Corrected exercises: maximum and minimum values (MATH 140)

The absolute maximum is the highest point of the WHOLE graph: $f(0) = 7$, at the left endpoint. The absolute minimum is the lowest point: $f(8) = 1$. Two different questions are hidden in each answer, the VALUE ($7$, $1$) and the PLACE ($x = 0$, $x = 8$); a marker who asks for the value and reads $x = 8$ gives no mark. Every other point of the graph lies strictly between $1$ and $7$: the smooth bump at $x = 2$ bottoms out at $2$, the corner at $x = 4$ peaks at $5$, and the right endpoint is at $4$.

Local maximum value: $f(4) = 5$, at the corner: on a small open interval around $4$, no value exceeds $5$. Local minimum values: $f(2) = 2$ and $f(8) = 1$; the second one is also the absolute minimum, and an absolute extremum reached at an INTERIOR point is always a local one. $f(0) = 7$ is NOT a local maximum value with Stewart's definition: there is no open interval around $0$ inside the domain $[0, 10]$, so the condition cannot even be tested. It is the absolute maximum, which is a statement about the whole domain. The same holds for $f(10) = 4$. Note also that the local minimum value $2$ is smaller than the local maximum value $5$ here, but nothing forces that order: local means compared with the neighbours only.

Horizontal tangents give $f'(c) = 0$: $c = 2$, $c = 6$ and $c = 8$. The corner at $x = 4$ has two different one-sided slopes, so $f'(4)$ does not exist: $4$ is a critical number too, and it is the one a student who only solves $f'(x) = 0$ misses. So the critical numbers in $(0, 10)$ are $2$, $4$, $6$, $8$. Local extrema at $2$ (min), $4$ (max), $8$ (min); none at $6$, where the graph comes down, flattens, and keeps coming down: $f(x) > 3$ just to the left of $6$ and $f(x) < 3$ just to the right. This is the whole chapter in one picture: every local extremum inside the interval sits at a critical number, but a critical number is only a CANDIDATE.

On $(0, 10)$, the value $7$ is no longer available: as $x \to 0^+$ the values $f(x)$ climb toward $7$, but every $f(x)$ with $0 < x \le 10$ is strictly less than $7$ (the graph decreases from $7$ on $[0, 2]$ and stays at or below $5$ beyond). There is no largest value, since any value below $7$ is beaten by a point closer to $0$: NO absolute maximum. The absolute minimum $f(8) = 1$ is still there, because $8$ is an interior point. No contradiction with the Extreme Value Theorem: its hypothesis is a CLOSED interval, and that hypothesis fails here. The theorem says nothing about open intervals; it neither promises nor forbids extrema there.

On $[2, 10]$ the candidates are the endpoints $2$ and $10$ and the critical numbers inside, $4$, $6$, $8$. Values: $f(2) = 2$, $f(4) = 5$, $f(6) = 3$, $f(8) = 1$, $f(10) = 4$. Absolute maximum $5$ at $x = 4$, absolute minimum $1$ at $x = 8$. The local maximum at $4$ became absolute because the only point higher than $5$, the left part of the graph near $x = 0$, is no longer in the domain. Absolute is always relative to a DOMAIN: change the interval, and the answer can change even though the graph did not.

By the power rule, $f'(x) = 6x^2 - 6x - 12 = 6(x^2 - x - 2) = 6(x - 2)(x + 1)$. A polynomial is differentiable everywhere, so there is no point where $f'$ fails to exist, and the critical numbers are the zeros: $x = -1$ and $x = 2$. Check by substitution: $f'(2) = 24 - 12 - 12 = 0$ and $f'(-1) = 6 + 6 - 12 = 0$. Nothing is said yet about maxima or minima: the question asks for the candidates only.

