Corrected exercises: monotonic functions and the First Derivative Test (MATH 203)

$f$ is a polynomial, so $f'$ exists everywhere and the critical points are the zeros of $f'$. The algebraic gesture: take out the common factor FIRST, $f'(x) = 3x^2 - 6x - 24 = 3(x^2 - 2x - 8)$, then factor the trinomial by looking for two numbers with product $-8$ and sum $-2$: $f'(x) = 3(x - 4)(x + 2)$. The factor $3$ is positive and never matters. On $(-\infty, -2)$ both brackets are negative, $f' > 0$; on $(-2, 4)$ only $x - 4$ is negative, $f' < 0$; on $(4, \infty)$ both are positive, $f' > 0$. By the corollary, $f$ is increasing on $(-\infty, -2]$, decreasing on $[-2, 4]$ and increasing on $[4, \infty)$ (the endpoints may be included since $f$ is continuous). First Derivative Test: $+$ to $-$ at $-2$, local maximum $f(-2) = -8 - 12 + 48 + 4 = 32$; $-$ to $+$ at $4$, local minimum $f(4) = 64 - 48 - 96 + 4 = -76$.

$g'(x) = 12x^3 + 12x^2 + 12x = 12x(x^2 + x + 1)$. Now the trinomial: its discriminant is $1^2 - 4(1)(1) = -3 < 0$, so it has no real root and it keeps the sign of its value at $0$, which is $1$: $x^2 + x + 1 > 0$ for every $x$ (completing the square, $\left(x + \frac{1}{2}\right)^2 + \frac{3}{4}$). This factor is therefore dropped from the sign, never from the derivative: the sign of $g'$ is the sign of $12x$. The only critical point is $0$, $g$ is decreasing on $(-\infty, 0]$ and increasing on $[0, \infty)$, and the First Derivative Test gives a local minimum $g(0) = -5$, and no local maximum at all. The classic loss: the student sees that $x^2 + x + 1 = 0$ has no solution, writes “$g'$ has no zero, no critical point”, and forgets the factor $12x$ that was taken out first.

$h'(x) = 5x^4 - 20x^3 + 15x^2 = 5x^2(x^2 - 4x + 3) = 5x^2(x - 1)(x - 3)$, with the lowest power $x^2$ taken out as a common factor. Critical points: $0$, $1$ and $3$. The sign chart of the solution figure takes each factor on each interval. $5x^2$ is positive except at $0$; $x - 1$ and $x - 3$ change sign at their zeros. So $h' > 0$ on $(-\infty, 0)$ and on $(0, 1)$, $h' < 0$ on $(1, 3)$, $h' > 0$ on $(3, \infty)$. Hence $h$ is increasing on $(-\infty, 1]$ (the two intervals share the point $0$), decreasing on $[1, 3]$, increasing on $[3, \infty)$. At $1$: $+$ to $-$, local maximum $h(1) = 1 - 5 + 5 = 1$. At $3$: $-$ to $+$, local minimum $h(3) = 243 - 405 + 135 = -27$. At $0$: $+$ on both sides, no local extremum, although $h'(0) = 0$.

Signs alternate only across a zero of ODD multiplicity. The factor $x^2$ is a square: it is positive on both sides of $0$ and cannot flip the sign of $h'$. The row written by rhythm, $- \mid + \mid - \mid +$, is wrong on $(-\infty, 0)$: a test value confirms it, $h'(-1) = 5(1)(-2)(-4) = 40 > 0$. The correct row is $+ \mid + \mid - \mid +$, and the point $0$ is a critical point without extremum, a horizontal tangent on a rising curve. The student's “local minimum at $0$” is a point that does not exist, and a marker takes the whole classification away. The safe method: one row per factor, as in the figure, or one test value in EVERY interval, never a pattern written in advance.

With $u = x^2 + 5x$ and $v = x - 4$: $f'(x) = \frac{(2x + 5)(x - 4) - (x^2 + 5x)(1)}{(x - 4)^2}$. Brackets first, then expand: $(2x + 5)(x - 4) = 2x^2 - 3x - 20$, and $-(x^2 + 5x) = -x^2 - 5x$. The numerator is $x^2 - 8x - 20 = (x - 10)(x + 2)$. The domain is $x \ne 4$, and $(x - 4)^2 > 0$ on it, so the sign of $f'$ is that of $(x - 10)(x + 2)$: positive for $x < -2$, negative on $(-2, 4)$ and on $(4, 10)$, positive for $x > 10$. The chart has THREE columns, $-2$, $4$ and $10$, the middle one being a double bar. So $f$ is increasing on $(-\infty, -2]$, decreasing on $[-2, 4)$ and on $(4, 10]$ (separately: never “on $[-2, 10]$”, which contains $4$), increasing on $[10, \infty)$. Local maximum $f(-2) = \frac{4 - 10}{-6} = 1$, local minimum $f(10) = \frac{100 + 50}{6} = 25$.

