Answer: Yes, because $f(0) = -1$ is the only local minimum and $f(x) > -1$ for all $x \neq 0$, confirming it is also the absolute minimum.

Analyze the function's behavior: $f(0) = \frac{-1}{1} = -1$. Since $f$ is decreasing on $(-\infty, 0)$ and increasing on $(0, \infty)$, the function reaches its lowest value at $x = 0$. Verify global behavior: $f(x) = \frac{x^2-1}{x^2+1} = 1 - \frac{2}{x^2+1}$. The term $\frac{2}{x^2+1}$ is maximized at $x = 0$ (where it equals 2), so $f(0) = 1 - 2 = -1$ is the smallest value. As $|x| \to \infty$, $f(x) \to 1$. Why the correct answer works: Since $f$ decreases to $-1$ and then increases forever (approaching 1), $f(0) = -1$ is both the only local minimum and the absolute minimum on the entire real line. Why distractors fail: Option A incorrectly implies absolute extrema cannot be found on open intervals — they can when there is only one critical point and the function increases away from it. Option C is backwards: $f(x) \to 1 > -1$, which confirms $f(0)$ is the lowest. Option D invents a requirement — the Second Derivative Test classifies local extrema, not absolute ones.

Answer: $x = 3$

Read the graph of $f'(x)$: The graph shows $f'(x) = (x+1)(x-3)$, which crosses zero at $x = -1$ and $x = 3$. Apply the First Derivative Test from the graph: At $x = -1$: $f'(x)$ goes from positive to negative → local maximum. At $x = 3$: $f'(x)$ goes from negative to positive → local minimum. Why distractors fail: Option A ($x = -1$) is a local maximum, not minimum. Option B ($x = 1$) is not a zero of $f'(x)$, so it is not a critical point. Option D ($x = 0$) is also not a critical point.

Answer: Local max at $x = 1$ and local min at $x = 3$

Factor the first derivative: $f'(x) = 5x^4 - 20x^3 + 15x^2 = 5x^2(x^2 - 4x + 3) = 5x^2(x-1)(x-3)$. Critical points: $x = 0$, $x = 1$, $x = 3$. Analyze sign changes: For $x < 0$: $5x^2 > 0$, $(x-1) < 0$, $(x-3) < 0$, product $> 0$. For $0 < x < 1$: $5x^2 > 0$, $(x-1) < 0$, $(x-3) < 0$, product $> 0$. At $x = 0$: no sign change → not an extremum. For $1 < x < 3$: $5x^2 > 0$, $(x-1) > 0$, $(x-3) < 0$, product $< 0$. At $x = 1$: sign changes $+$ to $-$ → local max. For $x > 3$: all factors positive → product $> 0$. At $x = 3$: sign changes $-$ to $+$ → local min. Why distractors fail: Option B incorrectly includes $x = 0$ as an extremum and reverses the types. Option C misses the local minimum at $x = 3$. Option D assigns incorrect types and wrongly labels $x = 0$ as an extremum.

Answer: $(1, \infty)$

Analyze concavity from the graph: The function shown is $f(x) = x^3 - 3x^2 - 9x + 10$. $f''(x) = 6x - 6 = 6(x - 1)$. So $f''(x) > 0$ when $x > 1$. Verify visually: Looking at the graph, for $x > 1$ the curve bends upward (concave up, bowl shape), while for $x < 1$ the curve bends downward (concave down). Why distractors fail: Option A gives the concave down interval. Options C and D are based on increasing/decreasing intervals (critical points at $x = -1$ and $x = 3$), not concavity intervals.

Answer: A local minimum at $x = c$

Apply the First Derivative Test: When $f'(x)$ changes from negative to positive at $c$, the function decreases before $c$ and increases after $c$. Why the correct answer works: A function that decreases then increases forms a valley — a local minimum — at the transition point $x = c$. Why distractors fail: Option A has the sign change reversed — positive to negative gives a local maximum. Option B is wrong because inflection points involve $f''(x)$ changing sign, not $f'(x)$. Option D is wrong because a sign change in $f'(x)$ does indicate an extremum.

