, $f'$ changes from negative to positive. By Theorem 3 (ii), $x = 2$ is a point of local minima, with local minimum value $f(2) = 0^4\cdot 3^3 = 0$⁠.

   Falling into 2 and rising out of it is a valley. It makes sense directly too: $(x-2)^4 \ge 0$ and, near $x = 2$⁠, $(x+1)^3 > 0$⁠, so $f(x) \ge 0 = f(2)$ for all $x$ close to 2.
   Theorem 3 (ii), Section 6.4. Lesson: derivative tests

6. (iii) Point of inflexion: $x = -1$

   On both sides of $-1$⁠, $f'(x) > 0$⁠: the factor $(x+1)^2$ touches 0 at $-1$ but does not change sign. So $f'$ does not change sign as $x$ increases through $-1$⁠, and by Theorem 3 (iii) $x = -1$ is neither a point of local maxima nor of local minima; the book calls such a point a point of inflexion. Here $f(-1) = 0$⁠.

   Directly: $f(x) < 0$ just left of $-1$ (the cube $(x+1)^3$ is negative) and $f(x) > 0$ just right of it, while $f(-1) = 0$⁠. A value that is below on one side and above on the other is neither a peak nor a valley. Going deeper: later courses define a point of inflexion as a point where the curve changes from bending down to bending up (where $f''$ changes sign). $x = -1$ is such a point too, and so are $x = \frac{2 \pm 3\sqrt2}{7}$ (about $-0.32$ and $0.89$⁠), where $f''(x)$⁠, which equals $6(x-2)^2(x+1)$ $\times\,(7x^2-4x-2)$⁠, changes sign. Those two are not critical points, so they are not what this question, built on Theorem 3, asks for.
   Theorem 3 (iii), Section 6.4: if $f'$ does not change sign through $c$⁠, then $c$ is neither a point of local maxima nor of local minima, "such a point is called point of inflection" (the book also spells it inflexion, as in Example 18). Lesson: derivative tests

7. Why not the Second Derivative Test at 2 and $-1$⁠?

   Differentiating $f'$ once more and factoring gives

   $$f''(x) = 6(x-2)^2(x+1)(7x^2-4x-2).$$

   At $x = 2$ and $x = -1$ this is 0, so by Theorem 4 (iii) the Second Derivative Test fails there, and we must use the First Derivative Test, as we did.

   Repeated factors in $f'$ (here $(x-2)^3$ and $(x+1)^2$⁠) leave a factor in $f''$ that vanishes at the same point. Whenever $f'$ has a repeated root, expect the second derivative test to be inconclusive there and go straight to the sign chart.
   Theorem 4 (Second Derivative Test), Section 6.4: the test fails if $f'(c) = 0$ and $f''(c) = 0$⁠. Lesson: derivative tests

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