Exercise 1: Reading extrema on a graph with Thomas's definitions: endpoints included

Definitions of Thomas, section 4.1. $f$ has an absolute maximum value on its domain $D$ at $c$ if $f(x) \le f(c)$ for ALL $x$ in $D$ . $f$ has a local maximum value at $c$ if $f(x) \le f(c)$ for all $x$ in $D$ lying in some open interval containing $c$ . Because only the points of $D$ are compared, an ENDPOINT can carry a local extremum: at $x = a$ the comparison is made on a half-open interval $[a, a + \delta)$ . A critical point is an INTERIOR point of the domain where $f'$ is zero or undefined.

The figure shows a function $f$ , continuous on $[-3, 4]$ , with $f(-3) = 2$ , $f(-2) = 3$ , $f(0) = -1$ , $f(1) = 2$ , $f(2) = 3$ , $f(3) = 4$ , $f(4) = 2$ . It has a horizontal tangent at $x = -2$ , a cusp at $x = 0$ , a corner at $x = 1$ (slope $2$ on the left, $\frac{1}{3}$ on the right), a vertical tangent at $x = 2$ and a corner at $x = 3$ . It is differentiable everywhere else in $(-3, 4)$ .

a) Find the absolute maximum and minimum values of $f$ on $[-3, 4]$ , and where they occur.
b) With Thomas's definition, list the local maximum values and the local minimum values of $f$ , endpoints included.
c) List the critical points of $f$ . For each, say whether $f' = 0$ or $f'$ is undefined, and whether $f$ has a local extremum there. Are $-3$ and $4$ critical points?
d) Now restrict $f$ to $[-1, 2]$ . Find its absolute extrema. What is new about the point $x = 2$ ?
e) Now restrict $f$ to the open interval $(-3, 3)$ . Does it have an absolute maximum? An absolute minimum? Does this contradict the Extreme Value Theorem?

Type your answers, the page tells you right or wrong 0/14

a)

Absolute maximum value

It occurs at

x =

Absolute minimum value

It occurs at

x =

b)

Number of local maximum values

Number of local minimum values

c)

Number of critical points

Critical points with no extremum

Is

x = -3

a critical point?

d)

Absolute maximum value

It occurs at

x =

Absolute minimum value

e)

Absolute maximum on

(-3, 3)

Absolute minimum value

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Answers

a) Absolute max $4$ at $x = 3$ ; absolute min $-1$ at $x = 0$
b) Local max $3$ at $x = -2$ and $4$ at $x = 3$ ; local min $2$ at $x = -3$ , $-1$ at $x = 0$ , $2$ at $x = 4$
c) $-2$ ( $f' = 0$ ); $0$ , $1$ , $2$ , $3$ ( $f'$ undefined); no extremum at $1$ and $2$ . Endpoints are not critical points.
d) Absolute max $3$ at the endpoint $x = 2$ ; absolute min $-1$ at $x = 0$
e) No absolute max (values approach $4$ ); absolute min $-1$ at $x = 0$ . No contradiction: the interval is not closed.

a) The highest point of the whole graph is the corner $(3, 4)$ and the lowest is the cusp $(0, -1)$ . Absolute maximum value $4$ , at $x = 3$ ; absolute minimum value $-1$ , at $x = 0$ . Each answer has a VALUE and a PLACE: the question asks for both, and writing max at $4$ when $4$ is the value costs the location mark. Every other point of the graph lies strictly between $-1$ and $4$ .

b) Local maximum values: $f(-2) = 3$ (smooth top) and $f(3) = 4$ (corner). Local minimum values: $f(0) = -1$ (cusp), and the two ENDPOINTS, $f(-3) = 2$ and $f(4) = 2$ . At $x = -3$ the graph rises right after the endpoint, so $f(x) \ge 2$ on some $[-3, -3 + \delta)$ : in Thomas's sense this is a local minimum. The same holds at $x = 4$ , approached from the left on $(4 - \delta, 4]$ . So two local maxima and three local minima. Thomas also gives a consequence of his definition: an absolute extremum is always a local one, which is why $f(3) = 4$ and $f(0) = -1$ appear in both lists. Some other textbooks refuse local extrema at endpoints; in MATH 203 follow Thomas, and write the word endpoint next to such an answer.

c) The critical points are INTERIOR points of $[-3, 4]$ where $f' = 0$ or $f'$ is undefined. $f'(-2) = 0$ (horizontal tangent). $f'$ is undefined at $0$ (cusp), $1$ and $3$ (corners: two different one-sided slopes) and $2$ (vertical tangent, infinite slope). Critical points: $-2$ , $0$ , $1$ , $2$ , $3$ . Extrema at $-2$ (max), $0$ (min), $3$ (max). NONE at $1$ and $2$ : at both, the graph keeps rising through the point, so there are values below $f(c)$ on the left and above it on the right. Undefined $f'$ nominates a point exactly like $f' = 0$ does, and it can be a false alarm exactly like $f' = 0$ can. The endpoints $-3$ and $4$ are NOT critical points, since they are not interior; they still carry the local minima of b), and they are still candidates for absolute extrema.

d) On $[-1, 2]$ the candidates are the endpoints $-1$ and $2$ , and the critical points inside $(-1, 2)$ , namely $0$ and $1$ . Values: $f(-1) = 3 - 1 = 2$ (from the first piece, $3 - (x + 2)^2$ ), $f(0) = -1$ , $f(1) = 2$ , $f(2) = 3$ . Absolute maximum $3$ at $x = 2$ , absolute minimum $-1$ at $x = 0$ . The point $x = 2$ was a critical point WITHOUT extremum on $[-3, 4]$ ; on $[-1, 2]$ it is an endpoint, and it carries the maximum. Absolute is always relative to the domain: the graph did not change, the list of candidates did.

e) On $(-3, 3)$ the corner $(3, 4)$ is no longer in the domain. As $x \to 3^-$ the values climb toward $4$ , but every value with $-3 < x < 3$ is strictly less than $4$ , and every value below $4$ is beaten by a point closer to $3$ : NO absolute maximum. The absolute minimum $f(0) = -1$ survives, because $0$ is still in the domain. No contradiction: Thomas's Theorem 1 requires a CLOSED interval, and when a hypothesis fails the theorem simply says nothing.

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Exercise 2: Critical points hide in the algebra: write f' as one factored fraction

Thomas's definition: a critical point of $f$ is an interior point of the domain of $f$ where $f' = 0$ or $f'$ is undefined. Both halves are read on the SAME object: $f'$ written as a single fraction, fully factored. The zeros of the numerator give $f' = 0$ ; the zeros of the denominator, where $f$ itself is defined, give $f'$ undefined; the domain of $f$ filters both.

The algebra is where the marks go: a negative exponent left as it is, a square root not put over a common denominator, a factor divided out of an equation. For each function, give the domain, rewrite $f'$ as one factored fraction, and list the critical points.

a) $f(x) = x^{4/5}(x - 9)$ .
b) $g(x) = x^2 + \frac{16}{x}$ .
c) $h(x) = (x + 2)\sqrt{3 - x}$ .
d) $k(x) = \sin^2 x + \cos x$ , critical points in the open interval $(0, 2\pi)$ .
e) $m(x) = x^2 \ln x$ .

