Exercise 1: Polynomials: factor f' completely, then read the sign factor by factor

Thomas states the tool as a corollary of the Mean Value Theorem, which MATH 203 uses without proving it: if $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$ , and $f'(x) > 0$ at each point of $(a, b)$ , then $f$ is increasing on $[a, b]$ ; if $f'(x) < 0$ there, $f$ is decreasing on $[a, b]$ .

First Derivative Test for Local Extrema: at a critical point $c$ where $f$ is continuous, if $f'$ changes from negative to positive, $f$ has a local minimum at $c$ ; from positive to negative, a local maximum; if $f'$ keeps its sign, no local extremum. Everything therefore rests on the SIGN of $f'$ , and a sign is read on a product, one factor at a time. Expanded, $f'$ tells you nothing.

a) Let $f(x) = x^3 - 3x^2 - 24x + 4$ . Find the intervals on which $f$ is increasing and decreasing, and the local maximum and minimum values of $f$ .
b) Let $g(x) = 3x^4 + 4x^3 + 6x^2 - 5$ . Factor $g'(x)$ as far as possible over the real numbers, and find the local extreme values of $g$ .
c) Let $h(x) = x^5 - 5x^4 + 5x^3$ . Find the critical points of $h$ and classify each one with the First Derivative Test.
d) For c), a student factors $h'$ correctly, then fills the sign row from the right, alternating: $+$ after $3$ , $-$ on $(1, 3)$ , $+$ on $(0, 1)$ , $-$ before $0$ , and announces a local minimum at $0$ . Correct the chart.

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Answers

a) $f'(x) = 3(x - 4)(x + 2)$ : increasing on $(-\infty, -2]$ and $[4, \infty)$ , decreasing on $[-2, 4]$ ; local max $f(-2) = 32$ , local min $f(4) = -76$
b) $g'(x) = 12x(x^2 + x + 1)$ , the quadratic has no real root; local min $g(0) = -5$ , no local max
c) $h'(x) = 5x^2(x - 1)(x - 3)$ : critical points $0$ , $1$ , $3$ ; local max $h(1) = 1$ , local min $h(3) = -27$ , nothing at $0$
d) $5x^2 > 0$ on both sides of $0$ : $h' > 0$ on $(-\infty, 0)$ AND on $(0, 1)$ , so no extremum at $0$ .

a) $f$ 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 $f(4) = 64 - 48 - 96 + 4 = -76$ .

b) $g'(x) = 12x^3 + 12x^2 + 12x = 12x(x^2 + x + 1)$ . 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.

c) $h'(x) = 5x^4 - 20x^3 + 15x^2 = 5x^2(x^2 - 4x + 3) = 5x^2(x - 1)(x - 3)$ , 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 $h'(0) = 0$ .

d) 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.

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Exercise 2: Quotients: the minus sign of the quotient rule, and the holes of the domain

For a quotient, $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$ , and the minus sign applies to the WHOLE product $uv'$ : write it with brackets before expanding. The denominator $v^2$ is positive wherever $f$ is defined, so the sign of $f'$ is the sign of the numerator, once it is factored.

A number where $f$ is not defined is not a critical point, yet it must be a column of the sign chart: $f'$ can change sign there with no extremum at all. The figure shows the graph of $f(x) = \frac{x^2 + 5x}{x - 4}$ and its vertical asymptote $x = 4$ .

a) Show that $f'(x) = \frac{(x - 10)(x + 2)}{(x - 4)^2}$ . Find the intervals on which $f$ is increasing and decreasing, and its local extreme values.
b) On the figure, the local maximum value is SMALLER than the local minimum value. Explain why this is not a contradiction.
c) Let $g(x) = \frac{x}{(x - 2)^2}$ . Show that $g'(x) = -\frac{x + 2}{(x - 2)^3}$ , find the intervals of increase and decrease, and the local extrema of $g$ .
d) A student starts a) with $f'(x) = \frac{(2x + 5)(x - 4) - x^2 + 5x}{(x - 4)^2} = \frac{x^2 + 2x - 20}{(x - 4)^2}$ . Find the error and say what it does to the answer.

Type your answers, the page tells you right or wrong 0/9

a)

Local maximum value

Local minimum value

f

increases on the right of ,

b)

The max is below the min because

c)

Local minimum value of

g

Number of local maxima of

g

Sign of

g'

on

(2, \infty)

d)

The error

Correct numerator at

x = 10

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Answers

a) Increasing on $(-\infty, -2]$ and $[10, \infty)$ , decreasing on $[-2, 4)$ and $(4, 10]$ ; local max $f(-2) = 1$ , local min $f(10) = 25$
b) Local means near the point: the two extrema lie on the two branches separated by the asymptote $x = 4$ .
c) Decreasing on $(-\infty, -2]$ and $(2, \infty)$ , increasing on $[-2, 2)$ ; local min $g(-2) = -\frac{1}{8}$ , no local max ( $2$ is not in the domain)
d) The minus sign was not applied to $5x$ : the numerator is $x^2 - 8x - 20$ , not $x^2 + 2x - 20$ , and the critical points $-1 \pm \sqrt{21}$ are wrong.

a) 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 $f(10) = \frac{100 + 50}{6} = 25$ .

b) 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.

c) 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.

d) 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 $0$ .

