Answer: $0$

Find the derivative and critical points: $g'(x) = 2\cos(x) + 1$. Setting $g'(x) = 0$ gives $\cos(x) = -\tfrac{1}{2}$, so $x = \tfrac{2\pi}{3}$ and $x = \tfrac{4\pi}{3}$ in $[0, 2\pi]$. Evaluate $g$ at all candidates: $g(0) = 0$. $g\!\left(\tfrac{2\pi}{3}\right) = 2\sin\!\left(\tfrac{2\pi}{3}\right) + \tfrac{2\pi}{3} = \sqrt{3} + \tfrac{2\pi}{3} \approx 3.83$. $g\!\left(\tfrac{4\pi}{3}\right) = 2\sin\!\left(\tfrac{4\pi}{3}\right) + \tfrac{4\pi}{3} = -\sqrt{3} + \tfrac{4\pi}{3} \approx 2.46$. $g(2\pi) = 0 + 2\pi \approx 6.28$. Identify the absolute minimum: The smallest value is $g(0) = 0$, so the absolute minimum value is $0$. Why distractors fail: Option B ($2\pi$) is the value at the right endpoint, the absolute maximum. Option C ($\tfrac{4\pi}{3} - \sqrt{3}$) is the value at the critical point $x = \tfrac{4\pi}{3}$, which is larger than $0$. Option D ($\tfrac{2\pi}{3} + \sqrt{3}$) is the value at $x = \tfrac{2\pi}{3}$.

Answer: Only Student B is correct; while the EVT does not apply, the absolute maximum can still be established by analyzing the function's behavior.

Evaluate Student A's claim: Student A is partially right that the EVT does not apply on an unbounded domain. However, they incorrectly conclude that no absolute maximum can exist. The EVT gives sufficient conditions, not necessary ones. Evaluate Student B's claim: Student B correctly observes that $e^{-x^2} \leq e^0 = 1$ for all real $x$ (since $-x^2 \leq 0$), and $f(0) = 1$. So the function attains its supremum. The absolute maximum exists and equals $1$. Why the correct answer works: Option B is correct: the EVT cannot be invoked here, but the absolute maximum still exists. Student A's error is in treating the EVT as a necessary condition. Why distractors fail: Option A is wrong because the absolute maximum does exist. Option C is internally contradictory for Student A's position — A claims no absolute max exists. Option D is false because $f$ is bounded above by $1$.

Answer: No; by the Candidates Test, the absolute minimum must occur at $x = 0$ with value $2$.

Apply the Candidates Test: For a continuous function on a closed interval, the absolute extrema must occur among the critical points and the endpoints. Here, the candidates are $x = 0$, $x = 1$, and $x = 3$ with values $2$, $5$, and $4$ respectively. Determine the absolute minimum: The smallest candidate value is $f(0) = 2$, so the absolute minimum is $2$ at $x = 0$. It is impossible for the absolute minimum to occur at a non-candidate point. Evaluate the colleague's claim: The colleague's premise — that the absolute minimum avoids both critical points and endpoints — contradicts a fundamental result. If $f$ is continuous on $[a,b]$ and the only critical point is $x=1$, then the absolute min must be among $\{f(0), f(1), f(3)\}$. Why distractors fail: Option A is wrong because absolute extrema of continuous functions on closed intervals can only occur at critical points or endpoints. Option C is wrong because $f(1) = 5$ is the largest candidate value, not the smallest. Option D is wrong because the EVT requires continuity (which is given), not differentiability everywhere.

Answer: The function must be continuous and the interval must be closed and bounded.

Recall the hypotheses of the EVT: The Extreme Value Theorem requires two conditions: (1) the function $f$ must be continuous on the interval, and (2) the interval must be closed and bounded, i.e., of the form $[a, b]$. Why the correct answer works: Option B correctly identifies both hypotheses: continuity of $f$ and a closed, bounded interval. Why distractors fail: Option A requires differentiability (too strong) and an open interval (wrong type). Option C requires integrability (not the relevant condition) and 'finite' alone does not imply closed. Option D imposes an increasing condition and a location constraint, neither of which is part of the EVT.

Answer: $t = 2$ with $C = 5$

Set up the derivative using the quotient rule: $C'(t) = \dfrac{20(t^2+4) - 20t \cdot 2t}{(t^2+4)^2} = \dfrac{20(4 - t^2)}{(t^2+4)^2}$. Find critical points: Setting the numerator to zero: $4 - t^2 = 0 \Rightarrow t = 2$ (taking $t \geq 0$). So the only critical point in $[0,10]$ is $t = 2$. Evaluate at candidates: $C(0) = 0$, $C(2) = \dfrac{40}{8} = 5$, $C(10) = \dfrac{200}{104} \approx 1.92$. The absolute maximum concentration is $5$ at $t = 2$. Why distractors fail: Option A gives $t=0$, which yields the minimum concentration. Option C gives $t=4$ with $C(4) = \tfrac{80}{20} = 4$, but $4 < 5$. Option D gives the right-endpoint value, which is approximately $1.92$.

Answer: Absolute maximum is $3$ at $x = -2$; absolute minimum is $0$ at $x = 1$.

Verify EVT applies: $h(x) = |x-1|$ is continuous on $[-2,3]$ (absolute value functions are continuous everywhere), so the EVT guarantees absolute extrema exist. Find critical points: For $x < 1$, $h(x) = 1-x$ so $h'(x) = -1$. For $x > 1$, $h(x) = x-1$ so $h'(x) = 1$. At $x = 1$, $h'$ does not exist (corner), so $x = 1$ is a critical point. There are no points where $h'(x) = 0$. Evaluate at candidates: $h(-2) = |-2-1| = 3$, $h(1) = 0$, $h(3) = |3-1| = 2$. The absolute maximum value is $3$ at $x = -2$ and the absolute minimum value is $0$ at $x = 1$. Why distractors fail: Option B incorrectly identifies the absolute maximum as $2$, which is $h(3)$, not the largest value. Option C gives the minimum as $1$ at $x=0$, but $h(1) = 0 < 1$. Option D is wrong because the EVT requires continuity, not differentiability; $h$ is continuous on $[-2,3]$.