Expand first, to differentiate term by term: $g(x) = 8x^{1/3} - x^{4/3}$, so $g'(x) = \frac{8}{3}x^{-2/3} - \frac{4}{3}x^{1/3}$. Factor out $\frac{4}{3}x^{-2/3}$: $g'(x) = \frac{4}{3}x^{-2/3}(2 - x) = \frac{4(2 - x)}{3x^{2/3}}$. The numerator vanishes at $x = 2$. The denominator vanishes at $x = 0$, where $g'$ does not exist, while $g(0) = 0$ is perfectly defined (the cube root of $0$ is $0$): $0$ IS a critical number. Critical numbers: $0$ and $2$. The graph has a vertical tangent at the origin, and a student who only solves $g'(x) = 0$ loses it.

By the quotient rule, $q'(x) = \frac{2x(x - 1) - x^2 \cdot 1}{(x - 1)^2} = \frac{x^2 - 2x}{(x - 1)^2} = \frac{x(x - 2)}{(x - 1)^2}$. Zeros of the numerator: $x = 0$ and $x = 2$, both in the domain. The denominator vanishes at $x = 1$, but $1$ is NOT in the domain of $q$ ($q(1)$ would divide by $0$), so $1$ is NOT a critical number. It matters later: a vertical asymptote is not a place where $q$ can have a maximum or a minimum, since $q$ has no value there. Critical numbers: $0$ and $2$.

By the chain rule with inner function $2\theta$, $(\sin 2\theta)' = 2\cos 2\theta$, so $s'(\theta) = -2\sin\theta + 2\cos 2\theta$. Use $\cos 2\theta = 1 - 2\sin^2\theta$ to get a single trigonometric function: $s'(\theta) = -2\sin\theta + 2 - 4\sin^2\theta = -2(2\sin^2\theta + \sin\theta - 1) = -2(2\sin\theta - 1)(\sin\theta + 1)$. So $\sin\theta = \frac{1}{2}$ or $\sin\theta = -1$. The first gives $\theta = \frac{\pi}{6} + 2k\pi$ and $\theta = \frac{5\pi}{6} + 2k\pi$; the second gives $\theta = \frac{3\pi}{2} + 2k\pi$, for every integer $k$. In $[0, 2\pi)$: $\frac{\pi}{6}$, $\frac{5\pi}{6}$, $\frac{3\pi}{2}$. The classic loss is to keep only $\frac{\pi}{6}$ from $\sin\theta = \frac{1}{2}$: the sine takes each value in $(-1, 1)$ twice per period.

Remove the absolute value by cases. For $|x| > 2$, $x^2 - 4 > 0$ and $k(x) = x^2 - 4$, so $k'(x) = 2x$, never $0$ there. For $|x| < 2$, $k(x) = 4 - x^2$ and $k'(x) = -2x$, which vanishes at $x = 0$. At $x = 2$ the two formulas give one-sided slopes $-4$ (from the left) and $4$ (from the right): $k'(2)$ does not exist, and by symmetry neither does $k'(-2)$. Critical numbers: $-2$, $0$, $2$. The corners of an absolute value sit exactly where the inside changes sign, and they are critical numbers even though no equation $k'(x) = 0$ will ever produce them.

$\frac{1}{x}$ is a rational function, continuous wherever it is defined, and $0 \notin [1, 3]$: it is continuous on the closed interval $[1, 3]$, and the EVT guarantees both extrema. To find them: $f'(x) = -\frac{1}{x^2}$ is never $0$ and exists on all of $[1, 3]$, so there is no critical number and the extrema are at the endpoints: $f(1) = 1$ is the absolute maximum, $f(3) = \frac{1}{3}$ the absolute minimum. Writing both hypotheses, with the reason for continuity ($0$ is outside the interval), is what earns the theorem's mark.

$\frac{1}{x}$ is continuous on $(0, 2]$, but the interval is not closed: the EVT does not apply. As $x \to 0^+$, $\frac{1}{x} \to \infty$, so the values are unbounded above and there is NO absolute maximum. The absolute minimum exists: for $0 < x \le 2$, $\frac{1}{x} \ge \frac{1}{2}$, with equality at $x = 2$. So the minimum $\frac{1}{2}$ is attained at $x = 2$. The missing endpoint $0$ is exactly where the maximum escapes.