Exercise 3: Fractional powers and roots: factor out the LOWEST power

When $f'$ is a sum of powers such as $x^{5/3}$ and $x^{-1/3}$ , or contains a square root, it becomes readable only as ONE fraction. The gesture: factor out the power with the smallest exponent (the most negative one), even if it is negative, then send it to the denominator. Nothing is ever divided away: a number where $f'$ does not exist but $f$ does is a critical point.

Thomas defines a local extremum at an ENDPOINT of the domain too: $f$ has a local minimum at an endpoint $c$ if $f(x) \ge f(c)$ for the points $x$ of the domain near $c$ . The sign of $f'$ next to the endpoint decides it.

a) Let $f(x) = x^{2/3}(x^2 - 16)$ . Show that $f'(x) = \frac{8(x - 2)(x + 2)}{3x^{1/3}}$ , and list ALL the critical points of $f$ .
b) Classify each critical point of $f$ with the First Derivative Test, and give the local extreme values exactly.
c) Let $g(x) = x\sqrt{6 - x}$ . Give its domain, show that $g'(x) = \frac{3(4 - x)}{2\sqrt{6 - x}}$ , and find all the local extrema of $g$ , endpoint included.
d) A student solves $f'(x) = 0$ in a) by multiplying $\frac{8}{3}x^{5/3} - \frac{32}{3}x^{-1/3} = 0$ by $x^{1/3}$ , finds $x = \pm 2$ and writes: the critical points are $-2$ and $2$ . What did this cost?

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Answers

a) Critical points $-2$ , $2$ ( $f' = 0$ ) and $0$ ( $f'$ undefined, $f(0) = 0$ )
b) Local min $f(\pm 2) = -12\sqrt[3]{4}$ at $x = -2$ and $x = 2$ ; local max $f(0) = 0$ , at a cusp
c) Domain $(-\infty, 6]$ ; increasing on $(-\infty, 4]$ , decreasing on $[4, 6]$ ; local max $g(4) = 4\sqrt 2$ , local min $g(6) = 0$ at the endpoint
d) The critical point $0$ was lost, and with it the only local maximum of $f$ .

a) 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.

b) 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.

c) 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.

d) 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.

Exercise 4: Reading f on the graph of f', when f' jumps

The figure shows the graph of the DERIVATIVE $f'$ of a function $f$ that is continuous on $[-4, 6]$ and differentiable at every point of $(-4, 6)$ except $x = 2$ . The graph of $f'$ is made of two straight pieces through grid points; the open circles at $(2, 2)$ and $(2, -3)$ mean that $f'(2)$ does not exist.

Only the SIGN of $f'$ , above or below the axis, speaks about the direction of $f$ . The height of $f'$ says how steep $f$ is, never how high $f$ is.

a) List the critical points of $f$ in $(-4, 6)$ , saying for each whether $f' = 0$ or $f'$ does not exist there.
b) On which intervals is $f$ increasing, decreasing?
c) Classify each critical point, and say whether $f$ has a local extremum at each endpoint $x = -4$ and $x = 6$ .
d) A student writes: the graph of $f'$ is lowest just to the right of $2$ , at height $-3$ , so $f$ has a local minimum at $x = 2$ . Correct.
e) Compare, when the figure allows it: $f(0)$ and $f(2)$ ; $f(3)$ and $f(5)$ ; $f(-4)$ and $f(-2)$ . Can the figure alone decide between $f(-4)$ and $f(6)$ ?

Type your answers, the page tells you right or wrong 0/9

a)

Number of critical points in

(-4, 6)

b)

f

increases on ,

f

decreases on the interval ,

c)

At

x = 2

,

f

has a

At

x = 5

,

f

has a

At the endpoint

x = -4

,

f

has a

d)

At

x = 2

, the graph of

f

has

e)

f(0)

against

f(2)

f(-4)

against

f(6)

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Answers

a) $-2$ and $5$ ( $f' = 0$ ), and $2$ ( $f'$ does not exist)
b) Decreasing on $[-4, -2]$ and $[2, 5]$ , increasing on $[-2, 2]$ and $[5, 6]$
c) Local min at $-2$ and at $5$ , local max at $2$ ; local max at both endpoints $-4$ and $6$
d) False: $f'$ goes from $+$ to $-$ at $2$ , so $f$ has a local MAXIMUM there, at a corner.
e) $f(0) < f(2)$ , $f(3) > f(5)$ , $f(-4) > f(-2)$ ; $f(-4)$ against $f(6)$ is not decided by the signs.

a) 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.

b) 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.

c) 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.

d) 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.

e) $f$ 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.

Exercise 5: Trigonometric functions on an interval: factor cos x, never divide by it

With trigonometric functions the algebra that costs marks is the identity used too late and the factor divided away. $\sin 2x = 2\sin x\cos x$ and $\cos 2x = 1 - 2\sin^2 x = 2\cos^2 x - 1$ turn $f'$ into a product; then EACH factor is set to zero on the interval.

The figure shows $f(x) = \cos 2x + 2\sin x$ on $[0, 2\pi]$ . The endpoints belong to the interval, so they can carry local extrema in Thomas's sense.

a) Show that $f'(x) = 2\cos x\,(1 - 2\sin x)$ and find the critical points of $f$ in $(0, 2\pi)$ .
b) Make the sign chart of $f'$ on $[0, 2\pi]$ and give all the local extreme values of $f$ , endpoints included, with the points where they occur.
c) A student writes $-4\sin x\cos x + 2\cos x = 0$ , divides by $2\cos x$ , gets $\sin x = \frac{1}{2}$ , and concludes that $f$ has exactly two critical points. What is lost?
d) Let $h(x) = 2x - \sin 2x$ on $[0, 2\pi]$ . Show that $h'(x) = 4\sin^2 x$ , and deduce that $h$ is increasing on $[0, 2\pi]$ although $h'$ vanishes inside the interval.