Answer: $(-\infty, 0)$ and $(2, \infty)$

Find the first derivative: $f'(x) = 3x^2 - 6x = 3x(x - 2)$. Setting $f'(x) = 0$ gives critical points $x = 0$ and $x = 2$. Determine sign of $f'(x)$ on each interval: Test points: For $x < 0$ (say $x = -1$): $f'(-1) = 3(−1)(−3) = 9 > 0$. For $0 < x < 2$ (say $x = 1$): $f'(1) = 3(1)(−1) = −3 < 0$. For $x > 2$ (say $x = 3$): $f'(3) = 3(3)(1) = 9 > 0$. Conclusion: $f$ is increasing where $f'(x) > 0$, which is on $(-\infty, 0)$ and $(2, \infty)$. Why distractors fail: Option B gives the interval where $f$ is decreasing. Option C ignores the critical point at $x = 0$. Option D ignores the decreasing interval $(0, 2)$.

Answer: $f(x)$ is increasing on $(a, b)$

Recall the meaning of the first derivative: The first derivative $f'(x)$ gives the rate of change of $f(x)$. When $f'(x) > 0$, the function's output is increasing as $x$ increases. Why the correct answer works: A positive first derivative on an interval means the function is strictly increasing on that interval — this is a direct consequence of the definition. Why distractors fail: Option A confuses the first derivative with the second derivative (concavity is determined by $f''(x)$). Option C is wrong because a local maximum requires $f'(x)$ to change from positive to negative, which cannot happen if $f'(x) > 0$ throughout. Option D is wrong because $f'(x) > 0$ means the function is increasing, not that the function values themselves are positive.

Answer: A local minimum at $x = c$

State the Second Derivative Test: If $f'(c) = 0$ and $f''(c) > 0$, the function is concave up at $c$. A horizontal tangent on a concave-up curve creates a local minimum. Why the correct answer works: $f''(c) > 0$ means the curve is concave up (bowl-shaped) at $c$. Combined with $f'(c) = 0$, this guarantees a local minimum. Why distractors fail: Option A reverses the condition — a local maximum would require $f''(c) < 0$. Option B is wrong because inflection points require a sign change of $f''$, not a positive value. Option D (saddle point) is a multivariable calculus concept and does not apply to single-variable functions in this context.

Answer: $(0, 4)$

Connect $f''(x)$ to concavity: $f(x)$ is concave down where $f''(x) < 0$. Read the graph: The graph of $f''(x) = x(x-4)$ is a parabola opening upward with roots at $x = 0$ and $x = 4$. Between the roots, $f''(x) < 0$, so $f$ is concave down on $(0, 4)$. Why distractors fail: Option A gives the intervals where $f''(x) > 0$ (concave up). Options C and D are based on the vertex $x = 2$ rather than the roots, and do not correctly represent where $f''(x) < 0$.

Answer: A point where $f'(x) = 0$ or $f'(x)$ is undefined, provided $f(x)$ is defined there

Define critical points: A critical point of $f$ is a value $c$ in the domain of $f$ where either $f'(c) = 0$ or $f'(c)$ does not exist. Why the correct answer works: This is the standard definition. Critical points include both where the tangent is horizontal ($f'(c)=0$) and where the derivative fails to exist (cusps, corners), as long as $f(c)$ itself is defined. Why distractors fail: Option A confuses zeros of the function (x-intercepts) with critical points. Option C describes candidates for inflection points, not critical points. Option D describes a local maximum specifically, which is just one possible behavior at a critical point.

Answer: The graph lies below all of its tangent lines on that interval.

Interpret concavity geometrically: Concavity describes the curvature of a graph. Concave down means the graph bends downward like an upside-down bowl; equivalently, $f''(x) < 0$. Why the correct answer works: When a curve is concave down, the tangent lines at any point lie above the curve. Equivalently, the graph lies below its tangent lines. This is the geometric hallmark of concave down behavior. Why distractors fail: Option B confuses concavity with monotonicity — a function can be concave down and still increasing (e.g., $f(x) = -x^2 + 10x$ near $x = 1$). Option C makes the same error: negative $f'(x)$ means decreasing, not concave down. Option D describes concave up, not concave down.

