For , on and at , with on . What occurs at ?
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AP Calculus AB Quiz
Practice First Derivative Test in AP Calculus AB with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.
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For J, J′(x)>0 on (−5,1) and J′(x)=0 at x=1, with J′(x)>0 on (1,4). What occurs at x=1?
This quiz focuses on First Derivative Test, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus AB.
Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.
For J, J′(x)>0 on (−5,1) and J′(x)=0 at x=1, with J′(x)>0 on (1,4). What occurs at x=1?
Explanation: This problem assesses the First Derivative Test. The First Derivative Test determines local extrema by examining sign changes in the first derivative around critical points. If f' changes from positive to negative at a point, it indicates a local maximum there. If it changes from negative to positive, it indicates a local minimum, and no sign change means no extremum. In this case, at x=1, J' is positive on both sides with J'(1)=0, so there is no sign change and thus no local extremum. A tempting distractor is choice A, which claims a local maximum at x=1, but this fails because the derivative remains positive on both sides, indicating continued increase without a peak. To apply this generally, always identify points where the derivative might be zero and check the signs immediately to the left and right for extrema.
The derivative v′(x) is negative on (−3,0) and positive on (0,2). What occurs at x=0?
Explanation: This problem involves the First Derivative Test to identify local extrema based on the sign of the derivative. The First Derivative Test indicates that a local minimum occurs where the derivative changes from negative to positive. Here, v'(x) is negative on (-3,0) and positive on (0,2), showing a sign change from negative to positive at x=0. This means the function decreases before x=0 and increases after, confirming a local minimum. A tempting distractor is choice B, which claims a local maximum at x=0, but that would require a positive to negative change, not observed here. Remember, to apply this test effectively, always check for sign changes around critical points where the derivative is zero or undefined.
The derivative z′(x) is positive on (−1,2) and negative on (2,8). Where does z have a local maximum?
Explanation: This problem involves the First Derivative Test to identify local extrema based on the sign of the derivative. The First Derivative Test states that a local maximum occurs where the derivative changes from positive to negative. Here, z'(x) is positive on (-1,2) and negative on (2,8), showing a sign change from positive to negative at x=2. This means the function increases before x=2 and decreases after, confirming a local maximum. A tempting distractor is choice A, which claims a local minimum at x=2, but that would require a negative to positive change, not observed here. Remember, to apply this test effectively, always check for sign changes around critical points where the derivative is zero or undefined.
The derivative satisfies k′(x)<0 on (−9,−4) and k′(x)>0 on (−4,−1). Where does k have a local minimum?
Explanation: This problem involves the First Derivative Test to identify local extrema based on the sign of the derivative. The First Derivative Test indicates that a local minimum occurs where the derivative changes from negative to positive. Here, k'(x) is negative on (-9,-4) and positive on (-4,-1), showing a sign change from negative to positive at x=-4. This means the function decreases before x=-4 and increases after, confirming a local minimum. A tempting distractor is choice B, which claims a local maximum at x=-4, but that would require a positive to negative change, not observed here. Remember, to apply this test effectively, always check for sign changes around critical points where the derivative is zero or undefined.
For N, N′(x)<0 on (−4,−1), N′(x)>0 on (−1,3), and N′(x)<0 on (3,6). Where is a local maximum?
Explanation: This problem assesses the First Derivative Test. The First Derivative Test determines local extrema by examining sign changes in the first derivative around critical points. If f' changes from positive to negative at a point, it indicates a local maximum there. If it changes from negative to positive, it indicates a local minimum, and no sign change means no extremum. Here, at x=3, N' changes from positive on (-1,3) to negative on (3,6), indicating a local maximum, while at x=-1 it changes from negative to positive, suggesting a minimum. A tempting distractor is choice A, which claims a local maximum at x=-1, but this fails because the sign change there is from negative to positive, indicating a minimum instead. To apply this generally, always identify points where the derivative might be zero and check the signs immediately to the left and right for extrema.
For K, K′(x)<0 on (−5,1) and K′(x)=0 at x=1, with K′(x)<0 on (1,4). What occurs at x=1?
Explanation: This problem assesses the First Derivative Test. The First Derivative Test determines local extrema by examining sign changes in the first derivative around critical points. If f' changes from positive to negative at a point, it indicates a local maximum there. If it changes from negative to positive, it indicates a local minimum, and no sign change means no extremum. In this case, at x=1, K' is negative on both sides with K'(1)=0, so there is no sign change and thus no local extremum. A tempting distractor is choice A, which claims a local maximum at x=1, but this fails because the derivative remains negative on both sides, indicating continued decrease without a turn. To apply this generally, always identify points where the derivative might be zero and check the signs immediately to the left and right for extrema.
A function m has m′(x)>0 on (−1,4) and m′(x)<0 on (4,10). Where does m have a local maximum?