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Answers

a) $\frac{\pi}{6}$ , $\frac{\pi}{2}$ , $\frac{5\pi}{6}$ , $\frac{3\pi}{2}$
b) Local max $\frac{3}{2}$ at $\frac{\pi}{6}$ and $\frac{5\pi}{6}$ , and $1$ at the endpoint $2\pi$ ; local min $1$ at $0$ and at $\frac{\pi}{2}$ , $-3$ at $\frac{3\pi}{2}$
c) The zeros of $\cos x$ , $\frac{\pi}{2}$ and $\frac{3\pi}{2}$ , are lost, among them the lowest point $f\left(\frac{3\pi}{2}\right) = -3$ .
d) $h'(x) = 2 - 2\cos 2x = 4\sin^2 x \ge 0$ , zero only at $\pi$ inside: $h$ is increasing on $[0, 2\pi]$ , no local extremum inside

a) By the chain rule, $(\cos 2x)' = -2\sin 2x$ , so $f'(x) = -2\sin 2x + 2\cos x$ . The identity is used NOW, on the derivative: $\sin 2x = 2\sin x\cos x$ , so $f'(x) = -4\sin x\cos x + 2\cos x$ . Both terms contain $2\cos x$ : $f'(x) = 2\cos x\,(1 - 2\sin x)$ . It is zero when $\cos x = 0$ , that is $x = \frac{\pi}{2}$ or $\frac{3\pi}{2}$ , or when $\sin x = \frac{1}{2}$ , that is $x = \frac{\pi}{6}$ or $\frac{5\pi}{6}$ . $f'$ exists everywhere: these four numbers are all the critical points in $(0, 2\pi)$ .

b) Sign of each factor. $\cos x > 0$ on $\left[0, \frac{\pi}{2}\right)$ and $\left(\frac{3\pi}{2}, 2\pi\right]$ , $< 0$ between. $1 - 2\sin x > 0$ when $\sin x < \frac{1}{2}$ , that is on $\left[0, \frac{\pi}{6}\right)$ and $\left(\frac{5\pi}{6}, 2\pi\right]$ . Product: $+$ on $\left(0, \frac{\pi}{6}\right)$ , $-$ on $\left(\frac{\pi}{6}, \frac{\pi}{2}\right)$ , $+$ on $\left(\frac{\pi}{2}, \frac{5\pi}{6}\right)$ , $-$ on $\left(\frac{5\pi}{6}, \frac{3\pi}{2}\right)$ , $+$ on $\left(\frac{3\pi}{2}, 2\pi\right)$ . First Derivative Test: local maxima $f\left(\frac{\pi}{6}\right) = \cos\frac{\pi}{3} + 2 \cdot \frac{1}{2} = \frac{3}{2}$ and $f\left(\frac{5\pi}{6}\right) = \cos\frac{5\pi}{3} + 1 = \frac{3}{2}$ ; local minima $f\left(\frac{\pi}{2}\right) = \cos\pi + 2 = 1$ and $f\left(\frac{3\pi}{2}\right) = \cos 3\pi - 2 = -3$ . Endpoints: $f$ increases right after $0$ , so local minimum $f(0) = 1$ ; $f$ increases up to $2\pi$ , so local maximum $f(2\pi) = 1$ . The figure confirms the alternation max, min, max, min between the endpoints.

c) Dividing by $2\cos x$ is only allowed where $\cos x \ne 0$ , and precisely the solutions of $\cos x = 0$ are solutions of the equation: $\frac{\pi}{2}$ and $\frac{3\pi}{2}$ disappear. The student then finds two maxima and no minimum inside the interval, which the figure refutes at once, and misses $f\left(\frac{3\pi}{2}\right) = -3$ , the lowest point of the graph. Correct gesture: factor, $2\cos x\,(1 - 2\sin x) = 0$ , then “ $\cos x = 0$ OR $\sin x = \frac{1}{2}$ ”, each factor on the interval.

d) $h'(x) = 2 - 2\cos 2x = 2(1 - \cos 2x)$ , and with $\cos 2x = 1 - 2\sin^2 x$ : $h'(x) = 2 \cdot 2\sin^2 x = 4\sin^2 x$ . It is a square: $h'(x) \ge 0$ everywhere, and in $(0, 2\pi)$ it is zero only at $x = \pi$ . So $h' > 0$ on $(0, \pi)$ and on $(\pi, 2\pi)$ : by the corollary $h$ is increasing on $[0, \pi]$ and on $[\pi, 2\pi]$ , hence on $[0, 2\pi]$ . The critical point $\pi$ is not an extremum, since the sign does not change; the only local extrema are at the endpoints, a local minimum $h(0) = 0$ and a local maximum $h(2\pi) = 4\pi$ . A derivative that vanishes at isolated points does not stop a function from increasing.

Exercise 6: How many solutions? Monotonic pieces, then the Intermediate Value Theorem

On an interval where $f$ is continuous and strictly monotonic, $f$ takes each value AT MOST once: two different points give two different values. The Intermediate Value Theorem then gives AT LEAST one solution of $f(x) = k$ when $k$ lies between the values of $f$ at the ends of the piece. Together: exactly one solution per piece whose range contains $k$ .