Answer: Local max at $x = 1$, local min at $x = 2$

Find critical points: $f'(x) = 6x^2 - 18x + 12 = 6(x^2 - 3x + 2) = 6(x-1)(x-2)$. Critical points at $x = 1$ and $x = 2$. Compute the second derivative: $f''(x) = 12x - 18$. Apply the Second Derivative Test: $f''(1) = 12(1) - 18 = -6 < 0$, so $x = 1$ is a local maximum. $f''(2) = 12(2) - 18 = 6 > 0$, so $x = 2$ is a local minimum. Why distractors fail: Option B reverses the classification. Options C and D incorrectly assign the same type to both critical points, ignoring the different signs of $f''$.

Answer: The claim is incorrect — $f''(x) = \frac{2(1 - x^2)}{(x^2+1)^2}$, which changes sign at $x = \pm 1$, giving two inflection points.

Compute the second derivative: $f'(x) = \frac{2x}{x^2+1}$. $f''(x) = \frac{2(x^2+1) - 2x \cdot 2x}{(x^2+1)^2} = \frac{2 - 2x^2}{(x^2+1)^2} = \frac{2(1-x^2)}{(x^2+1)^2}$. Find where $f''$ changes sign: The numerator $2(1-x^2) = 0$ when $x = \pm 1$. For $|x| < 1$: $f''(x) > 0$ (concave up). For $|x| > 1$: $f''(x) < 0$ (concave down). So $f''$ changes sign at $x = -1$ and $x = 1$. Conclusion: There are two inflection points at $x = -1$ and $x = 1$, and the function is concave up near $x = 0$. The student's claim is incorrect. Why distractors fail: Options A and D wrongly accept the claim. The graph clearly shows a concave-up region near $x = 0$. Option C is wrong because $f''(0) = 2 > 0$, so $x = 0$ is in the concave-up region, not an inflection point.

Answer: Local maximum at $x = 1$

Find the critical point: $f'(x) = e^{-x} - xe^{-x} = e^{-x}(1 - x)$. Setting $f'(x) = 0$: since $e^{-x} > 0$ always, we need $1 - x = 0$, giving $x = 1$. Compute the second derivative: $f''(x) = -e^{-x}(1 - x) - e^{-x} = e^{-x}(x - 2)$. Apply the Second Derivative Test: $f''(1) = e^{-1}(1 - 2) = -e^{-1} < 0$. Since $f''(1) < 0$, the function is concave down at $x = 1$, so there is a local maximum. Why distractors fail: Option A claims a local minimum, but $f''(1) < 0$ rules this out. Option C is wrong because inflection points are not detected by the Second Derivative Test in this way. Option D is wrong because the test is conclusive when $f''(c) \neq 0$.

Answer: $x = -1$ and $x = 1$

Compute the second derivative: $f'(x) = 4x^3 - 12x + 8$. $f''(x) = 12x^2 - 12 = 12(x^2 - 1) = 12(x-1)(x+1)$. Find candidates and check sign changes: $f''(x) = 0$ at $x = -1$ and $x = 1$. Test: $f''(-2) = 12(3) > 0$, $f''(0) = -12 < 0$, $f''(2) = 12(3) > 0$. The sign changes at both $x = -1$ and $x = 1$. Conclusion: Since $f''(x)$ changes sign at both $x = -1$ and $x = 1$, both are inflection points. Why distractors fail: Option A is wrong because $f''(0) \neq 0$. Options C and D each miss one of the two inflection points.

Answer: $f''(x)$ changes sign

Define an inflection point: An inflection point is a point on the curve where the concavity changes — that is, the curve switches from concave up to concave down or vice versa. Why the correct answer works: Concavity is determined by $f''(x)$. A change in concavity requires $f''(x)$ to change sign (from positive to negative or negative to positive), which is the precise condition for an inflection point. Why distractors fail: Option A describes critical points, not inflection points. Option C describes extrema, not inflection points. Option D is a common misconception: $f''(x) = 0$ is necessary but not sufficient — for example, $f(x) = x^4$ has $f''(0) = 0$ but no inflection point at $x = 0$ because $f''(x)$ does not change sign there.

