First Derivative Test
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AP Calculus AB › First Derivative Test
The derivative $z'(x)$ is positive on $( -1,2)$ and negative on $(2,8)$. Where does $z$ have a local maximum?
At $x=2$, $z$ has a local minimum.
There is no local maximum.
At $x=-1$, $z$ has a local maximum.
At $x=2$, $z$ has a local maximum.
At $x=8$, $z$ has 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 $B'(x)>0$ on $( -2,3)$ and $B'(x)<0$ on $(3,7)$. Where does $B$ have a local maximum?
There is no local maximum.
At $x=7$, $B$ has a local maximum.
At $x=3$, $B$ has a local minimum.
At $x=3$, $B$ has a local maximum.
At $x=-2$, $B$ has 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=3, B' changes from positive on (-2,3) to negative on (3,7), confirming a local maximum. A tempting distractor is choice C, which claims a local minimum at x=3, 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 $N$, $N'(x)<0$ on $( -4,-1)$, $N'(x)>0$ on $(-1,3)$, and $N'(x)<0$ on $(3,6)$. Where is a local maximum?
At $x=3$, $N$ has a local maximum.
At $x=3$, $N$ has a local minimum.
At $x=-4$, $N$ has a local maximum.
At $x=-1$, $N$ has a local maximum.
There is no 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 a differentiable function $f$, $f'(x)>0$ on $(-3,1)$ and $f'(x)<0$ on $(1,4)$. Where does $f$ have a local maximum?
At $x=1$, $f$ has a local minimum.
At $x=1$, $f$ has a local maximum.
At $x=-3$, $f$ has a local maximum.
At $x=4$, $f$ has a local maximum.
There is no local extremum.
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. In this case, f'(x) is positive on (-3,1) and negative on (1,4), indicating a sign change from positive to negative at x=1. This confirms a local maximum at x=1, as the function increases before and decreases after this point. A tempting distractor is choice D, which suggests a local minimum at x=1, but that would require a change from negative to positive, 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.
The derivative $s'(x)$ is positive for $x<3$ and negative for $x>3$. Where does $s$ have a local extremum?
At $x=3$, $s$ has a local maximum.
At $x=0$, $s$ has a local minimum.
At $x=0$, $s$ has a local maximum.
At $x=3$, $s$ has a local minimum.
There is no local extremum.
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. In this case, s'(x) is positive for x<3 and negative for x>3, indicating a sign change from positive to negative at x=3. This confirms a local maximum at x=3, as the function increases before and decreases after this point. A tempting distractor is choice A, which suggests a local minimum at x=3, but that would require a change from negative to positive, 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.
A function $m$ has $m'(x)>0$ on $( -1,4)$ and $m'(x)<0$ on $(4,10)$. Where does $m$ have a local maximum?
At $x=4$, $m$ has a local maximum.
There is no local extremum.
At $x=4$, $m$ has a local minimum.
At $x=-1$, $m$ has a local maximum.
At $x=10$, $m$ has 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.
On $(-8,-3)$, $f'(x)>0$; on $(-3,2)$, $f'(x)<0$; on $(2,6)$, $f'(x)>0$. Where is a local minimum?
At $x=-3$, $f$ has a local minimum.
There is no local minimum.
At $x=-8$, $f$ has a local minimum.
At $x=2$, $f$ has a local maximum.
At $x=2$, $f$ has 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, f'(x) >0 on (-8,-3), <0 on (-3,2), and >0 on (2,6), showing - to + at x=2. This confirms a local minimum at x=2. A tempting distractor is choice C, claiming a max at x=2, but the sign change is - to +, indicating a min instead. Remember, to apply this test effectively, always check for sign changes around critical points where the derivative is zero or undefined.
A differentiable function $b$ has $b'(x)<0$ on $( -7,-2)$ and $b'(x)>0$ on $(-2,5)$. Where is a local minimum?
At $x=-7$, $b$ has a local minimum.
At $x=-2$, $b$ has a local minimum.
At $x=5$, $b$ has a local minimum.
At $x=-2$, $b$ has a local maximum.
There is no 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, b'(x) is negative on (-7,-2) and positive on (-2,5), 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 B, 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 $v'(x)$ is negative on $( -3,0)$ and positive on $(0,2)$. What occurs at $x=0$?
At $x=-3$, $v$ has a local minimum.
At $x=0$, $v$ has no local extremum.
At $x=0$, $v$ has a local minimum.
At $x=0$, $v$ has a local maximum.
At $x=2$, $v$ has 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 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.
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?
There is no local maximum.
At $x=2$, $P$ has a local minimum.
At $x=-2$, $P$ has a local maximum.
At $x=2.5$, $P$ has a local maximum.
At $x=2$, $P$ has 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.