A local maximum is only larger than the values NEAR it, and a local minimum only smaller than the values near it. Here the maximum at $x = -2$ lives on the left branch and the minimum at $x = 10$ on the right branch; between them $f$ is not even defined at $4$, where the left branch falls to $-\infty$ and the right branch comes down from $+\infty$. Nothing forces the two heights to be in any order. On an interval where $f$ is continuous, two consecutive extrema of opposite types are ordered; across an asymptote, no rule applies.

Quotient rule with $u = x$, $v = (x - 2)^2$, $v' = 2(x - 2)$ (chain rule): $g'(x) = \frac{(x - 2)^2 - x \cdot 2(x - 2)}{(x - 2)^4}$. The gesture: do NOT expand. Take out the common factor $(x - 2)$ of the numerator and cancel it once: $g'(x) = \frac{(x - 2)\left[(x - 2) - 2x\right]}{(x - 2)^4} = \frac{-x - 2}{(x - 2)^3} = -\frac{x + 2}{(x - 2)^3}$. This time the denominator is a CUBE, so it changes sign at $2$. For $x < -2$: $-(x + 2) > 0$ over a negative cube, $g' < 0$. On $(-2, 2)$: $-(x + 2) < 0$ over a negative cube, $g' > 0$. For $x > 2$: negative over positive, $g' < 0$. So $g$ is decreasing on $(-\infty, -2]$, increasing on $[-2, 2)$, decreasing on $(2, \infty)$. Local minimum $g(-2) = \frac{-2}{16} = -\frac{1}{8}$. At $2$, $g'$ goes from $+$ to $-$, but $2$ is not in the domain: no maximum, the graph climbs to $+\infty$ on both sides.

The student wrote $- x^2 + 5x$ instead of $-(x^2 + 5x) = -x^2 - 5x$: the minus sign was applied to the first term of $uv'$ only. The numerator becomes $2x^2 - 3x - 20 - x^2 + 5x = x^2 + 2x - 20$, whose zeros are $-1 \pm \sqrt{21}$, about $-5.58$ and $3.58$: two false critical points, a false chart, and every conclusion of a) lost, although the quotient rule itself was known. A check catches it, but not at $x = 0$: there the faulty term $5x$ vanishes and both formulas give $-\frac{20}{16} = -\frac{5}{4}$. At $x = 10$ the true value is $0$ and the student's is $\frac{100}{36} = \frac{25}{9}$. Test the simplified derivative against the unsimplified one at two easy points, one of them away from $0$.

Expand first to have pure powers: $f(x) = x^{8/3} - 16x^{2/3}$, so $f'(x) = \frac{8}{3}x^{5/3} - \frac{32}{3}x^{-1/3}$. The exponents are $\frac{5}{3}$ and $-\frac{1}{3}$; the smaller one is $-\frac{1}{3}$, and factoring it out SUBTRACTS it from each exponent: $x^{5/3} = x^{-1/3} \cdot x^{2}$. So $f'(x) = \frac{8}{3}x^{-1/3}(x^2 - 4) = \frac{8(x - 2)(x + 2)}{3x^{1/3}}$. The domain of $f$ is all real numbers (a cube root accepts negatives). Critical points: the zeros of the numerator, $-2$ and $2$, AND $0$, where the denominator vanishes, $f'(0)$ does not exist, but $f(0) = 0$ exists.

The cube root $x^{1/3}$ has the sign of $x$, and $(x - 2)(x + 2)$ is positive outside $[-2, 2]$, negative inside. For $x < -2$: $+$ over $-$, $f' < 0$. On $(-2, 0)$: $-$ over $-$, $f' > 0$. On $(0, 2)$: $-$ over $+$, $f' < 0$. For $x > 2$: $f' > 0$. First Derivative Test: local minima at $-2$ and $2$, with $f(\pm 2) = 4^{1/3}(4 - 16) = -12\sqrt[3]{4}$ (about $-19.05$), and a local maximum at $0$, $f(0) = 0$. The maximum sits exactly where $f'$ does not exist: as $x \to 0^-$, $f'$ is large and positive, as $x \to 0^+$ large and negative, so the graph has a CUSP at the origin, as in the solution figure. The symmetry is a free check: $f(-x) = f(x)$, the two minima must have the same value.