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Answers

a) $f'(x) = \frac{9(x - 4)}{5x^{1/5}}$ ; critical points $0$ ( $f'$ undefined) and $4$
b) $g'(x) = \frac{2(x - 2)(x^2 + 2x + 4)}{x^2}$ ; critical point $2$ only ( $0$ not in the domain)
c) $h'(x) = \frac{4 - 3x}{2\sqrt{3 - x}}$ ; critical point $\frac{4}{3}$ only ( $3$ is an endpoint of the domain)
d) $k'(x) = \sin x(2\cos x - 1)$ ; critical points $\frac{\pi}{3}$ , $\pi$ , $\frac{5\pi}{3}$
e) $m'(x) = x(2\ln x + 1)$ ; critical point $e^{-1/2} \approx 0.6065$ ( $0$ not in the domain)

a) Domain: all real numbers, since the fifth root is defined for every real number. Expand first: $f(x) = x^{9/5} - 9x^{4/5}$ , so $f'(x) = \frac{9}{5}x^{4/5} - \frac{36}{5}x^{-1/5}$ . The algebraic gesture: factor out the LOWEST power, $x^{-1/5}$ , even though it is negative: $f'(x) = \frac{9}{5}x^{-1/5}(x - 4) = \frac{9(x - 4)}{5x^{1/5}}$ . Check the factoring by distributing: $x^{-1/5} \cdot x = x^{4/5}$ . Numerator zero at $x = 4$ ; denominator zero at $x = 0$ , where $f(0) = 0$ is defined and $0$ is interior: $f'(0)$ is undefined and $0$ IS a critical point. Critical points: $0$ and $4$ . A student who sets $\frac{9}{5}x^{4/5} - \frac{36}{5}x^{-1/5} = 0$ and multiplies through by $x^{1/5}$ finds $x = 4$ and never sees $0$ .

b) Domain: $x \ne 0$ . Rewrite $\frac{16}{x} = 16x^{-1}$ before differentiating: $g'(x) = 2x - 16x^{-2} = 2x - \frac{16}{x^2}$ . Common denominator $x^2$ : $g'(x) = \frac{2x^3 - 16}{x^2} = \frac{2(x^3 - 8)}{x^2} = \frac{2(x - 2)(x^2 + 2x + 4)}{x^2}$ , by the difference of cubes. The quadratic factor has discriminant $4 - 16 < 0$ , so it never vanishes. Numerator zero: $x = 2$ . Denominator zero: $x = 0$ , NOT in the domain of $g$ , so not a critical point. Critical point: $2$ only. The frequent error is $-16x^{-2}$ written as $16x^{-2}$ or as $-\frac{16}{x}$ : the power rule on $x^{-1}$ gives $-x^{-2}$ .

c) Domain: $3 - x \ge 0$ , that is $(-\infty, 3]$ . Product rule and chain rule (inner function $3 - x$ ): $h'(x) = \sqrt{3 - x} + (x + 2) \cdot \frac{-1}{2\sqrt{3 - x}}$ . Common denominator $2\sqrt{3 - x}$ : multiply the first term by $\frac{2\sqrt{3 - x}}{2\sqrt{3 - x}}$ , which turns $\sqrt{3 - x}$ into $\frac{2(3 - x)}{2\sqrt{3 - x}}$ . So $h'(x) = \frac{2(3 - x) - (x + 2)}{2\sqrt{3 - x}} = \frac{4 - 3x}{2\sqrt{3 - x}}$ for $x < 3$ . Numerator zero: $x = \frac{4}{3}$ , interior. Denominator zero: $x = 3$ , where $h(3) = 0$ is defined, but $3$ is an ENDPOINT of the domain, not an interior point: with Thomas's definition it is not a critical point. Critical point: $\frac{4}{3}$ only. The point $3$ is not forgotten for all that: on any closed interval ending at $3$ it is a candidate as an endpoint.

d) Chain rule on $\sin^2 x = (\sin x)^2$ : $k'(x) = 2\sin x\cos x - \sin x$ . Factor $\sin x$ : $k'(x) = \sin x(2\cos x - 1)$ . It exists everywhere, so only the zeros count. $\sin x = 0$ in $(0, 2\pi)$ : $x = \pi$ ( $0$ and $2\pi$ are excluded). $\cos x = \frac{1}{2}$ in $(0, 2\pi)$ : $x = \frac{\pi}{3}$ and $x = \frac{5\pi}{3}$ . Critical points: $\frac{\pi}{3}$ , $\pi$ , $\frac{5\pi}{3}$ . The costly move is to write $2\sin x\cos x = \sin x$ and DIVIDE by $\sin x$ : it leaves $\cos x = \frac{1}{2}$ and silently erases $x = \pi$ . Never divide an equation by a factor that can be zero; factor it out and set each factor to zero.

e) Domain: $x > 0$ . Product rule: $m'(x) = 2x\ln x + x^2 \cdot \frac{1}{x} = 2x\ln x + x = x(2\ln x + 1)$ . The factor $x$ vanishes at $0$ , which is NOT in the domain: rejected. $2\ln x + 1 = 0$ gives $\ln x = -\frac{1}{2}$ , so $x = e^{-1/2} = \frac{1}{\sqrt e} \approx 0.6065$ . Critical point: $e^{-1/2}$ only. Write the exact value first; the calculator gives the decimal only after the equation is solved by hand.

Exercise 3: The Extreme Value Theorem: check the hypotheses, then see what survives

Thomas, Theorem 1 (the Extreme Value Theorem). If $f$ is continuous on a closed interval $[a, b]$ , then $f$ attains both an absolute maximum value $M$ and an absolute minimum value $m$ in $[a, b]$ .

Two hypotheses, CONTINUOUS and FINITE CLOSED interval, and one conclusion: the extrema EXIST. The theorem does not say where they are, and when a hypothesis fails it says nothing at all. For each case, name the hypothesis that fails (if any), then find the absolute extrema that exist. The figure shows the function $k$ of part c).

a) $f(x) = \sqrt{x}$ on $[0, 4]$ .
b) $f(x) = \tan x$ on $\left[-\frac{\pi}{4}, \frac{\pi}{2}\right)$ .
c) $k(x) = \frac{4 - x^2}{2 - x}$ for $0 \le x < 2$ , and $k(2) = 1$ , on $[0, 2]$ .
d) $f(x) = \frac{6x}{x^2 + 9}$ on $(-\infty, \infty)$ . Hint: compute $1 - f(x)$ as one fraction.
e) Without computing anything, does $g(x) = x^5 - 3x^3 + e^x\cos x$ have an absolute maximum on $[-2, 3]$ ?