Explanation: This problem involves the First Derivative Test to identify local extrema based on the sign of the derivative. The First Derivative Test states that a local maximum occurs where the derivative changes from positive to negative. Here, m'(x) is positive on (-1,4) and negative on (4,10), showing a sign change from positive to negative at x=4. This means the function increases before x=4 and decreases after, confirming a local maximum. A tempting distractor is choice A, which claims a local minimum at x=4, but that would require a negative to positive change, not observed here. Remember, to apply this test effectively, always check for sign changes around critical points where the derivative is zero or undefined.
A function u has u′(x)>0 on (−3,0) and u′(x)<0 on (0,2). What occurs at x=0?
Explanation: This problem involves the First Derivative Test to identify local extrema based on the sign of the derivative. The First Derivative Test states that a local maximum occurs where the derivative changes from positive to negative. Here, u'(x) is positive on (-3,0) and negative on (0,2), showing a sign change from positive to negative at x=0. This means the function increases before x=0 and decreases after, confirming a local maximum. A tempting distractor is choice A, which claims a local minimum at x=0, but that would require a negative to positive change, not observed here. Remember, to apply this test effectively, always check for sign changes around critical points where the derivative is zero or undefined.
On (−8,−3), f′(x)>0; on (−3,2), f′(x)<0; on (2,6), f′(x)>0. Where is a local maximum?
Explanation: This problem involves the First Derivative Test to identify local extrema based on the sign of the derivative. The First Derivative Test states that a local maximum occurs where the derivative changes from positive to negative. Here, f'(x) >0 on (-8,-3) and <0 on (-3,2), showing a + to - change at x=-3; then <0 to >0 at x=2, which is a min. Thus, the local maximum is at x=-3. A tempting distractor is choice A, suggesting a max at x=2, but that's actually a min due to - to + change. Remember, to apply this test effectively, always check for sign changes around critical points where the derivative is zero or undefined.
The derivative c′(x) is negative on (2,3) and negative on (3,9). What local extremum occurs at x=3?
Explanation: This problem involves the First Derivative Test to identify local extrema based on the sign of the derivative. The First Derivative Test requires a sign change in the derivative to indicate an extremum; no change means none. Here, c'(x) is negative on (2,3) and negative on (3,9), showing no sign change at x=3. Thus, the function is decreasing on both sides, so no local extremum at x=3. A tempting distractor is choice C, suggesting a local minimum at x=3, but without a sign change from negative to positive, this is incorrect. Remember, to apply this test effectively, always check for sign changes around critical points where the derivative is zero or undefined.
A differentiable function r has r′(x)<0 on (1,5) and r′(x)>0 on (5,9). Where is a local minimum?
Explanation: This problem involves the First Derivative Test to identify local extrema based on the sign of the derivative. The First Derivative Test indicates that a local minimum occurs where the derivative changes from negative to positive. Here, r'(x) is negative on (1,5) and positive on (5,9), showing a sign change from negative to positive at x=5. This means the function decreases before x=5 and increases after, confirming a local minimum. A tempting distractor is choice C, which claims a local maximum at x=5, but that would require a positive to negative change, not observed here. Remember, to apply this test effectively, always check for sign changes around critical points where the derivative is zero or undefined.
The sign chart shows f′(x)<0 on (−4,2) and f′(x)>0 on (2,7). Where does f have a local minimum?
Explanation: This problem involves the First Derivative Test to identify local extrema based on the sign of the derivative. The First Derivative Test states that a local minimum occurs where the derivative changes from negative to positive. In this case, f'(x) is negative on (-4,2) and positive on (2,7), indicating a sign change from negative to positive at x=2. This confirms a local minimum at x=2, as the function decreases before and increases after this point. A tempting distractor is choice A, which suggests a local maximum at x=2, but that would require a change from positive to negative, which does not occur here. Remember, to apply this test effectively, always check for sign changes around critical points where the derivative is zero or undefined.
For g on [1,9], g′(x)>0 on (1,6) and g′(x)<0 on (6,9). Where does g have a local maximum?
Explanation: This problem assesses the First Derivative Test. The First Derivative Test determines local extrema by examining sign changes in the first derivative around critical points. If f' changes from positive to negative at a point, it indicates a local maximum there. If it changes from negative to positive, it indicates a local minimum, and no sign change means no extremum. In this case, at x=6, g' changes from positive on (1,6) to negative on (6,9), confirming a local maximum. A tempting distractor is choice A, which claims a local minimum at x=6, but this fails because the sign change is from positive to negative, not the reverse. To apply this generally, always identify points where the derivative might be zero and check the signs immediately to the left and right for extrema.
A function g has g′(x)<0 on (0,2) and g′(x)>0 on (2,6). Where does g have a local minimum?