Answer: $5$

Find the derivative and critical points: $f'(x) = -3x^2 + 6x = -3x(x - 2)$. Setting $f'(x) = 0$ gives $x = 0$ and $x = 2$. Both lie in $[0, 4]$. Evaluate $f$ at all candidates: $f(0) = 1$, $f(2) = -8 + 12 + 1 = 5$, $f(4) = -64 + 48 + 1 = -15$. Identify the absolute maximum: Comparing the values $1, 5, -15$, the largest is $5$ at $x = 2$. So the absolute maximum value is $5$. Why distractors fail: Option A ($1$) is $f(0)$, a candidate value but not the largest. Option C ($-15$) is $f(4)$, the absolute minimum. Option D ($3$) does not correspond to any candidate value.

Answer: Absolute maximum at $x = 5$; absolute minimum at $x = -3$

Evaluate $f$ at the endpoints and approximate critical points: We compute: $f(-3) = 0.05(81) - 0.5(9) + 0.3(-3) + 2 = 4.05 - 4.5 - 0.9 + 2 = 0.65$. $f(5) = 0.05(625) - 0.5(25) + 0.3(5) + 2 = 31.25 - 12.5 + 1.5 + 2 = 22.25$. Critical points yield local extrema near $x \approx -2.2$, $x \approx 0.15$, and $x \approx 3.05$, but their $f$-values are between roughly $0.73$ and $2.1$. Compare all candidate values: The largest value is $f(5) = 22.25$ (absolute maximum) and the smallest is $f(-3) = 0.65$ (absolute minimum). Both occur at endpoints. Why distractors fail: Option B correctly identifies the absolute maximum at $x=5$ but incorrectly places the minimum at an interior local minimum whose value (~$0.73$) exceeds $f(-3) = 0.65$. Option C misidentifies the local max near $x \approx 0.15$ as the absolute max, but $f(0.15) \approx 2.05 \ll 22.25$. Option D reverses the roles entirely.

Answer: All critical points of $f$ in $(a, b)$ together with the endpoints $x = a$ and $x = b$

Define the Candidates Test: The Candidates Test says: to find the absolute max and min of a continuous function on $[a,b]$, evaluate $f$ at every critical point in $(a,b)$ and at both endpoints $a$ and $b$. The largest value is the absolute max; the smallest is the absolute min. Why the correct answer works: Option C includes all critical points (where $f'(x)=0$ or $f'(x)$ does not exist) AND the endpoints, which is exactly the candidate list. Why distractors fail: Option A omits endpoints and points where $f'$ does not exist. Option B omits interior critical points. Option D uses inflection points (where concavity changes), which are not the candidates for absolute extrema.

Answer: They failed to include $x = 0$, where $f'(x)$ does not exist, as a critical point.

Identify all critical points: $f'(x) = \tfrac{2}{3}x^{-1/3}$. This derivative is never zero, but it is undefined at $x = 0$. Since $x = 0 \in [-8, 27]$ and $f$ is continuous there, $x = 0$ is a critical point. Evaluate the missed candidate: $f(0) = 0^{2/3} = 0$. Since $0 < 4 < 9$, the actual absolute minimum is $0$ at $x = 0$, not $4$ at $x = -8$. Why the correct answer works: Option B correctly identifies the error: the student forgot that critical points include places where the derivative does not exist, not just where it equals zero. Why distractors fail: Option A is wrong — $f(-8) = (-8)^{2/3} = (\sqrt[3]{-8})^2 = 4$ is well-defined. Option C is wrong — the Candidates Test is the standard method for absolute extrema on closed intervals. Option D is wrong — $x^{2/3}$ is continuous on all of $\mathbb{R}$.

Answer: The function attains both an absolute maximum value and an absolute minimum value on $[a, b]$.

State the Extreme Value Theorem: The EVT guarantees that a function continuous on a closed interval $[a, b]$ attains both an absolute maximum value and an absolute minimum value somewhere on that interval. Why the correct answer works: Option B correctly captures the conclusion of the EVT: the function attains (achieves) both a greatest and a least value on the interval. Why distractors fail: Option A says 'exactly one' of each, but a function could attain its maximum at multiple points (e.g., $f(x) = \sin(x)$ on $[0, 2\pi]$). Option C is not guaranteed by EVT — a constant function is continuous on $[a,b]$ but has no critical point where $f'=0$ in the interior. Option D confuses continuity with differentiability; a continuous function need not be differentiable (e.g., $f(x)=|x|$).

Answer: The Extreme Value Theorem requires a closed interval, not an open one, so its conclusion cannot be applied here.

Check the hypotheses of the EVT: While $f(x) = x^2$ is indeed continuous everywhere, the EVT requires the domain to be a closed interval $[a, b]$. The interval $(0, 4)$ is open, so the EVT does not apply. Investigate whether an absolute min actually exists: On $(0, 4)$, $f(x) = x^2 > 0$ for all $x$ in the domain, and $\inf f = 0$. But there is no $x \in (0,4)$ with $f(x) = 0$, so the infimum is not attained. The function has no absolute minimum on this open interval. Why distractors fail: Option A is false — $x^2$ is continuous everywhere. Option C is irrelevant; having no critical point in the interior doesn't by itself mean no extremum exists (endpoints could provide extrema on closed intervals). Option D is wrong because $x = 0$ is not in the open interval $(0, 4)$.