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Answers

a) Hypotheses hold; max $2$ at $x = 4$ , min $0$ at $x = 0$
b) Not closed; no absolute max, absolute min $-1$ at $x = -\frac{\pi}{4}$
c) Not continuous at $2$ ; no absolute max, absolute min $1$ at $x = 2$
d) Not a finite closed interval, yet max $1$ at $x = 3$ and min $-1$ at $x = -3$
e) Yes: $g$ is continuous on the closed interval $[-2, 3]$ , so the EVT guarantees it.

a) $\sqrt x$ is continuous on its domain $[0, \infty)$ , hence on $[0, 4]$ (at $0$ , continuity from the right is what the closed interval requires). The EVT applies. $f'(x) = \frac{1}{2\sqrt x}$ is never $0$ , and it is undefined only at $0$ , an endpoint: no critical point in $(0, 4)$ . Candidates: the endpoints. $f(0) = 0$ , $f(4) = 2$ . Absolute maximum $2$ at $x = 4$ , absolute minimum $0$ at $x = 0$ . The vertical tangent at $0$ does not make $0$ a critical point, since it is not interior, but $0$ is on the list anyway as an endpoint.

b) $\tan x = \frac{\sin x}{\cos x}$ is continuous wherever $\cos x \ne 0$ , which holds on $\left[-\frac{\pi}{4}, \frac{\pi}{2}\right)$ ; but the interval is NOT closed, the right end $\frac{\pi}{2}$ is missing. The EVT does not apply. As $x \to \frac{\pi}{2}^-$ , $\sin x \to 1$ and $\cos x \to 0^+$ , so $\tan x \to +\infty$ : the values are unbounded above, NO absolute maximum. The tangent is increasing on $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ (a property of its graph, section 1.3), so its smallest value on the interval is at the left end: absolute minimum $\tan\left(-\frac{\pi}{4}\right) = -1$ , at $x = -\frac{\pi}{4}$ .

c) First the algebra: for $x \ne 2$ , $\frac{4 - x^2}{2 - x} = \frac{(2 - x)(2 + x)}{2 - x} = 2 + x$ . So on $[0, 2)$ , $k(x) = 2 + x$ , whose values fill $[2, 4)$ , and $k(2) = 1$ . The interval $[0, 2]$ is closed, but $\lim_{x\to 2^-} k(x) = 4 \ne 1 = k(2)$ : $k$ is NOT continuous at $2$ , and the EVT does not apply. The value $4$ is approached (hollow dot on the figure) and never taken: NO absolute maximum. The lowest value is $k(2) = 1$ , below every value $2 + x \ge 2$ : absolute minimum $1$ at $x = 2$ (full dot). Without the factorization, the formula looks undefined at $2$ and the whole question is misread.

d) $f$ is continuous on $(-\infty, \infty)$ , since $x^2 + 9 > 0$ , but the interval is not a finite closed interval: the EVT is silent. Follow the hint: $1 - \frac{6x}{x^2 + 9} = \frac{x^2 - 6x + 9}{x^2 + 9} = \frac{(x - 3)^2}{x^2 + 9} \ge 0$ , with equality only at $x = 3$ . So $f(x) \le 1 = f(3)$ for every $x$ : absolute maximum $1$ at $x = 3$ . In the same way, $f(x) + 1 = \frac{x^2 + 6x + 9}{x^2 + 9} = \frac{(x + 3)^2}{x^2 + 9} \ge 0$ : absolute minimum $-1$ at $x = -3$ . The algebraic gesture is to recognize the perfect square $x^2 - 6x + 9$ in the numerator. Both extrema exist although the theorem does not apply: its hypotheses are SUFFICIENT, not necessary.

e) Yes. $x^5 - 3x^3$ is a polynomial, $e^x$ and $\cos x$ are continuous everywhere, and sums and products of continuous functions are continuous: $g$ is continuous on the closed interval $[-2, 3]$ . By the Extreme Value Theorem, $g$ attains an absolute maximum (and an absolute minimum) there. The theorem is an EXISTENCE result: it answers this question in one sentence, without a derivative, and without saying where the maximum is.

Exercise 4: The closed interval method on polynomials and a quotient, and a calculator that lies

Thomas's method for the absolute extrema of a function $f$ CONTINUOUS on a finite CLOSED interval: evaluate $f$ at all critical points and endpoints, then take the largest and smallest of these values.

Write the method in four lines: (1) continuity on $[a, b]$ , with its reason; (2) $f'$ factored, critical points in $(a, b)$ , the others written rejected; (3) a table of values of $f$ , not of $f'$ ; (4) the conclusion, value AND location.

a) $f(x) = 2x^3 - 9x^2 + 12x - 1$ on $[0, 3]$ .
b) $g(x) = x^4 - 8x^2 + 3$ on $[-1, 3]$ .
c) $h(x) = \frac{x}{x^2 + 4}$ on $[0, 5]$ .
d) A student finds the absolute extrema of $p(x) = x^4 - 2x^2$ on $[-2, 1]$ . She gets $p'(x) = 4x(x - 1)(x + 1)$ , lists the candidates $-2$ , $-1$ , $0$ , $1$ , types each value on her calculator and gets $p(-2) = -24$ , $p(-1) = -1$ , $p(0) = 0$ , $p(1) = -1$ . She concludes: absolute maximum $0$ at $x = 0$ , absolute minimum $-24$ at $x = -2$ . Find the error and give the correct answer.

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Answers

a) Max $8$ at $x = 3$ ; min $-1$ at $x = 0$
b) Max $12$ at $x = 3$ ; min $-13$ at $x = 2$ ( $x = -2$ rejected)
c) Max $\frac{1}{4}$ at $x = 2$ ; min $0$ at $x = 0$ ( $x = -2$ rejected)
d) $-2^4$ was read as $-(2^4)$ : $p(-2) = 16 - 8 = 8$ . Max $8$ at $x = -2$ ; min $-1$ at $x = -1$ and $x = 1$

a) $f$ is a polynomial, continuous on $[0, 3]$ . $f'(x) = 6x^2 - 18x + 12 = 6(x^2 - 3x + 2) = 6(x - 1)(x - 2)$ : critical points $1$ and $2$ , both in $(0, 3)$ . Values: $f(0) = -1$ , $f(1) = 2 - 9 + 12 - 1 = 4$ , $f(2) = 16 - 36 + 24 - 1 = 3$ , $f(3) = 54 - 81 + 36 - 1 = 8$ . Absolute maximum $8$ at $x = 3$ , absolute minimum $-1$ at $x = 0$ . Both critical points lose to the endpoints: $f(1) = 4$ is a local maximum and $f(2) = 3$ a local minimum, neither absolute. The derivative only fills the list; the values decide.

b) $g$ is a polynomial, continuous on $[-1, 3]$ . $g'(x) = 4x^3 - 16x = 4x(x^2 - 4) = 4x(x - 2)(x + 2)$ . Zeros $0$ , $2$ , $-2$ ; the last is outside $(-1, 3)$ : REJECTED. Values: $g(-1) = 1 - 8 + 3 = -4$ , $g(0) = 3$ , $g(2) = 16 - 32 + 3 = -13$ , $g(3) = 81 - 72 + 3 = 12$ . Absolute maximum $12$ at $x = 3$ , absolute minimum $-13$ at $x = 2$ . Note that $g(-2) = -13$ too: keeping $-2$ gives the right minimum VALUE at a location that is not in the interval, and costs the location mark. The factoring gesture: take out the common factor $4x$ first, then the difference of squares; dividing $4x^3 = 16x$ by $x$ loses the critical point $0$ .