The figure shows $p(x) = 2x^3 + 3x^2 - 36x$ and a horizontal line $y = k$ . The number of solutions of $p(x) = k$ is the number of times the line meets the curve, and the point is to count them without trusting the drawing.

a) Find the intervals on which $p$ is increasing and decreasing, and its local extreme values.
b) Give $\lim_{x \to -\infty} p(x)$ and $\lim_{x \to \infty} p(x)$ , then the set of values taken by $p$ on each of the three intervals of a).
c) According to the value of the constant $k$ , give the number of real solutions of $p(x) = k$ .
d) Solve $p(x) = 0$ exactly, and check that the three roots fall one in each interval of a), as c) predicts.
e) For which values of $k$ does $p(x) = k$ have exactly one solution in $[-3, 2]$ ? Justify.

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a) $p'(x) = 6(x + 3)(x - 2)$ : increasing on $(-\infty, -3]$ and $[2, \infty)$ , decreasing on $[-3, 2]$ ; local max $p(-3) = 81$ , local min $p(2) = -44$
b) $-\infty$ and $+\infty$ ; values $(-\infty, 81]$ , then $[-44, 81]$ , then $[-44, \infty)$
c) $k > 81$ or $k < -44$ : one solution; $k = 81$ or $k = -44$ : two; $-44 < k < 81$ : three
d) $x = 0$ and $x = \frac{-3 \pm 3\sqrt{33}}{4}$ , about $-5.06$ and $3.56$
e) Exactly when $-44 \le k \le 81$

a) $p'(x) = 6x^2 + 6x - 36 = 6(x^2 + x - 6) = 6(x + 3)(x - 2)$ , common factor $6$ taken out first. Positive outside $[-3, 2]$ , negative inside. So $p$ is increasing on $(-\infty, -3]$ , decreasing on $[-3, 2]$ , increasing on $[2, \infty)$ . Local maximum $p(-3) = -54 + 27 + 108 = 81$ , local minimum $p(2) = 16 + 12 - 72 = -44$ . The substitution is where the marks go: $2(-3)^3 = -54$ , and $-36(-3) = +108$ , two signs to keep.

b) The leading term $2x^3$ dominates (chapter 5): $p(x) \to -\infty$ as $x \to -\infty$ and $p(x) \to +\infty$ as $x \to \infty$ . On $(-\infty, -3]$ , $p$ is continuous and increasing from $-\infty$ up to $81$ : by the Intermediate Value Theorem it takes every value of $(-\infty, 81]$ . On $[-3, 2]$ it decreases from $81$ to $-44$ : it takes every value of $[-44, 81]$ . On $[2, \infty)$ it increases from $-44$ to $+\infty$ : values $[-44, \infty)$ . On each piece, strict monotonicity says that each of these values is taken exactly once.

c) Count the pieces whose set of values contains $k$ . If $k > 81$ : only the third piece, one solution. If $k = 81$ : the first piece (at $x = -3$ ), the second (the same point $x = -3$ !) and the third. Careful, $x = -3$ belongs to the first two pieces, so it counts once: two solutions, $x = -3$ and one in $(2, \infty)$ . If $-44 < k < 81$ : one solution in each piece, and none at a shared endpoint, so three. If $k = -44$ : $x = 2$ (shared by the last two pieces) and one in $(-\infty, -3)$ , two solutions. If $k < -44$ : first piece only, one solution. The shared endpoints are the trap: counting a solution twice gives “three” at $k = 81$ .

d) $p(x) = x(2x^2 + 3x - 36)$ , and the quadratic formula gives $x = \frac{-3 \pm \sqrt{9 + 288}}{4} = \frac{-3 \pm \sqrt{297}}{4} = \frac{-3 \pm 3\sqrt{33}}{4}$ , since $297 = 9 \cdot 33$ . With the calculator, $\sqrt{33} \approx 5.745$ : the roots are about $-5.06$ , $0$ and $3.56$ . The value $k = 0$ lies in $(-44, 81)$ , so c) announces three solutions, one per piece: indeed $-5.06 < -3$ , $-3 < 0 < 2$ and $3.56 > 2$ .

e) On $[-3, 2]$ , $p$ is continuous and strictly decreasing from $p(-3) = 81$ to $p(2) = -44$ . If $-44 \le k \le 81$ , the Intermediate Value Theorem gives at least one solution in $[-3, 2]$ , and strict monotonicity at most one: exactly one. If $k > 81$ or $k < -44$ , $k$ is not between the values taken on $[-3, 2]$ , so there is none. Answer: exactly when $-44 \le k \le 81$ . The Intermediate Value Theorem alone would only have given “at least one”; the sign of $p'$ is what turns it into “exactly one”.

Exercise 7: Monotonicity as a tool: one-to-one functions and inequalities

The corollary is a tool, not only a description. A function with $f' > 0$ on an interval is increasing there, hence one-to-one, hence invertible (chapter 2). And to prove $F(x) > G(x)$ for $x > a$ , study the difference $d = F - G$ : if $d$ is continuous on $[a, \infty)$ , $d' > 0$ on $(a, \infty)$ and $d(a) = 0$ , then $d(x) > d(a) = 0$ for $x > a$ .