Answer: No, because $f(x) = x^4$ has $f'(0) = 0$ and $f''(0) = 0$ but $x = 0$ is a local minimum, not an inflection point.

Evaluate the claim: The claim is false. $f'(c) = 0$ and $f''(c) = 0$ together do not guarantee an inflection point. A sign change of $f''$ is required. Provide a counterexample: For $f(x) = x^4$: $f'(0) = 0$, $f''(x) = 12x^2$, so $f''(0) = 0$. But $f''(x) \geq 0$ everywhere — no sign change occurs, and $x = 0$ is a local minimum. Why distractors fail: Options A and B both incorrectly validate the claim. Option D invents a false restriction — inflection points can occur where $f'(x) = 0$ (e.g., $f(x) = x^3$ at $x = 0$).

Answer: Local maximum

Find and analyze $h'(x)$: Using the product rule with fractional exponents, $h'(x) = \frac{2}{3}x^{-1/3}(6-x)^{1/3} + x^{2/3}\cdot\frac{1}{3}(6-x)^{-2/3}(-1)$. After simplification, $h'(x) = \frac{4 - x}{x^{1/3}(6-x)^{2/3}}$. Determine the sign of $h'(x)$ near $x = 4$: The numerator is $4 - x$: positive for $x < 4$, negative for $x > 4$ (near $x = 4$). The denominator $x^{1/3}(6-x)^{2/3}$ is positive for $0 < x < 6$. So $h'(x)$ changes from positive to negative at $x = 4$. Conclude: Since $h'$ changes from positive to negative, $x = 4$ is a local maximum by the First Derivative Test. Why distractors fail: Option A has the sign change reversed. Option C is incorrect because there is a definitive sign change. Option D is incorrect because the First Derivative Test is sufficient — the Second Derivative Test is not required.

Answer: $f(x) = (x - 1)^4$

Identify the requirements: We need: (1) $f'(1) = 0$ (critical point), (2) $f''(1) = 0$ (Second Derivative Test inconclusive), (3) $f'$ changes from negative to positive at $x = 1$ (local minimum via First Derivative Test). Check $f(x) = (x-1)^4$: $f'(x) = 4(x-1)^3$, so $f'(1) = 0$. $f''(x) = 12(x-1)^2$, so $f''(1) = 0$ — the Second Derivative Test is inconclusive. For $x < 1$: $(x-1)^3 < 0$, so $f'(x) < 0$ (decreasing). For $x > 1$: $(x-1)^3 > 0$, so $f'(x) > 0$ (increasing). Sign change: negative to positive → local minimum. ✓ Why distractors fail: Option A: $f'(x) = 3(x-1)^2 \geq 0$ always — no sign change, so no local extremum. Option C: $f''(x) = 2 \neq 0$ at $x = 1$, so the Second Derivative Test is conclusive (not inconclusive). Option D: $f'(x) = -4(x-1)^3$ changes from positive to negative — that's a local maximum, not minimum.

Answer: The test is inconclusive because $g''(0) = 0$; the First Derivative Test should be used instead, revealing no extremum at $x = 0$.

Check the conditions: $g'(x) = 3x^2$, so $g'(0) = 0$. $g''(x) = 6x$, so $g''(0) = 0$. The Second Derivative Test requires $f''(c) \neq 0$ to be conclusive. Apply an alternative method: Using the First Derivative Test: $g'(x) = 3x^2 \geq 0$ for all $x$, and $g'$ does not change sign at $x = 0$. Therefore $x = 0$ is neither a local max nor a local min. Why distractors fail: Options A and B incorrectly claim the Second Derivative Test is conclusive. Option D jumps to a conclusion — while $x = 0$ does happen to be an inflection point for $x^3$, saying the test being inconclusive means it 'must' be an inflection point is a logical error.

