Corrected exercises: extreme values of functions (MATH 203)

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 $4$.

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.

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.

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.

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.

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 $0$.

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 $-x^{-2}$.

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.

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.

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.

$\sqrt x$ 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.

$\tan x = \frac{\sin x}{\cos x}$ 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 $x = -\frac{\pi}{4}$.

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.

$f$ 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.

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.

$f$ 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.

$g$ 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 $0$.

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 $x = 0$.

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.

$x^{2/3} = \left(x^{1/3}\right)^2$ 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.

$5 - x \ge 0$ 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.