The corollary needs continuity on the CLOSED interval, starting value included: the value $d(a)$ is half of the proof.

a) Let $f(x) = x^3 + 3x - 2$ . Show that $f$ is increasing on $(-\infty, \infty)$ , hence has an inverse, and compute $\left(f^{-1}\right)'(2)$ .
b) Prove that $\ln(1 + x) > \frac{x}{1 + x}$ for every $x > 0$ .
c) Prove that $2\sqrt x > 3 - \frac{1}{x}$ for every $x > 1$ .
d) A student proves b) in one line: “ $g'(x) > 0$ for $x > 0$ , so $g(x) > 0$ for $x > 0$ .” What is missing? Show with an example that the line proves nothing.

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a) $f'(x) = 3x^2 + 3 > 0$ : increasing, one-to-one; $f(1) = 2$ , $\left(f^{-1}\right)'(2) = \frac{1}{f'(1)} = \frac{1}{6}$
b) $g(x) = \ln(1 + x) - \frac{x}{1 + x}$ , $g'(x) = \frac{x}{(1 + x)^2} > 0$ on $(0, \infty)$ , $g(0) = 0$
c) $d(x) = 2\sqrt x + \frac{1}{x} - 3$ , $d'(x) = \frac{x^{3/2} - 1}{x^2} > 0$ for $x > 1$ , $d(1) = 0$
d) The starting value $g(0) = 0$ and the continuity of $g$ at $0$ ; $g - 1$ has the same derivative and $g(x) - 1 < 0$ for small $x > 0$ .

a) $f'(x) = 3x^2 + 3 = 3(x^2 + 1)$ , and $x^2 + 1 \ge 1$ : $f'(x) > 0$ for every $x$ . By the corollary, $f$ is increasing on every interval $[a, b]$ , hence on the whole line; so two different numbers have different images, $f$ is one-to-one and has an inverse $f^{-1}$ . To differentiate the inverse at $2$ (chapter 12), first find which $x$ has $f(x) = 2$ : $x^3 + 3x - 2 = 2$ gives $x^3 + 3x - 4 = 0$ , with the obvious root $x = 1$ (and no other, $f$ being one-to-one). So $f^{-1}(2) = 1$ and $\left(f^{-1}\right)'(2) = \frac{1}{f'(1)} = \frac{1}{6}$ . Without the monotonicity, “the” $x$ with $f(x) = 2$ would not be known to be unique.

b) Let $g(x) = \ln(1 + x) - \frac{x}{1 + x}$ on $[0, \infty)$ , continuous there. The derivative of $\frac{x}{1 + x}$ is $\frac{(1 + x) - x}{(1 + x)^2} = \frac{1}{(1 + x)^2}$ . So $g'(x) = \frac{1}{1 + x} - \frac{1}{(1 + x)^2}$ , and the gesture is the common denominator $(1 + x)^2$ : $g'(x) = \frac{(1 + x) - 1}{(1 + x)^2} = \frac{x}{(1 + x)^2}$ , which is $> 0$ for $x > 0$ . By the corollary, $g$ is increasing on $[0, \infty)$ , and for $x > 0$ , $g(x) > g(0) = \ln 1 - 0 = 0$ . That is the inequality. Left as two separate fractions, the sign of $g'$ is invisible.

c) Let $d(x) = 2\sqrt x + \frac{1}{x} - 3$ on $[1, \infty)$ , continuous there, with $d(1) = 2 + 1 - 3 = 0$ . $d'(x) = \frac{1}{\sqrt x} - \frac{1}{x^2}$ . Common denominator $x^2$ , using $\frac{1}{\sqrt x} = \frac{x^{3/2}}{x^2}$ (since $x^{3/2} \cdot x^{-2} = x^{-1/2}$ ): $d'(x) = \frac{x^{3/2} - 1}{x^2}$ . For $x > 1$ , $x^{3/2} > 1$ , so $d'(x) > 0$ . By the corollary, $d$ is increasing on $[1, \infty)$ , and $d(x) > d(1) = 0$ for $x > 1$ : $2\sqrt x + \frac{1}{x} > 3$ , which is the inequality. The exponent rule $x^{3/2} \cdot x^{-2} = x^{-1/2}$ is exactly the kind of step the department's algebra tutorials exist for: check it before relying on it.

d) A positive derivative says that $g$ goes UP, not that it is positive: an increasing function can be negative everywhere. Take $k(x) = g(x) - 1$ : $k'(x) = g'(x) > 0$ on $(0, \infty)$ , yet $k(x)$ is close to $g(0) - 1 = -1$ for small $x > 0$ , so $k(x) < 0$ there. The one-line proof applies word for word to $k$ and “proves” a false statement. What is missing: $g$ is continuous on $[0, \infty)$ (so the corollary covers the closed interval, point $0$ included), and $g(0) = 0$ . Then $g(x) > g(0) = 0$ . On a paper, the value at the starting point is written explicitly, or the proof earns part marks at best.

Exercise 8: Five statements to correct

Each statement below was written on a MATH 203 assignment, and each is false as written. Say what is wrong, give a counterexample, and write a correct statement.

a) If $f$ is differentiable and increasing on $(a, b)$ , then $f'(x) > 0$ for every $x$ in $(a, b)$ .
b) If $f'$ changes from positive to negative at $c$ , then $f$ has a local maximum at $c$ .
c) If $f$ is increasing on an interval $I$ and $f(x) \ne 0$ on $I$ , then $\frac{1}{f}$ is increasing on $I$ .
d) If $f'(x) = \frac{x + 1}{x - 2}$ , then $f'(x) > 0$ exactly when $x + 1 > 0$ , so $f$ is increasing on $(-1, \infty)$ .
e) If $f'(x) > 0$ for every $x$ in $(0, 1)$ , then $f(0) < f(1)$ .