Answer: $f(x) = x^4 + x^2$

Check each candidate systematically: We need: (1) exactly one local minimum at $x = 0$, (2) no local maximum, (3) concave up for all $x$ (i.e., $f''(x) > 0$ everywhere, or $f''(x) \geq 0$ with equality only at isolated points). Analyze $f(x) = x^4 + x^2$: $f'(x) = 4x^3 + 2x = 2x(2x^2 + 1)$. Since $2x^2 + 1 > 0$ always, $f'(x) = 0$ only at $x = 0$. For $x < 0$: $f'(x) < 0$ (decreasing). For $x > 0$: $f'(x) > 0$ (increasing). So $x = 0$ is the only critical point and it is a local minimum. ✓ No local maximum. ✓ $f''(x) = 12x^2 + 2 \geq 2 > 0$ for all $x$, so $f$ is concave up everywhere. ✓ Why distractors fail: Option B: $f'(x) = 4x^3 - 8x = 4x(x^2-2)$, giving critical points at $x = 0, \pm\sqrt{2}$. Sign analysis shows $x = 0$ is a local max and $x = \pm\sqrt{2}$ are local minima — fails conditions 1 and 2. Option C: $f'(x) = -4x^3$, so $x = 0$ is a local max (not min), and $f''(x) = -12x^2 \leq 0$ (concave down) — fails all conditions. Option D: $f'(x) = 4x^3 - 4x = 4x(x^2-1)$, giving critical points at $x = 0, \pm 1$. At $x = 0$: local max; at $x = \pm 1$: local minima — fails conditions 1 and 2.

Answer: Concave up, because $f''(0) > 0$

Compute the derivatives: Using the quotient rule: $f'(x) = \frac{6x}{(x^2+3)^2}$. Differentiating again: $f''(x) = \frac{6(x^2+3)^2 - 6x \cdot 2(x^2+3)(2x)}{(x^2+3)^4} = \frac{6(3 - 3x^2)}{(x^2+3)^3} = \frac{18(1 - x^2)}{(x^2+3)^3}$. Evaluate at $x = 0$: $f''(0) = \frac{18(1)}{27} = \frac{2}{3} > 0$, so $f$ is concave up at $x = 0$. Visual confirmation: Looking at the graph near $x = 0$, the curve has a bowl shape (opens upward), confirming concave up behavior. Why distractors fail: Option A has the wrong sign for $f''(0)$. Option C is wrong because $f''(0) \neq 0$. Option D confuses concavity with monotonicity — $f'(0) = 0$ (neither increasing nor decreasing), and the claim about concavity from monotonicity is incorrect regardless.

Answer: No, because $f''(x) = 12x^2 \geq 0$ for all $x$, so $f''$ does not change sign at $x = 0$.

Identify the misconception: The student assumes that $f''(c) = 0$ is sufficient for an inflection point. In fact, it is only necessary — the second derivative must also change sign. Verify with $f(x) = x^4$: $f''(x) = 12x^2$. At $x = 0$, $f''(0) = 0$. But $12x^2 \geq 0$ for all $x$, so $f''$ never changes sign. The function is concave up on both sides of $x = 0$. Why distractors fail: Options A and B both perpetuate the misconception that $f''(c) = 0$ is sufficient. Option D invents a false rule — inflection points are not restricted to where $f'(x) = 0$.

Answer: If $f'(x)$ changes from positive to negative at the critical point, the function rises then falls, creating a peak.

Understand the First Derivative Test mechanism: The First Derivative Test classifies critical points by examining the sign of $f'(x)$ on either side of the critical point. Why the correct answer works: When $f'(x)$ changes from positive (increasing) to negative (decreasing), the function rises on the left and falls on the right, so the critical point is a local maximum. This is the core logic of the First Derivative Test. Why distractors fail: Option A is insufficient — $f'(x) = 0$ alone does not guarantee a maximum (e.g., $f(x) = x^3$ at $x = 0$). Option C describes the Second Derivative Test, not the First Derivative Test. Option D describes a global (absolute) maximum, not a local maximum, and uses the wrong test.