The square root needs $6 - x \ge 0$: the domain is $(-\infty, 6]$. Product rule: $g'(x) = \sqrt{6 - x} + x \cdot \frac{-1}{2\sqrt{6 - x}}$, for $x < 6$. Common denominator $2\sqrt{6 - x}$: the first term becomes $\frac{2(6 - x)}{2\sqrt{6 - x}}$, since $\sqrt{6 - x} \cdot \sqrt{6 - x} = 6 - x$. So $g'(x) = \frac{12 - 2x - x}{2\sqrt{6 - x}} = \frac{3(4 - x)}{2\sqrt{6 - x}}$. The denominator is positive on $(-\infty, 6)$: $g' > 0$ for $x < 4$, $g' < 0$ on $(4, 6)$. Critical points: $4$ ($g' = 0$) and $6$ ($g'$ undefined, and an endpoint). $g$ is increasing on $(-\infty, 4]$ and decreasing on $[4, 6]$. Local maximum $g(4) = 4\sqrt 2$. At the endpoint $6$, $g$ decreases INTO it, so $g(x) > g(6) = 0$ for $x$ of the domain near $6$: local minimum $0$ at the endpoint, with a vertical tangent.

Multiplying by $x^{1/3}$ is legitimate only for $x \ne 0$, and precisely at $0$ the derivative does not exist: the multiplication silently deleted a critical point. It was not a harmless one: $0$ is the only local maximum of $f$, and its value $0$ is part of the expected answer. The correct gesture is the factorization of a), which keeps $x^{1/3}$ visible in the denominator, so the chart shows the column $0$ and its sign change. The same rule holds for any factor: never multiply or divide an equation $f'(x) = 0$ by an expression that can be zero or undefined; factor it and look at it.

A critical point is a number of the domain where $f'$ is zero or does not exist. The graph of $f'$ meets the axis at $x = -2$ (first piece) and at $x = 5$ (second piece): $f'(-2) = f'(5) = 0$. At $x = 2$ both pieces end with an open circle: $f'(2)$ does not exist, while $f(2)$ exists because $f$ is continuous on $[-4, 6]$. So the critical points are $-2$, $2$ and $5$. Forgetting $2$ is the classic loss: the eye looks for crossings of the axis, and at $2$ the graph of $f'$ does not cross it, it JUMPS over it.

Read the sign: below the axis on $[-4, -2)$, above on $(-2, 2)$, below on $(2, 5)$, above on $(5, 6]$. By the corollary, applied on each interval where $f$ is continuous and differentiable inside: $f$ is decreasing on $[-4, -2]$, increasing on $[-2, 2]$, decreasing on $[2, 5]$ and increasing on $[5, 6]$. The point $2$ may close both intervals $[-2, 2]$ and $[2, 5]$ because $f$ is continuous at $2$; the corollary only needs differentiability INSIDE each interval.

At $-2$: $-$ to $+$, local minimum. At $2$: $+$ to $-$, local maximum (the First Derivative Test only needs $f$ continuous at $2$, not differentiable there). At $5$: $-$ to $+$, local minimum. Endpoints: $f$ is decreasing on $[-4, -2]$, so $f(x) < f(-4)$ for $x$ just to the right of $-4$: local maximum at $-4$. $f$ is increasing on $[5, 6]$, so $f(x) < f(6)$ just to the left of $6$: local maximum at $6$. A table of local extrema on a closed interval that stops at the interior points is missing two lines.

The student read the graph of $f'$ as if it were the graph of $f$. The value $-3$ is the slope of $f$ just after $2$: there $f$ is falling steeply, it does not sit at a bottom. What happens at $2$ is read on the SIGN: $f'$ is positive just before $2$ (values near $2$) and negative just after (values near $-3$), so $f$ rises then falls: a local MAXIMUM at $2$. Since the two one-sided slopes are different, $2$ and $-3$, the graph of $f$ has a corner there, a peak with two different tangent directions.

$f$ is increasing on $[-2, 2]$ and $0 < 2$: $f(0) < f(2)$, strictly since $f' > 0$ on $(0, 2)$. $f$ is decreasing on $[2, 5]$ and $3 < 5$: $f(3) > f(5)$. $f$ is decreasing on $[-4, -2]$: $f(-4) > f(-2)$. But $-4$ and $6$ are separated by increases AND decreases: the signs say where $f$ goes, not how far. Deciding would require measuring the net change of $f$ from the areas between the graph of $f'$ and the axis, which is the business of MATH 205, not of the First Derivative Test.