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a) False: $(x - 1)^3$ is increasing on $(0, 2)$ with $f'(1) = 0$ . True: $f'(x) \ge 0$ on $(a, b)$ .
b) False: $\frac{1}{x^2}$ at $0$ , not in the domain. True if $c$ is in the domain and $f$ is continuous at $c$ .
c) False: $f(x) = x$ on $(0, \infty)$ , $\frac{1}{x}$ decreases. True: $\frac{1}{f}$ is DEcreasing, $\left(\frac{1}{f}\right)' = -\frac{f'}{f^2}$ .
d) False: $f'(0) = -\frac{1}{2}$ . True: $f' > 0$ on $(-\infty, -1)$ and on $(2, \infty)$ .
e) False without continuity at the ends: $f(x) = x$ on $[0, 1)$ , $f(1) = -1$ . True if $f$ is continuous on $[0, 1]$ .

a) FALSE. $f(x) = (x - 1)^3$ is increasing on $(0, 2)$ : $f'(x) = 3(x - 1)^2 > 0$ except at $1$ , and a derivative zero at one isolated point does not stop the increase. Yet $f'(1) = 0$ , not $> 0$ . The corollary goes in one direction only: $f' > 0$ implies increasing, not the converse. Correct statement: if $f$ is differentiable and increasing on $(a, b)$ , then $f'(x) \ge 0$ for every $x$ in $(a, b)$ , since every difference quotient $\frac{f(x + h) - f(x)}{h}$ is positive and a limit of positive numbers is $\ge 0$ .

b) FALSE. $f(x) = \frac{1}{x^2}$ has $f'(x) = -\frac{2}{x^3}$ , positive for $x < 0$ and negative for $x > 0$ : the sign goes from $+$ to $-$ at $0$ . But $0$ is not in the domain, and near $0$ the function grows without bound: no maximum. The First Derivative Test is stated at a critical point $c$ , a point OF THE DOMAIN where $f$ is continuous. Correct statement: if $c$ is in the domain, $f$ is continuous at $c$ and $f'$ changes from positive to negative at $c$ , then $f$ has a local maximum at $c$ . On a sign chart, a number outside the domain gets a double bar and no verdict.

c) FALSE. $f(x) = x$ is increasing and nonzero on $(0, \infty)$ , and $\frac{1}{f(x)} = \frac{1}{x}$ is decreasing there. If $f$ is differentiable, the derivative shows why: $\left(\frac{1}{f}\right)' = -\frac{f'}{f^2}$ , with $f^2 > 0$ , so $\frac{1}{f}$ has the sign OPPOSITE to $f'$ . Correct statement: if $f$ is increasing and never zero on an interval $I$ , then $\frac{1}{f}$ is decreasing on $I$ . The same holds without differentiability, since for $u < v$ with $f(u)$ and $f(v)$ of the same sign, $\frac{1}{f(u)} > \frac{1}{f(v)}$ ; on an interval, a continuous $f$ that is never zero keeps one sign.

d) FALSE. At $x = 0$ , $x + 1 = 1 > 0$ but $f'(0) = \frac{1}{-2} = -\frac{1}{2} < 0$ . The student multiplied the inequality $\frac{x + 1}{x - 2} > 0$ by $x - 2$ without knowing its sign; for $x < 2$ that multiplication reverses the inequality. The quotient is positive when numerator and denominator have the SAME sign: both positive for $x > 2$ , both negative for $x < -1$ . Correct statement: $f' > 0$ on $(-\infty, -1)$ and on $(2, \infty)$ , $f' < 0$ on $(-1, 2)$ , so $f$ is increasing on $(-\infty, -1]$ and on $(2, \infty)$ , separately. An inequality is multiplied only by a quantity of KNOWN sign; otherwise, a sign chart.

e) FALSE. Let $f(x) = x$ for $0 \le x < 1$ and $f(1) = -1$ . Then $f'(x) = 1 > 0$ on $(0, 1)$ , yet $f(1) = -1 < 0 = f(0)$ . The corollary asks that $f$ be continuous on the CLOSED interval $[0, 1]$ , and this $f$ jumps at $1$ . Correct statement: if $f$ is continuous on $[0, 1]$ and $f'(x) > 0$ on $(0, 1)$ , then $f$ is increasing on $[0, 1]$ , so $f(0) < f(1)$ . Differentiability at the endpoints is NOT needed; continuity there is.

Exercise 9: Weekly sales after a launch: when do they rise, when do they fall?

A studio releases a video game. Its weekly sales, in thousands of copies per week, $t$ weeks after the launch, are modelled by $S(t) = \frac{720t}{t^2 + 36}$ for $t \ge 0$ . The figure shows the graph of $S$ .

A sales report is written in words: “sales rise during the first ... weeks, peak at ... thousand copies per week, then fall.” The sign of $S'$ writes that sentence, and the algebra of the quotient rule decides whether it is right.

a) Show that $S'(t) = \frac{720(6 - t)(6 + t)}{(t^2 + 36)^2}$ .
b) Write the sales report: when do weekly sales rise, when do they fall, which week is best, and what are the best weekly sales?
c) Compute $S(3)$ and $S(12)$ . The marketing team says: sales were the same in week $3$ and in week $12$ , so nothing happened in between. Correct them.
d) During which period are weekly sales at least $36$ thousand copies? How many weeks is that?
e) Compute $S'(3)$ and $S'(12)$ , with units, and interpret each in one sentence.