c) The denominator $x^2 + 4$ never vanishes, so $h$ is continuous on $[0, 5]$ . Quotient rule: $h'(x) = \frac{1 \cdot (x^2 + 4) - x \cdot 2x}{(x^2 + 4)^2} = \frac{4 - x^2}{(x^2 + 4)^2} = \frac{(2 - x)(2 + x)}{(x^2 + 4)^2}$ . The sign trap is in the numerator: the minus applies to ALL of $x \cdot 2x$ , giving $x^2 + 4 - 2x^2 = 4 - x^2$ . Zeros $\pm 2$ ; keep $2$ , reject $-2$ . Values: $h(0) = 0$ , $h(2) = \frac{2}{8} = \frac{1}{4}$ , $h(5) = \frac{5}{29} \approx 0.1724$ . Absolute maximum $\frac{1}{4}$ at $x = 2$ , absolute minimum $0$ at $x = 0$ .

d) The derivative and the list of candidates are correct: $-1$ and $0$ are the critical points in $(-2, 1)$ , and $-2$ , $1$ are the endpoints. The error is in ONE value: the student typed -2^4 on the calculator, which computes $-(2^4) = -16$ , because the power is done before the negation. So she got $-16 - 2 \cdot 4 = -24$ . The correct value is $p(-2) = (-2)^4 - 2(-2)^2 = 16 - 8 = 8$ : type the brackets. Correct table: $p(-2) = 8$ , $p(-1) = 1 - 2 = -1$ , $p(0) = 0$ , $p(1) = -1$ . Absolute maximum $8$ at $x = -2$ ; absolute minimum $-1$ , attained at $x = -1$ AND $x = 1$ . A five-second check would have caught it: $p(x) = (x^2 - 1)^2 - 1 \ge -1$ for every $x$ , so a value of $-24$ is impossible.

Exercise 5: The closed interval method with roots and fractional exponents

With a root or a fractional exponent, the critical points where $f'$ is UNDEFINED appear, and they are exactly the ones a student who only solves $f'(x) = 0$ never finds. The algebraic gesture of this exercise: rewrite $f'$ as one fraction, by factoring out the lowest power or by a common denominator, before reading anything.

The calculator gesture: a scientific calculator often answers ERROR to a fractional power of a negative number, such as $(-4)^{2/3}$ . The number exists: compute it as $\left((-4)^2\right)^{1/3} = 16^{1/3}$ , or as $\left((-4)^{1/3}\right)^2$ . Give exact values, then decimals to 4 places.

a) $f(x) = x^{2/3}(5 - 2x)$ on $[-1, 2]$ .
b) $g(x) = x^2\sqrt{5 - x}$ on $[-1, 5]$ .
c) For the function $g$ of part b): is $x = 5$ a critical point of $g$ in Thomas's sense? Is it a candidate in the closed interval method? And $x = 0$ ?
d) $h(x) = \left(x^2 - 4\right)^{2/3}$ on $[-1, 3]$ .

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Answers

a) Max $7$ at $x = -1$ ; min $0$ at $x = 0$ (where $f'$ is undefined)
b) Max $16$ at $x = 4$ ; min $0$ at $x = 0$ and $x = 5$
c) $5$ : not a critical point (endpoint), but a candidate. $0$ : a critical point ( $g'(0) = 0$ ) and a candidate.
d) Max $5^{2/3} \approx 2.9240$ at $x = 3$ ; min $0$ at $x = 2$

a) $x^{2/3} = \left(x^{1/3}\right)^2$ is defined and continuous for every real $x$ , so $f$ is continuous on $[-1, 2]$ . Expand: $f(x) = 5x^{2/3} - 2x^{5/3}$ , so $f'(x) = \frac{10}{3}x^{-1/3} - \frac{10}{3}x^{2/3}$ . Factor out the lowest power $x^{-1/3}$ : $f'(x) = \frac{10}{3}x^{-1/3}(1 - x) = \frac{10(1 - x)}{3x^{1/3}}$ . Numerator zero at $x = 1$ ; denominator zero at $x = 0$ , where $f(0) = 0$ exists: $0$ is a critical point with $f'$ undefined (a cusp). Values: $f(-1) = (-1)^{2/3}(5 + 2) = 1 \cdot 7 = 7$ , since $(-1)^{2/3} = \left((-1)^{1/3}\right)^2 = (-1)^2 = 1$ ; $f(0) = 0$ ; $f(1) = 1 \cdot 3 = 3$ ; $f(2) = 2^{2/3} \cdot 1 \approx 1.5874$ . Absolute maximum $7$ at $x = -1$ , absolute minimum $0$ at $x = 0$ . Forget the cusp, and the minimum becomes $1.5874$ , which is false; trust a calculator ERROR at $x = -1$ , and the maximum disappears.

b) $5 - x \ge 0$ on $[-1, 5]$ , so $g$ is defined and continuous there. Product and chain rules: $g'(x) = 2x\sqrt{5 - x} + x^2 \cdot \frac{-1}{2\sqrt{5 - x}}$ . Common denominator $2\sqrt{5 - x}$ : $g'(x) = \frac{4x(5 - x) - x^2}{2\sqrt{5 - x}} = \frac{20x - 5x^2}{2\sqrt{5 - x}} = \frac{5x(4 - x)}{2\sqrt{5 - x}}$ for $x < 5$ . Zeros in $(-1, 5)$ : $x = 0$ and $x = 4$ . Values: $g(-1) = \sqrt 6 \approx 2.4495$ , $g(0) = 0$ , $g(4) = 16 \cdot 1 = 16$ , $g(5) = 25 \cdot 0 = 0$ . Absolute maximum $16$ at $x = 4$ ; absolute minimum $0$ , attained TWICE, at $x = 0$ and $x = 5$ . Report both places. The figure of the solution shows the four candidates.

c) $x = 5$ is an endpoint of $[-1, 5]$ and of the domain $(-\infty, 5]$ of $g$ . Thomas defines a critical point as an INTERIOR point where $f'$ is zero or undefined, so $5$ is NOT a critical point, although $g'(5)$ does not exist. It IS a candidate: the method evaluates $f$ at all critical points AND endpoints, and here $5$ even carries half of the minimum. $x = 0$ is interior with $g'(0) = 0$ : a critical point, and a candidate. In Thomas's vocabulary, the list of the method is the critical points plus the endpoints; the word critical is not stretched to cover the endpoints.

d) $x^2 - 4$ is a polynomial and $u \mapsto u^{2/3}$ is continuous for every real $u$ , so $h$ is continuous on $[-1, 3]$ . Chain rule (inner function $x^2 - 4$ ): $h'(x) = \frac{2}{3}\left(x^2 - 4\right)^{-1/3} \cdot 2x = \frac{4x}{3\left(x^2 - 4\right)^{1/3}}$ . Numerator zero: $x = 0$ . Denominator zero: $x^2 = 4$ , $x = \pm 2$ ; keep $2$ , which is in $(-1, 3)$ , reject $-2$ . At $2$ , $h(2) = 0$ exists: critical point with $h'$ undefined. Values: $h(-1) = (-3)^{2/3} = 9^{1/3} \approx 2.0801$ ; $h(0) = (-4)^{2/3} = 16^{1/3} \approx 2.5198$ ; $h(2) = 0$ ; $h(3) = 5^{2/3} \approx 2.9240$ . Absolute maximum $5^{2/3} \approx 2.9240$ at $x = 3$ , absolute minimum $0$ at $x = 2$ . Two of the four values need the calculator gesture of the intro: typing (-4)^(2/3) often returns ERROR, and a value that the calculator refuses is still a value of $h$ . Here dropping $x = 0$ and $x = -1$ would leave the answer unchanged by luck; in part a), dropping $x = -1$ loses the maximum.

