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.

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.

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.

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=1/2, k'(1/2)=0 marks a critical point, and k''(1/2)=3, positive, indicates concave up. Concave up implies a minimum point, as the graph curves upward like a bowl. Thus, there is a local minimum at x=1/2, with function values rising away from it. A tempting distractor is choice D, which assumes it must be an inflection point, but positive f'' confirms a minimum, not a concavity change. 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.

The Second Derivative Test classifies critical points by checking the second derivative's sign at locations where the first derivative is zero. If negative, like b''(3) = -1/5 < 0, the function is concave down, indicating a peak. This means the point is higher than its neighbors, marking a local maximum. Therefore, x = 3 is a local maximum. Some might think negative implies concave up, but that's incorrect; negative denotes concave down. Ensure the critical point is isolated and the function smooth to apply the test across various functions.

The Second Derivative Test is applied to critical points to determine their nature through the second derivative. When the second derivative is zero, as with c''(-1) = 0, concavity doesn't clearly indicate a max or min. It could be either or neither, like an inflection point or flat spot. Thus, the test is inconclusive here. A distractor suggesting it's necessarily an inflection point fails because zero second derivative alone doesn't confirm a concavity change; further checks are needed. Always confirm if higher derivatives or other tests like the first derivative test can resolve inconclusive cases.

Employing the Second Derivative Test, positive derivatives indicate minima. r''(-5/4) = 16/3 > 0 means concave up, local minimum. Graph bottoms locally. So, x = -5/4 is a local minimum. Needing f value is unnecessary; test uses derivatives only. Confirm differentiability for application strategy.

The Second Derivative Test assesses concavity for critical point classification. A negative second derivative, g''(-9) = -11/3 < 0, shows concave down, indicating a local maximum. Values decrease from this peak. Therefore, x = -9 is a local maximum. The idea that negative implies concave up is false; it's the opposite. Always check if the first derivative changes sign or use alternatives if inconclusive.

This problem applies the Second Derivative Test with a negative second derivative. We have r'(0) = 0 (critical point) and r''(0) = -9 < 0. According to the Second Derivative Test, when f'(c) = 0 and f''(c) < 0, the function has a local maximum at x = c. The negative second derivative indicates concave down behavior, creating a hill shape at x = 0. Choice A reverses the relationship, incorrectly pairing negative concavity with a minimum. For test success, remember the mnemonic: "negative second derivative = frown shape = maximum" and "positive second derivative = smile shape = minimum."