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Answers

a) Numerator $720(t^2 + 36) - 720t \cdot 2t = 720(36 - t^2)$
b) Rise on $[0, 6]$ , fall for $t \ge 6$ ; best week $t = 6$ , $S(6) = 60$ thousand copies per week
c) $S(3) = S(12) = 48$ ; sales rose to the peak of $60$ at week $6$ , then fell back
d) $2 \le t \le 18$ , a period of $16$ weeks
e) $S'(3) = 9.6$ and $S'(12) = -2.4$ thousand copies per week, per week

a) Quotient rule with $u = 720t$ , $v = t^2 + 36$ : $S'(t) = \frac{720(t^2 + 36) - 720t \cdot 2t}{(t^2 + 36)^2}$ . Take out the common factor $720$ BEFORE expanding: the numerator is $720\left(t^2 + 36 - 2t^2\right) = 720(36 - t^2) = 720(6 - t)(6 + t)$ . Expanding first gives $720t^2 + 25920 - 1440t^2$ , the same thing with three times more chances of an arithmetic slip.

b) For $t \ge 0$ : $720 > 0$ , $6 + t > 0$ and $(t^2 + 36)^2 > 0$ , so the sign of $S'$ is the sign of $6 - t$ : positive for $0 \le t < 6$ , negative for $t > 6$ . By the corollary, $S$ is increasing on $[0, 6]$ and decreasing on $[6, \infty)$ , and by the First Derivative Test the local maximum is at $t = 6$ , with $S(6) = \frac{4320}{72} = 60$ . The report: weekly sales rise during the first six weeks, peak in week $6$ at $60$ thousand copies per week, then decline week after week. Since $S$ increases up to $6$ and decreases after, the peak is also the best week overall.

c) $S(3) = \frac{2160}{45} = 48$ and $S(12) = \frac{8640}{180} = 48$ : the same weekly sales. But $S$ is not constant between them: it increases on $[3, 6]$ up to $60$ , then decreases on $[6, 12]$ back to $48$ . Equal values at two times say nothing about what happens between; only the sign of $S'$ does. There is even a pattern: $S\left(\frac{36}{t}\right) = \frac{720 \cdot 36/t}{36^2/t^2 + 36} = \frac{720t}{36 + t^2} = S(t)$ , so every weekly level below the peak is reached twice, once on the way up (before week $6$ ) and once on the way down, at times whose product is $36$ .

d) Solve $\frac{720t}{t^2 + 36} \ge 36$ . Multiplying by $t^2 + 36$ is allowed here because it is POSITIVE, so the inequality keeps its direction: $720t \ge 36t^2 + 1296$ , divide by $36$ : $20t \ge t^2 + 36$ , that is $t^2 - 20t + 36 \le 0$ , $(t - 2)(t - 18) \le 0$ , so $2 \le t \le 18$ . Weekly sales are at least $36$ thousand from week $2$ to week $18$ , a period of $16$ weeks. Consistent with c): $2 \times 18 = 36$ , and the peak week $6$ lies inside.

e) $S'(3) = \frac{720(3)(9)}{45^2} = \frac{19440}{2025} = 9.6$ and $S'(12) = \frac{720(-6)(18)}{180^2} = \frac{-77760}{32400} = -2.4$ , in thousand copies per week, per week. In week $3$ , weekly sales are growing at about $9600$ copies per week each week; in week $12$ they are shrinking by about $2400$ copies per week each week. The same level of sales, $48$ thousand, hides two opposite movements: the sign of $S'$ is what the report must say.

Exercise 10: A final exam question: a logarithmic family and the parameter that makes the minimum touch zero

For a real constant $a$ , let $f_a(x) = x^2 - a\ln x$ on $(0, \infty)$ . The figure shows $f_a$ for $a = 2$ , $a = 2e$ and $a = 8$ , with the lowest point of each curve.

The algebra of the chapter in its logarithmic version: one fraction over $x$ , the domain $x > 0$ used to drop factors and reject roots, and the laws of logarithms to simplify the values.

a) For $a = 8$ : write $f_8'(x)$ as one factored fraction, find the intervals of increase and decrease and the local extreme value of $f_8$ , exactly.
b) Show that if $a \le 0$ , $f_a$ is increasing on $(0, \infty)$ and has no local extremum.
c) Let $a > 0$ . Show that $f_a$ has exactly one local extremum, a minimum at $x = \sqrt{\frac{a}{2}}$ , with value $m(a) = \frac{a}{2}\left(1 - \ln\frac{a}{2}\right)$ .
d) Find the value of $a > 0$ for which the minimum value is $0$ , and deduce an inequality valid for every $x > 0$ .
e) Study the function $m$ on $(0, \infty)$ with the First Derivative Test. For which $a$ is the minimum value $m(a)$ the largest, and what is it?