Explanation: This problem involves the First Derivative Test to identify local extrema based on the sign of the derivative. The First Derivative Test indicates that a local minimum occurs where the derivative changes from negative to positive. Here, g'(x) is negative on (0,2) and positive on (2,6), showing a sign change from negative to positive at x=2. This means the function decreases before x=2 and increases after, confirming a local minimum. A tempting distractor is choice D, which claims a local maximum at x=2, but that would require a positive to negative change, not observed here. Remember, to apply this test effectively, always check for sign changes around critical points where the derivative is zero or undefined.
The derivative satisfies h′(x)<0 on (−5,−1) and h′(x)<0 on (−1,3). What local extremum occurs at x=−1?
Explanation: This problem involves the First Derivative Test to identify local extrema based on the sign of the derivative. The First Derivative Test requires a sign change in the derivative to indicate an extremum; no change means none. Here, h'(x) is negative on (-5,-1) and negative on (-1,3), showing no sign change at x=-1. Thus, the function is decreasing on both sides, so no local extremum at x=-1. A tempting distractor is choice B, suggesting a local minimum at x=-1, but without a sign change from negative to positive, this is incorrect. Remember, to apply this test effectively, always check for sign changes around critical points where the derivative is zero or undefined.
For Z, Z′(x)>0 on (−3,−2) and Z′(x)<0 on (−2,−1). Where does Z have a local maximum?
Explanation: This problem assesses the First Derivative Test. The First Derivative Test determines local extrema by examining sign changes in the first derivative around critical points. If f' changes from positive to negative at a point, it indicates a local maximum there. If it changes from negative to positive, it indicates a local minimum, and no sign change means no extremum. In this case, at x=-2, Z' changes from positive on (-3,-2) to negative on (-2,-1), confirming a local maximum. A tempting distractor is choice A, which claims a local minimum at x=-2, but this fails because the sign change is from positive to negative, not the reverse. To apply this generally, always identify points where the derivative might be zero and check the signs immediately to the left and right for extrema.
A differentiable function E has E′(x)>0 on (−9,−7) and E′(x)<0 on (−7,−3). Where does E have a local maximum?
Explanation: This problem assesses the First Derivative Test. The First Derivative Test determines local extrema by examining sign changes in the first derivative around critical points. If f' changes from positive to negative at a point, it indicates a local maximum there. If it changes from negative to positive, it indicates a local minimum, and no sign change means no extremum. In this case, at x=-7, E' changes from positive on (-9,-7) to negative on (-7,-3), confirming a local maximum. A tempting distractor is choice A, which claims a local minimum at x=-7, but this fails because the sign change is from positive to negative, not the reverse. To apply this generally, always identify points where the derivative might be zero and check the signs immediately to the left and right for extrema.
The derivative Y′(x) is positive on (0,4) and negative on (4,12). Where does Y have a local maximum?
Explanation: This problem assesses the First Derivative Test. The First Derivative Test determines local extrema by examining sign changes in the first derivative around critical points. If f' changes from positive to negative at a point, it indicates a local maximum there. If it changes from negative to positive, it indicates a local minimum, and no sign change means no extremum. In this case, at x=4, Y' changes from positive on (0,4) to negative on (4,12), confirming a local maximum. A tempting distractor is choice C, which claims a local minimum at x=4, but this fails because the sign change is from positive to negative, not the reverse. To apply this generally, always identify points where the derivative might be zero and check the signs immediately to the left and right for extrema.
For p, p′(x)>0 on (−2,0) and p′(x)>0 on (0,5). What happens at x=0?
Explanation: This problem involves the First Derivative Test to identify local extrema based on the sign of the derivative. The First Derivative Test requires a sign change in the derivative to indicate an extremum; no change means none. Here, p'(x) is positive on (-2,0) and positive on (0,5), showing no sign change at x=0. Thus, the function is increasing on both sides, so no local extremum at x=0. A tempting distractor is choice B, suggesting a local minimum at x=0, but without a sign change from negative to positive, this is incorrect. Remember, to apply this test effectively, always check for sign changes around critical points where the derivative is zero or undefined.
A function P satisfies P′(x)>0 on (−2,2) and P′(x)<0 on (2,2.5). Where does P have a local maximum?
Explanation: This problem assesses the First Derivative Test. The First Derivative Test determines local extrema by examining sign changes in the first derivative around critical points. If f' changes from positive to negative at a point, it indicates a local maximum there. If it changes from negative to positive, it indicates a local minimum, and no sign change means no extremum. In this case, at x=2, P' changes from positive on (-2,2) to negative on (2,2.5), confirming a local maximum. A tempting distractor is choice D, which claims a local minimum at x=2, but this fails because the sign change is from positive to negative, not the reverse. To apply this generally, always identify points where the derivative might be zero and check the signs immediately to the left and right for extrema.