Exercise 6: Transcendental functions: exact candidates, then the calculator compares

With trigonometric, exponential, logarithmic and inverse trigonometric functions, the critical points are solved BY HAND (the calculator does not solve equations), and the calculator then evaluates the finite list. Set it in RADIAN mode before the first value: $\sin\frac{\pi}{6}$ must display $0.5$ .

For each function: continuity on the interval, $f'$ factored, critical points inside, a table of exact values with their decimals to 4 places, the conclusion.

a) $f(x) = \sin x\,(1 + \cos x)$ on $\left[0, \frac{5\pi}{3}\right]$ . Hint: rewrite $f'$ with $\cos x$ only.
b) $g(x) = x e^{-x^2/2}$ on $[-1, 2]$ .
c) $h(x) = \ln\left(x^2 + 1\right) - x$ on $[0, 3]$ .
d) $k(x) = \arctan x - \frac{x}{2}$ on $[0, 3]$ .
e) $m(x) = \frac{\ln x}{\sqrt x}$ on $\left[1, e^3\right]$ .

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Answers

a) Max $\frac{3\sqrt 3}{4} \approx 1.2990$ at $x = \frac{\pi}{3}$ ; min $-\frac{3\sqrt 3}{4} \approx -1.2990$ at $x = \frac{5\pi}{3}$
b) Max $e^{-1/2} \approx 0.6065$ at $x = 1$ ; min $-e^{-1/2} \approx -0.6065$ at $x = -1$
c) Max $0$ at $x = 0$ ; min $\ln 10 - 3 \approx -0.6974$ at $x = 3$ (the critical point $1$ is neither)
d) Max $\frac{\pi}{4} - \frac{1}{2} \approx 0.2854$ at $x = 1$ ; min $\arctan 3 - \frac{3}{2} \approx -0.2510$ at $x = 3$
e) Max $\frac{2}{e} \approx 0.7358$ at $x = e^2$ ; min $0$ at $x = 1$

a) $f$ is a product of continuous functions. Product rule: $f'(x) = \cos x(1 + \cos x) + \sin x(-\sin x) = \cos x + \cos^2 x - \sin^2 x$ . The identity $\sin^2 x = 1 - \cos^2 x$ turns it into a quadratic in $\cos x$ : $f'(x) = 2\cos^2 x + \cos x - 1 = (2\cos x - 1)(\cos x + 1)$ . In the OPEN interval $\left(0, \frac{5\pi}{3}\right)$ : $\cos x = \frac{1}{2}$ gives $x = \frac{\pi}{3}$ only, since the other solution $\frac{5\pi}{3}$ is the endpoint, not an interior point; $\cos x = -1$ gives $x = \pi$ . Values: $f(0) = 0$ ; $f\left(\frac{\pi}{3}\right) = \frac{\sqrt 3}{2} \cdot \frac{3}{2} = \frac{3\sqrt 3}{4} \approx 1.2990$ ; $f(\pi) = 0$ ; $f\left(\frac{5\pi}{3}\right) = -\frac{\sqrt 3}{2} \cdot \frac{3}{2} = -\frac{3\sqrt 3}{4} \approx -1.2990$ . Absolute maximum $\frac{3\sqrt 3}{4}$ at $x = \frac{\pi}{3}$ , absolute minimum $-\frac{3\sqrt 3}{4}$ at the endpoint $\frac{5\pi}{3}$ , where $f'$ happens to vanish too: in Thomas it is an endpoint, not a critical point, and it is on the list either way. The critical point $\pi$ is neither: a candidate the table dismisses.

b) $g$ is continuous everywhere. Product rule and chain rule (inner function $-\frac{x^2}{2}$ , derivative $-x$ ): $g'(x) = e^{-x^2/2} + x \cdot (-x)e^{-x^2/2} = e^{-x^2/2}\left(1 - x^2\right)$ . Factor the exponential out: it is never $0$ , so the zeros are those of $1 - x^2$ , $x = \pm 1$ . Only $1$ is interior; $-1$ is an endpoint, evaluated anyway. Values: $g(-1) = -e^{-1/2} \approx -0.6065$ , $g(1) = e^{-1/2} \approx 0.6065$ , $g(2) = 2e^{-2} \approx 0.2707$ . Absolute maximum $e^{-1/2}$ at $x = 1$ , absolute minimum $-e^{-1/2}$ at $x = -1$ . Writing $e^{-x^2/2} = 0$ as an equation to solve is a lost line: an exponential is never zero.

c) $x^2 + 1 > 0$ , so $h$ is continuous on $[0, 3]$ . Chain rule: $h'(x) = \frac{2x}{x^2 + 1} - 1$ . Common denominator: $h'(x) = \frac{2x - x^2 - 1}{x^2 + 1} = -\frac{(x - 1)^2}{x^2 + 1}$ . One critical point, $x = 1$ . Values: $h(0) = \ln 1 - 0 = 0$ , $h(1) = \ln 2 - 1 \approx -0.3069$ , $h(3) = \ln 10 - 3 \approx -0.6974$ . Absolute maximum $0$ at $x = 0$ , absolute minimum $\ln 10 - 3$ at $x = 3$ . The critical point $1$ is neither: its value lies between the two endpoint values. The factorization shows why: $h' \le 0$ on both sides of $1$ , a horizontal tangent in the middle of a descent.

d) $k$ is continuous everywhere. $k'(x) = \frac{1}{1 + x^2} - \frac{1}{2} = \frac{2 - (1 + x^2)}{2(1 + x^2)} = \frac{1 - x^2}{2(1 + x^2)}$ . Critical point in $(0, 3)$ : $x = 1$ ( $-1$ rejected). Values: $k(0) = 0$ ; $k(1) = \frac{\pi}{4} - \frac{1}{2} \approx 0.2854$ ; $k(3) = \arctan 3 - \frac{3}{2} \approx 1.2490 - 1.5 = -0.2510$ . Absolute maximum $\frac{\pi}{4} - \frac{1}{2}$ at $x = 1$ , absolute minimum $\arctan 3 - \frac{3}{2}$ at $x = 3$ . The calculator trap: in DEGREE mode, $\arctan 3$ displays $71.5651$ , and $k(3) \approx 70.07$ becomes a false maximum. Every formula of calculus, and $(\arctan x)' = \frac{1}{1 + x^2}$ first of all, assumes radians.

e) On $\left[1, e^3\right]$ , $x > 0$ , so $m$ is continuous. Quotient rule: $m'(x) = \frac{\frac{1}{x}\sqrt x - \ln x \cdot \frac{1}{2\sqrt x}}{x}$ , a complex fraction. The gesture: multiply numerator and denominator by $2\sqrt x$ . Since $\frac{1}{x}\sqrt x \cdot 2\sqrt x = 2$ , this gives $m'(x) = \frac{2 - \ln x}{2x\sqrt x} = \frac{2 - \ln x}{2x^{3/2}}$ . Zero when $\ln x = 2$ , $x = e^2 \approx 7.3891$ , which lies in $\left(1, e^3\right)$ . Values: $m(1) = 0$ ; $m\left(e^2\right) = \frac{2}{e} \approx 0.7358$ ; $m\left(e^3\right) = \frac{3}{e^{3/2}} \approx 0.6694$ . Absolute maximum $\frac{2}{e}$ at $x = e^2$ , absolute minimum $0$ at $x = 1$ .