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Answers

a) $f_8'(x) = \frac{2(x - 2)(x + 2)}{x}$ : decreasing on $(0, 2]$ , increasing on $[2, \infty)$ ; local min $f_8(2) = 4 - 8\ln 2$
b) $f_a'(x) = \frac{2x^2 - a}{x} > 0$ on $(0, \infty)$ when $a \le 0$
c) $f_a'(x) = \frac{2\left(x - \sqrt{a/2}\right)\left(x + \sqrt{a/2}\right)}{x}$ , $-$ then $+$ ; $m(a) = \frac{a}{2} - \frac{a}{2}\ln\frac{a}{2}$
d) $a = 2e$ ; $x^2 \ge 2e\ln x$ for every $x > 0$ , with equality only at $x = \sqrt e$
e) $m'(a) = -\frac{1}{2}\ln\frac{a}{2}$ : $m$ increases on $(0, 2]$ , decreases on $[2, \infty)$ ; largest value $m(2) = 1$

a) $f_8'(x) = 2x - \frac{8}{x}$ . One fraction over $x$ : $f_8'(x) = \frac{2x^2 - 8}{x} = \frac{2(x^2 - 4)}{x} = \frac{2(x - 2)(x + 2)}{x}$ . On the domain $x > 0$ , both $x$ and $x + 2$ are positive: the sign of $f_8'$ is the sign of $x - 2$ . The root $-2$ is not in the domain and is rejected, not tested. So $f_8$ is decreasing on $(0, 2]$ and increasing on $[2, \infty)$ , with a local minimum $f_8(2) = 4 - 8\ln 2$ , about $-1.545$ , the lowest point of the blue curve.

b) $f_a'(x) = 2x - \frac{a}{x} = \frac{2x^2 - a}{x}$ . If $a \le 0$ , then $-a \ge 0$ and $2x^2 - a \ge 2x^2 > 0$ for $x > 0$ , while the denominator $x$ is positive: $f_a'(x) > 0$ on $(0, \infty)$ . By the corollary $f_a$ is increasing on every closed interval inside $(0, \infty)$ , hence on $(0, \infty)$ , and without a sign change there is no local extremum. For $a = 0$ , $f_0(x) = x^2$ on $(0, \infty)$ : its famous minimum at $0$ is outside the domain.

c) For $a > 0$ , $2x^2 - a = 2\left(x^2 - \frac{a}{2}\right) = 2\left(x - \sqrt{\frac{a}{2}}\right)\left(x + \sqrt{\frac{a}{2}}\right)$ . On $(0, \infty)$ the factors $x + \sqrt{a/2}$ and $x$ are positive, so $f_a'$ has the sign of $x - \sqrt{a/2}$ : negative, then positive. The only critical point in the domain is $\sqrt{a/2}$ , and the First Derivative Test gives a local minimum there, the only local extremum. Its value: $f_a\left(\sqrt{a/2}\right) = \frac{a}{2} - a\ln\sqrt{\frac{a}{2}}$ , and the law of logarithms $\ln\sqrt u = \frac{1}{2}\ln u$ turns it into $\frac{a}{2} - \frac{a}{2}\ln\frac{a}{2} = \frac{a}{2}\left(1 - \ln\frac{a}{2}\right)$ . Check with $a = 8$ : $4(1 - \ln 4) = 4 - 8\ln 2$ , as in a).

d) Since $a > 0$ , $m(a) = 0$ exactly when $1 - \ln\frac{a}{2} = 0$ , that is $\frac{a}{2} = e$ , $a = 2e$ . For this value, $f_{2e}$ is decreasing on $\left(0, \sqrt e\right]$ and increasing on $\left[\sqrt e, \infty\right)$ (since $\sqrt{a/2} = \sqrt e$ ), so $f_{2e}(x) \ge f_{2e}(\sqrt e) = 0$ for every $x > 0$ , with equality only at $\sqrt e$ : that is, $x^2 \ge 2e\ln x$ for every $x > 0$ . On the figure, the orange curve touches the axis at $\left(\sqrt e, 0\right)$ .

e) $m(a) = \frac{a}{2} - \frac{a}{2}\ln\frac{a}{2}$ . Product rule on the second term, with $\left(\ln\frac{a}{2}\right)' = \frac{1}{a}$ : $m'(a) = \frac{1}{2} - \left(\frac{1}{2}\ln\frac{a}{2} + \frac{a}{2} \cdot \frac{1}{a}\right) = -\frac{1}{2}\ln\frac{a}{2}$ . It is positive when $\frac{a}{2} < 1$ and negative when $\frac{a}{2} > 1$ : $m$ is increasing on $(0, 2]$ and decreasing on $[2, \infty)$ , so its largest value is $m(2) = 1 \cdot (1 - \ln 1) = 1$ . The green curve of the figure, $a = 2$ , has the highest lowest point of the whole family, $(1, 1)$ . The First Derivative Test has just been applied to a function of the PARAMETER, which is the same test on a different variable.

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

Course recap

Part A: the basics (/50)

Exercise 1: Polynomials: factor f' completely, then read the sign factor by factor

Exercise 2: Quotients: the minus sign of the quotient rule, and the holes of the domain

Exercise 3: Fractional powers and roots: factor out the LOWEST power

Exercise 4: Reading f on the graph of f', when f' jumps

Exercise 5: Trigonometric functions on an interval: factor cos x, never divide by it

Part B: problems and reasoning (/50)

Exercise 6: How many solutions? Monotonic pieces, then the Intermediate Value Theorem

Exercise 7: Monotonicity as a tool: one-to-one functions and inequalities

Exercise 8: Five statements to correct

Exercise 9: Weekly sales after a launch: when do they rise, when do they fall?

Exercise 10: A final exam question: a logarithmic family and the parameter that makes the minimum touch zero

See also

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