$x^2$ is continuous, but $(-1, 2)$ is open. For $-1 < x < 2$ we have $0 \le x^2 < 4$: the values come as close to $4$ as we like when $x \to 2^-$, but $4$ itself would need $x = 2$ (or $x = -2$), which is not in the interval. So there is NO absolute maximum. The absolute minimum $0$ is attained at $x = 0$, an interior point. Answering max $4$ at $x = 2$ is the most common error on this type of question: a value that is approached but never taken is not a maximum.

The interval $[0, 2]$ is closed, but $f$ is not continuous at $x = 1$: the left-hand limit is $\lim_{x\to 1^-}(x + 1) = 2$ while $f(1) = 0$. The EVT does not apply. On $[0, 1)$ the values fill $[1, 2)$, and on $[1, 2]$ they fill $[0, 1]$. The value $2$ is approached from the left of $1$ but never taken (hollow dot): NO absolute maximum. The absolute minimum is $0$, at $x = 1$ (full dot). On the figure, the top of the left segment ends in a hole, and that hole is the missing maximum.

The interval is open, so the EVT does not apply, yet both extrema exist: $\cos x \le 1$ with equality at $x = 0$ and $x = 2\pi$, both in $(-\pi, 3\pi)$, and $\cos x \ge -1$ with equality at $x = \pi$ (the other solutions $-\pi$ and $3\pi$ are excluded, but $\pi$ is enough). The theorem is an IMPLICATION: continuous on a closed interval implies extrema exist. When a hypothesis fails, the theorem is silent, and the extrema may or may not exist, as b), c), d) and e) show. Its converse, if the extrema exist then $f$ is continuous on a closed interval, is false.

$f'(x) = 3x^2$, so $f'(0) = 0$: $0$ is a critical number. But for $x < 0$, $x^3 < 0 = f(0)$, and for $x > 0$, $x^3 > 0 = f(0)$: on EVERY open interval around $0$ there are values below and values above $f(0)$, so $f(0)$ is neither a local maximum nor a local minimum. This refutes the CONVERSE of Fermat's theorem, if $f'(c) = 0$ then $f$ has a local extremum at $c$. The theorem goes one way only: from extremum to $f'(c) = 0$.

$|x| \ge 0 = g(0)$ for every $x$, so $g(0) = 0$ is a minimum, local and absolute. The derivative at $0$ is the limit of $\frac{|h|}{h}$ as $h \to 0$: it equals $1$ for $h > 0$ and $-1$ for $h < 0$, so the one-sided limits differ and $g'(0)$ does not exist. No contradiction: Fermat's theorem assumes that $f'(c)$ EXISTS, and here it does not. This is why the definition of a critical number includes the points where $f'$ does not exist: they are candidates Fermat's theorem cannot see.

No contradiction. Fermat's theorem is about LOCAL extrema, and with Stewart's definition a local extremum needs an open interval around $c$ inside the domain. At the endpoint $x = 1$ of $[0, 1]$ there is none, so $h(1)$ is an absolute maximum but not a local one, and the theorem does not apply. This is why the closed interval method evaluates the endpoints SEPARATELY: they can carry an extremum without being critical numbers.

$p$ is a polynomial, so $p'(x) = 5x^4 + 3x^2 + 2$ exists everywhere. Since $x^4 \ge 0$ and $x^2 \ge 0$, $p'(x) \ge 2 > 0$ for every $x$: $p'$ is never $0$, and $p$ has no critical number at all. By Fermat's theorem, a local extremum at $c$ would force $p'(c) = 0$ (since $p'(c)$ exists). There is no such $c$, so $p$ has no local maximum and no local minimum. This is the correct use of the theorem: in its CONTRAPOSITIVE form, if $f'(c)$ exists and is not $0$, then $f$ has no local extremum at $c$.