Exercise 7: Reasoning with Thomas's two theorems: no critical point, a false alarm, constants to find

Thomas, Theorem 2 (the First Derivative Theorem for Local Extreme Values). If $f$ has a local maximum or minimum value at an INTERIOR point $c$ of its domain, and if $f'$ is defined at $c$ , then $f'(c) = 0$ . It works in one direction only: it tells where interior extrema CAN be, never where they are.

In this chapter, a critical point is confirmed or dismissed by comparing values: by an inequality, or by the closed interval method on a small interval around it. The figure shows $y = x + \sin x$ on $[0, 2\pi]$ and its horizontal tangent at $x = \pi$ .

a) Find the critical points of $p(x) = x + \sin x$ in $(0, 2\pi)$ . Show that $p$ has no local extremum at $x = \pi$ , using the inequality $\sin h < h$ for $h > 0$ .
b) Show that $r(x) = 2x - \cos x$ has no critical point at all. Deduce the absolute extrema of $r$ on $[0, \pi]$ without any comparison of interior values.
c) Find the constants $a$ and $b$ so that $f(x) = x^3 + ax^2 + b$ satisfies $f'(2) = 0$ and $f(2) = 5$ . Then apply the closed interval method on $[1, 3]$ to show that $f(2) = 5$ is a local minimum value.
d) Find the absolute extrema of $q(x) = \frac{x^2 - 1}{x^2 + 1}$ on $(-\infty, \infty)$ , if they exist. Hint: write $q(x) = 1 - \frac{2}{x^2 + 1}$ .

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Answers

a) Only $x = \pi$ ; $p(\pi + h) > \pi > p(\pi - h)$ for $0 < h < \pi$ : no extremum
b) $r'(x) = 2 + \sin x \ge 1$ ; on $[0, \pi]$ , min $-1$ at $x = 0$ , max $2\pi + 1 \approx 7.2832$ at $x = \pi$
c) $a = -3$ , $b = 9$ ; $f(1) = 7$ , $f(2) = 5$ , $f(3) = 9$ : min $5$ on $[1, 3]$ at an interior point, so a local minimum
d) Absolute min $-1$ at $x = 0$ ; no absolute max (values approach $1$ )

a) $p'(x) = 1 + \cos x$ exists everywhere and vanishes when $\cos x = -1$ : in $(0, 2\pi)$ , only $x = \pi$ . Now compare values near $\pi$ , with $p(\pi) = \pi$ . For $0 < h < \pi$ : $p(\pi + h) = \pi + h + \sin(\pi + h) = \pi + h - \sin h$ , and since $\sin h < h$ , this is greater than $\pi$ . Likewise $p(\pi - h) = \pi - h + \sin h < \pi$ . So on every open interval around $\pi$ there are values above and below $p(\pi)$ : NO local extremum at $\pi$ . The identities $\sin(\pi \pm h) = \mp\sin h$ are the algebraic step; the figure shows the curve flattening at $(\pi, \pi)$ and climbing on.

b) $r'(x) = 2 + \sin x$ . Since $\sin x \ge -1$ , $r'(x) \ge 1 > 0$ for every $x$ : $r'$ is defined everywhere and never $0$ , so $r$ has NO critical point. By Theorem 2 read backwards (contrapositive), $r$ has no local extremum at any interior point. On $[0, \pi]$ , $r$ is continuous, the EVT guarantees the extrema, and the list of candidates reduces to the two endpoints: $r(0) = 0 - 1 = -1$ and $r(\pi) = 2\pi - (-1) = 2\pi + 1 \approx 7.2832$ . Absolute minimum $-1$ at $x = 0$ , absolute maximum $2\pi + 1$ at $x = \pi$ .

c) $f'(x) = 3x^2 + 2ax$ , so $f'(2) = 12 + 4a = 0$ gives $a = -3$ . Then $f(2) = 8 - 12 + b = 5$ gives $b = 9$ : $f(x) = x^3 - 3x^2 + 9$ . These two equations only express a NECESSARY condition from Theorem 2; they do not prove an extremum. So check with values: $f'(x) = 3x^2 - 6x = 3x(x - 2)$ , whose only zero in $(1, 3)$ is $2$ . $f$ is continuous on $[1, 3]$ , and $f(1) = 1 - 3 + 9 = 7$ , $f(2) = 5$ , $f(3) = 27 - 27 + 9 = 9$ . The absolute minimum of $f$ on $[1, 3]$ is $5$ , attained at the INTERIOR point $2$ , so $f(x) \ge f(2)$ on the open interval $(1, 3)$ around $2$ : $f(2) = 5$ is a local minimum value.

d) Long division, or adding and subtracting $1$ : $\frac{x^2 - 1}{x^2 + 1} = \frac{(x^2 + 1) - 2}{x^2 + 1} = 1 - \frac{2}{x^2 + 1}$ . Since $x^2 + 1 \ge 1$ , $0 < \frac{2}{x^2 + 1} \le 2$ , with $2$ reached only at $x = 0$ . So $-1 \le q(x) < 1$ : the absolute minimum is $q(0) = -1$ . The values approach $1$ as $|x| \to \infty$ but never reach it, since $\frac{2}{x^2 + 1} > 0$ : NO absolute maximum. The interval is not a finite closed interval, so the EVT was never going to help: the rewriting did the work. Check with Theorem 2: $q'(x) = \frac{4x}{(x^2 + 1)^2}$ vanishes only at $0$ , the minimum.

Exercise 8: Five statements to correct

Each statement below was written by a student on a MATH 203 assignment, and each is false. Say what is wrong, give a counterexample, and write the correct statement, with the definitions of Thomas.

a) An endpoint of the interval can never be a local extremum.
b) If $f'(c)$ does not exist, then $f$ has a local maximum or a local minimum at $c$ .
c) To find the absolute maximum of $f$ on $[a, b]$ , evaluate $f'$ at the critical points and at the endpoints, and take the largest value.
d) A function continuous on $[a, b]$ attains its absolute maximum at exactly one point.
e) My calculator gives an error for $(-8)^{2/3}$ , so $x = -8$ is not in the domain of $f(x) = x^{2/3}$ and cannot be a candidate.

