This problem requires the Second Derivative Test to classify critical points of a function. The test identifies a relative minimum when s'(c) = 0 and s''(c) > 0, showing concave up. Here, x = -2 is a critical point, and s''(-2) = 9 > 0 indicates upward concavity. This positive concavity forms a local trough where the graph bends up. A tempting distractor is choice B, a relative maximum, but positive second derivatives denote minima, not maxima. Ensure the point is indeed critical and the second derivative non-zero to qualify for the test's conclusive application.

The Second Derivative Test is a method to classify critical points of a function by examining the sign of the second derivative at those points. At x=-6, M'(-6)=0 marks a critical point, and M''(-6)=-1, negative, indicates concave down. This suggests a local maximum, as the graph peaks there. The negative value ensures decreasing function away from the point. A tempting distractor is choice B, which claims M''(-6)<0 implies concave up, but it means concave down. To apply the Second Derivative Test effectively, always ensure the first derivative is zero and check if the second derivative is non-zero; if it is zero, switch to the First Derivative Test for classification.

Using the Second Derivative Test, positive second derivatives classify minima. ℓ''(1/4) = 3/8 > 0 shows concave up, local minimum. The graph curves upward here. Thus, x = 1/4 is a local minimum. Claiming positive implies maximum reverses the rule, hence fails. Confirm f' = 0 first for eligibility in any function.

The Second Derivative Test is a method to classify critical points of a function by examining the sign of the second derivative at those points. At x=6, t'(6)=0 is a critical point, and t''(6)=1/3, positive, indicates concave up. Concave up means the graph forms a minimum point. Thus, there is a local minimum at x=6, with values increasing away. A tempting distractor is choice A, which assumes it must be an inflection point, but positive f'' confirms a minimum. To apply the Second Derivative Test effectively, always ensure the first derivative is zero and check if the second derivative is non-zero; if it is zero, switch to the First Derivative Test for classification.

Through the Second Derivative Test, critical points are classified via second derivative signs. Positive, as in f''(12) = 1/2 > 0, means concave up and a local minimum. The graph bottoms out here locally. Thus, x = 12 is a local minimum. Assuming it's an inflection point without concavity change evidence is a mistake. For eligibility, confirm twice differentiability to apply the test reliably in problems.

This problem assesses your understanding of the Second Derivative Test for classifying critical points. A negative second derivative at a critical point means the graph is concave down, indicating a local maximum. This downward curvature shows the point is higher than nearby points. Given u'(-1) = 0 and u''(-1) = -2, negative, there is a local maximum at x = -1. Choosing neither might seem appealing if overlooking the sign, but the negative value definitively indicates a maximum. For transferable strategy, always confirm the second derivative's sign after verifying the critical point, and note that zero requires alternative analysis.

This problem requires the Second Derivative Test to classify critical points of a function. The test states that F'(c) = 0 and F''(c) > 0 imply a relative minimum due to concave up. For F, F'(π) = 0 marks the critical point, and F''(π) = 2 > 0 shows upward concavity. This positive value means the graph forms a local minimum at x = π. A tempting distractor is choice A, a relative maximum, but this would require a negative second derivative instead. A transferable strategy is to first locate critical points where the first derivative is zero, then check the second derivative's sign for classification.

This question tests the Second Derivative Test with f''(c) = -1. We have u'(1) = 0 (critical point) and u''(1) = -1 < 0. The Second Derivative Test states that when f'(c) = 0 and f''(c) < 0, the function has a local maximum at x = c. The negative second derivative means the function is concave down at x = 1, forming a peak. Choice A incorrectly pairs negative concavity with a minimum, which is the opposite of the correct relationship. Remember: the sign of f''(c) determines the extremum type—negative means maximum (concave down), positive means minimum (concave up).

This problem requires applying the Second Derivative Test to classify a critical point. Since f'(2) = 0, we have a critical point at x = 2. The Second Derivative Test states that if f'(c) = 0 and f''(c) > 0, then f has a local minimum at x = c. Here, f''(2) = 5 > 0, which means the function is concave up at x = 2, creating a valley shape and thus a local minimum. Choice A incorrectly claims a maximum despite positive concavity, which would require f''(2) < 0. When using the Second Derivative Test, remember: positive second derivative means concave up (local min), negative means concave down (local max).

The Second Derivative Test is a method to classify critical points of a function by examining the sign of the second derivative at those points. At x=-π, G'(-π)=0 indicates a critical point, and G''(-π)=5, positive, shows concave up. This upward curve suggests a local minimum. The function forms a valley there, increasing away. A tempting distractor is choice B, which incorrectly states G''(-π)>0 implies a local maximum, but positive means minimum. To apply the Second Derivative Test effectively, always ensure the first derivative is zero and check if the second derivative is non-zero; if it is zero, switch to the First Derivative Test for classification.