$u'(x) = 4x^3 - 12x^2 = 4x^2(x - 3)$: critical numbers $0$ and $3$, both with $u' = 0$. Near $0$, write $u(x) = x^3(x - 4)$. For $x$ close to $0$, $x - 4 < 0$, so $u(x)$ has the sign opposite to $x^3$: $u(x) > 0$ for small $x < 0$ and $u(x) < 0$ for small $x > 0$, while $u(0) = 0$. So $u(0)$ is not a local extremum, exactly like $x^3$ in a). At $3$: $u(3) = 81 - 108 = -27$. Expand $(x - 3)^2(x^2 + 2x + 3) = (x^2 - 6x + 9)(x^2 + 2x + 3) = x^4 - 4x^3 + 27$, so $u(x) + 27 = (x - 3)^2(x^2 + 2x + 3)$. Both factors are $\ge 0$, since $x^2 + 2x + 3 = (x + 1)^2 + 2 > 0$. Hence $u(x) \ge -27 = u(3)$ for all $x$: $-27$ is the absolute minimum value, reached at $x = 3$ only. Two critical numbers with $u' = 0$, one extremum: the figure shows the flat point at the origin and the bottom of the curve at $(3, -27)$.

$f$ is a polynomial, continuous on $[0, 4]$. $f'(x) = 3x^2 - 12x + 9 = 3(x - 1)(x - 3)$: critical numbers $1$ and $3$, both in $(0, 4)$. Values: $f(0) = 2$, $f(1) = 1 - 6 + 9 + 2 = 6$, $f(3) = 27 - 54 + 27 + 2 = 2$, $f(4) = 64 - 96 + 36 + 2 = 6$. Absolute maximum $6$, attained TWICE, at $x = 1$ and $x = 4$; absolute minimum $2$, attained at $x = 0$ and $x = 3$. The question where is answered by ALL the places: stopping at the first $6$ found loses the half mark for the location.

$g$ is a polynomial, continuous on $[0, 3]$. $g'(x) = 3x^2 - 12 = 3(x - 2)(x + 2)$: the zeros are $\pm 2$, but only $2$ lies in $(0, 3)$; $-2$ is REJECTED. Values: $g(0) = 0$, $g(2) = 8 - 24 = -16$, $g(3) = 27 - 36 = -9$. Absolute maximum $0$ at $x = 0$, absolute minimum $-16$ at $x = 2$. The trap bites here: $g(-2) = 16$, and a student who keeps $-2$ in the list answers max $16$, a value the function never takes on $[0, 3]$. Write the word rejected, with its reason, next to every solution of $g'(x) = 0$ that falls outside the interval.

$h$ is the composition of a polynomial and the absolute value, both continuous, so $h$ is continuous on $[0, 4]$. Since $x^2 - 2x - 3 = (x - 3)(x + 1)$, it is negative on $[0, 3)$ and positive on $(3, 4]$. On $(0, 3)$: $h(x) = 3 + 2x - x^2$, $h'(x) = 2 - 2x$, zero at $x = 1$. On $(3, 4)$: $h(x) = x^2 - 2x - 3$, $h'(x) = 2x - 2 \ne 0$. At $x = 3$ the one-sided slopes are $-4$ and $4$: $h'(3)$ does not exist, so $3$ is critical. Values: $h(0) = 3$, $h(1) = 4$, $h(3) = 0$, $h(4) = 5$. Absolute maximum $5$ at $x = 4$, absolute minimum $0$ at $x = 3$. Forgetting the corner gives min $3$: the answer is then wrong, because the minimum IS the corner. The figure of the solution shows it.

$r$ is continuous on $[1, 4]$ since its only discontinuity, $x = 0$, is outside. $r'(x) = 1 - \frac{4}{x^2} = \frac{x^2 - 4}{x^2}$: zeros $\pm 2$, keep $2$, reject $-2$; $r'$ is undefined only at $0$, not in the domain of $r$, so not critical. Values: $r(1) = 5$, $r(2) = 2 + 2 = 4$, $r(4) = 4 + 1 = 5$. Absolute maximum $5$ at $x = 1$ and $x = 4$, absolute minimum $4$ at $x = 2$.