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Answers

a) False in Thomas: $\sqrt x$ on $[0, 4]$ has a local minimum at the endpoint $0$ . Endpoints are never CRITICAL points.
b) False: $x^{1/3}$ at $0$ . An undefined $f'$ only makes $c$ a critical point, a candidate.
c) False: $x^3 - 3x$ on $[0, 2]$ gives $9$ from $f'(2)$ ; the true maximum is $f(2) = 2$ . Evaluate $f$ , not $f'$ .
d) False: $\sin x$ on $[0, 4\pi]$ reaches $1$ at $\frac{\pi}{2}$ and $\frac{5\pi}{2}$ . The EVT gives existence, not uniqueness.
e) False: $(-8)^{2/3} = \left((-8)^{1/3}\right)^2 = 4$ . The domain of $x^{2/3}$ is all real numbers.

a) FALSE with the definitions of Thomas. $f(x) = \sqrt x$ on $[0, 4]$ : $f(x) \ge 0 = f(0)$ for all $x$ in $[0, \delta)$ , so $f(0) = 0$ is a local minimum value at the endpoint $0$ , and $f(4) = 2$ is a local maximum value at the endpoint $4$ . Thomas compares $f(c)$ with the values of $f$ at the points of its DOMAIN near $c$ , which at an endpoint means a half-open interval. Correct statement: an endpoint can be a local extremum, but it is never a critical point, since critical points are interior points. It is always a candidate for an absolute extremum.

b) FALSE. $f(x) = x^{1/3}$ : $f'(x) = \frac{1}{3x^{2/3}}$ is undefined at $0$ (vertical tangent), yet $x^{1/3} < 0$ for $x < 0$ and $x^{1/3} > 0$ for $x > 0$ , so $f(0) = 0$ is neither a local maximum nor a local minimum. Correct statement: if $f'(c)$ does not exist and $c$ is an interior point of the domain, then $c$ is a CRITICAL POINT, that is, a candidate; whether it carries an extremum must be decided separately. Undefined is no more a verdict than zero.

c) FALSE. Take $f(x) = x^3 - 3x$ on $[0, 2]$ : $f'(x) = 3x^2 - 3$ , critical point $1$ . The student's list is $f'(0) = -3$ , $f'(1) = 0$ , $f'(2) = 9$ , and the answer max $9$ . But the question is about the VALUES of $f$ : $f(0) = 0$ , $f(1) = -2$ , $f(2) = 8 - 6 = 2$ . Absolute maximum $2$ at $x = 2$ , absolute minimum $-2$ at $x = 1$ . At a critical point $f'$ is $0$ or undefined by definition, so a table of $f'$ carries no information. Correct statement: evaluate $f$ at the critical points and the endpoints, and take the largest value OF $f$ .

d) FALSE. $f(x) = \sin x$ is continuous on $[0, 4\pi]$ and reaches its absolute maximum value $1$ at $x = \frac{\pi}{2}$ AND at $x = \frac{5\pi}{2}$ ; a constant function reaches it at every point. Correct statement (Theorem 1): a function continuous on a closed interval $[a, b]$ attains an absolute maximum value at LEAST once. The theorem guarantees existence, not uniqueness, which is why a complete answer lists every place where the maximum value occurs.

e) FALSE. $x^{2/3} = \left(x^{1/3}\right)^2$ , and the cube root of a negative number exists: $(-8)^{1/3} = -2$ , so $(-8)^{2/3} = (-2)^2 = 4$ . Many scientific calculators refuse a fractional power of a negative base, because they compute it with logarithms; the number exists all the same. Correct statement: the domain of $x^{2/3}$ is all real numbers, and at a negative $x$ it is computed as $\left(x^{1/3}\right)^2$ or $\left(x^2\right)^{1/3}$ . On an interval such as $[-8, 1]$ , $x = -8$ is an endpoint and must be evaluated: $f(-8) = 4$ is in fact the absolute maximum there.

Exercise 9: The temperature over a day: warmest, coldest, and when it warms fastest

A weather station fits the outdoor temperature on a spring day by $T(t) = 10 - 1.92t + 0.3t^2 - 0.01t^3$ , in degrees Celsius, where $t$ is the time in hours after midnight, $0 \le t \le 20$ (until 8 p.m.). The figure shows the model.

The decimal coefficients are part of the question: factor them out before solving anything. A calculator is allowed for the values; give temperatures to 2 decimals.

a) Explain why $T$ has an absolute maximum and an absolute minimum on $[0, 20]$ , then find its critical points.
b) Find the highest and lowest temperatures of the day between midnight and 8 p.m., and the clock times at which they occur.
c) The rate of change $T'(t)$ is in degrees per hour. Find the absolute maximum and minimum of $T'$ on $[0, 20]$ : when does the air warm fastest, and when does it cool fastest?
d) A student uses the model on $[0, 24]$ . Find the absolute minimum of $T$ on $[0, 24]$ and comment on the model.
e) On $[0, 20]$ , how many local extrema does $T$ have, endpoints included as in Thomas? Justify the endpoints with the closed interval method on $[0, 4]$ and on $[16, 20]$ .

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Answers

a) $T$ is a polynomial, continuous on $[0, 20]$ ; $T'(t) = -0.03(t - 4)(t - 16)$ , critical points $4$ and $16$
b) Max $15.12$ °C at $t = 16$ (4 p.m.); min $6.48$ °C at $t = 4$ (4 a.m.)
c) Warms fastest at $t = 10$ (10 a.m.), $1.08$ °C/h; cools fastest at $t = 0$ and $t = 20$ , $-1.92$ °C/h
d) Min $-1.52$ °C at $t = 24$ : a drop of $13$ degrees after 8 p.m., the model is not reliable past $t = 20$
e) Four: local max at $t = 0$ and $t = 16$ , local min at $t = 4$ and $t = 20$

a) $T$ is a polynomial, so it is continuous on the closed interval $[0, 20]$ , and by the Extreme Value Theorem it attains an absolute maximum and an absolute minimum there. $T'(t) = -1.92 + 0.6t - 0.03t^2$ . The algebraic gesture: factor out $-0.03$ BEFORE solving, dividing each coefficient: $\frac{-1.92}{-0.03} = 64$ and $\frac{0.6}{-0.03} = -20$ . So $T'(t) = -0.03\left(t^2 - 20t + 64\right) = -0.03(t - 4)(t - 16)$ , since $4 \times 16 = 64$ and $4 + 16 = 20$ . Critical points: $t = 4$ and $t = 16$ , both in $(0, 20)$ . Applying the quadratic formula to $-0.03t^2 + 0.6t - 1.92 = 0$ directly gives the same roots, with three times as many chances of a sign slip.

b) Values: $T(0) = 10$ ; $T(4) = 10 - 7.68 + 4.8 - 0.64 = 6.48$ ; $T(16) = 10 - 30.72 + 76.8 - 40.96 = 15.12$ ; $T(20) = 10 - 38.4 + 120 - 80 = 11.6$ . The highest temperature is $15.12$ °C, at $t = 16$ , that is 4 p.m.; the lowest is $6.48$ °C, at $t = 4$ , that is 4 a.m. The answer to where is a clock time here, and a value of $t$ converted to it: $t = 16$ is not 4 o'clock without p.m.

c) $T'$ is a polynomial, continuous on $[0, 20]$ : apply the same method to $T'$ . Its derivative is $T''(t) = 0.6 - 0.06t$ , zero at $t = 10$ . Values of $T'$ : $T'(0) = -1.92$ ; $T'(10) = -0.03(6)(-6) = 1.08$ ; $T'(20) = -0.03(16)(4) = -1.92$ . The largest rate is $1.08$ °C per hour, at 10 a.m.: the air warms fastest in mid-morning. The smallest is $-1.92$ °C per hour, at $t = 0$ and $t = 20$ : the fastest cooling happens at both ends of the period. Here the closed interval method is applied to a DIFFERENT function, $T'$ , whose candidates are the zeros of $T''$ and the endpoints; nothing about the concavity of $T$ is needed.

d) On $[0, 24]$ the critical points are still $4$ and $16$ , and the new endpoint gives $T(24) = 10 - 46.08 + 172.8 - 138.24 = -1.52$ . Candidates: $T(0) = 10$ , $T(4) = 6.48$ , $T(16) = 15.12$ , $T(24) = -1.52$ . The absolute minimum becomes $-1.52$ °C at midnight at the end of the day, while $T(4) = 6.48$ stays a local minimum. Comment: the model predicts a drop of about $13$ degrees in four hours after 8 p.m., and a cubic with a negative leading coefficient eventually plunges without bound. The fit is valid on $[0, 20]$ only: an absolute extremum found at an endpoint where the model is extrapolated describes the formula, not the weather.

e) On $[0, 4]$ , $T$ is continuous and has no critical point inside $(0, 4)$ , so its extrema there are at the endpoints: $T(0) = 10$ is the maximum on $[0, 4]$ . Hence $T(t) \le T(0)$ on $[0, 4)$ , a half-open interval of the domain containing $0$ : a local maximum at the endpoint $0$ , in Thomas's sense. Similarly on $[16, 20]$ , no critical point inside, $T(16) = 15.12$ and $T(20) = 11.6$ : $T(20)$ is the minimum on $[16, 20]$ , a local minimum at the endpoint $20$ . Inside, $T(4) = 6.48$ is the absolute minimum on $[0, 20]$ at an interior point, hence a local minimum, and $T(16) = 15.12$ the absolute maximum, hence a local maximum. Four local extrema: maxima at $t = 0$ and $t = 16$ , minima at $t = 4$ and $t = 20$ .

Exercise 10: A final exam question: a parameter inside the formula, and a cubic to factor

Let $a > 0$ be a constant and $f(x) = x^3 - 3a^2x$ on the interval $[0, 2]$ .

The shape of a long final exam question: the list of candidates depends on $a$ , so the answer depends on $a$ , and every case must be argued. The algebra carries the marks: factor $f'$ , then factor a cubic in $a$ .

a) Find the critical points of $f$ in terms of $a$ . For which values of $a$ does one of them lie in $(0, 2)$ ?
b) Take $a = 1$ . Find the absolute maximum and minimum of $f$ on $[0, 2]$ .
c) Take $a = 3$ . Find the absolute maximum and minimum of $f$ on $[0, 2]$ .
d) Let $0 < a < 2$ . Show that $f(2) - f(a) = 2(a + 1)(a - 2)^2$ , and deduce the absolute minimum of $f$ on $[0, 2]$ . What is the double root of $a^3 - 3a^2 + 4$ ?
e) For which values of $a$ is the absolute maximum of $f$ on $[0, 2]$ attained at $x = 0$ rather than at $x = 2$ ? Give the boundary value to 4 decimals.

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Answers

a) $x = a$ and $x = -a$ ; only $x = a$ can be in $(0, 2)$ , when $0 < a < 2$
b) Max $2$ at $x = 2$ ; min $-2$ at $x = 1$
c) Max $0$ at $x = 0$ ; min $-46$ at $x = 2$
d) $f(2) - f(a) = 2(a^3 - 3a^2 + 4) = 2(a + 1)(a - 2)^2 \ge 0$ : min $-2a^3$ at $x = a$ ; double root $a = 2$
e) $a \ge \frac{2}{\sqrt 3} \approx 1.1547$ (tie at $x = 0$ and $x = 2$ when $a = \frac{2}{\sqrt 3}$ )

a) $f$ is a polynomial, continuous on $[0, 2]$ for every $a$ . $f'(x) = 3x^2 - 3a^2 = 3(x - a)(x + a)$ : critical points $x = a$ and $x = -a$ . Since $a > 0$ , $-a < 0$ is never in $(0, 2)$ : rejected in every case. And $x = a$ lies in $(0, 2)$ exactly when $0 < a < 2$ . Treat $a$ as a number while differentiating: the derivative of $3a^2x$ with respect to $x$ is $3a^2$ , not $6ax$ .

b) $a = 1$ : $f(x) = x^3 - 3x$ , candidates $0$ , $1$ , $2$ . $f(0) = 0$ , $f(1) = -2$ , $f(2) = 8 - 6 = 2$ . Absolute maximum $2$ at $x = 2$ , absolute minimum $-2$ at $x = 1$ .

c) $a = 3$ : the critical point $3$ is outside $(0, 2)$ , rejected, and so is $-3$ . Only the endpoints remain: $f(0) = 0$ , $f(2) = 8 - 3 \cdot 9 \cdot 2 = 8 - 54 = -46$ . Absolute maximum $0$ at $x = 0$ , absolute minimum $-46$ at $x = 2$ . A student who evaluates at $x = 3$ gets $f(3) = 27 - 81 = -54$ , a value $f$ never takes on $[0, 2]$ .

d) For $0 < a < 2$ the candidates are $0$ , $a$ , $2$ , with $f(0) = 0$ , $f(a) = a^3 - 3a^3 = -2a^3$ and $f(2) = 8 - 6a^2$ . Then $f(2) - f(a) = 8 - 6a^2 + 2a^3 = 2\left(a^3 - 3a^2 + 4\right)$ . To factor the cubic, look for an integer root among the divisors of $4$ : $a = -1$ gives $-1 - 3 + 4 = 0$ . Division by $a + 1$ gives $a^3 - 3a^2 + 4 = (a + 1)\left(a^2 - 4a + 4\right) = (a + 1)(a - 2)^2$ ; its double root is $a = 2$ . For $a > 0$ both factors are $\ge 0$ , so $f(2) \ge f(a)$ ; and $f(0) = 0 > -2a^3 = f(a)$ . The absolute minimum is $-2a^3$ , at $x = a$ . For $a = 1$ this gives $-2$ , as in b).

e) For $0 < a < 2$ the maximum is the larger of $f(0) = 0$ and $f(2) = 8 - 6a^2$ (the interior candidate gives the minimum by d)). $f(0) \ge f(2)$ means $6a^2 \ge 8$ , $a^2 \ge \frac{4}{3}$ , $a \ge \frac{2}{\sqrt 3} = \frac{2\sqrt 3}{3} \approx 1.1547$ . For $a \ge 2$ , by c), the only candidates are the endpoints and $f(2) = 8 - 6a^2 < 0$ , so the maximum is again at $x = 0$ . Conclusion: the absolute maximum $0$ is at $x = 0$ when $a \ge \frac{2}{\sqrt 3}$ (at both endpoints in the tie $a = \frac{2}{\sqrt 3}$ ), and it is $8 - 6a^2$ at $x = 2$ when $0 < a < \frac{2}{\sqrt 3}$ . The figure of the solution shows the two regimes, $a = 1$ and $a = 1.5$ .

Corrected exercises: extreme values of functions (MATH 203)

Course recap

Part A: the basics (/50)

Exercise 1: Reading extrema on a graph with Thomas's definitions: endpoints included

Exercise 2: Critical points hide in the algebra: write f' as one factored fraction

Exercise 3: The Extreme Value Theorem: check the hypotheses, then see what survives

Exercise 4: The closed interval method on polynomials and a quotient, and a calculator that lies

Exercise 5: The closed interval method with roots and fractional exponents

Part B: problems and reasoning (/50)

Exercise 6: Transcendental functions: exact candidates, then the calculator compares

Exercise 7: Reasoning with Thomas's two theorems: no critical point, a false alarm, constants to find

Exercise 8: Five statements to correct

Exercise 9: The temperature over a day: warmest, coldest, and when it warms fastest

Exercise 10: A final exam question: a parameter inside the formula, and a cubic to factor

See also

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