First Derivative Test - AP Calculus BC

For $f$ , $f'(x)>0$ on $( -1,4)$ , $f'(4)=0$ , and $f'(x)<0$ on $(4,6)$ ; where is a local maximum?
Answer: A local maximum at $x=4$ The First Derivative Test locates extrema by analyzing derivative sign changes around critical points. Given f'(x) > 0 on (-1, 4), f'(4) = 0, and f'(x) < 0 on (4, 6), the derivative changes from positive to negative at x = 4. This positive-to-negative transition indicates the function increases before x = 4 and decreases afterward, establishing a local maximum at x = 4. Choice A incorrectly identifies a minimum at x = 4, but the sign pattern (+ to -) specifically indicates a maximum, not a minimum. When f' changes from positive to negative at a critical point, that point represents the function's local peak.
For $v$ , $v'(x)>0$ on $( -6,-1)$ and $v'(x)<0$ on $(-1,2)$ ; where is a local maximum?
Answer: A local maximum at $x=-1$ The First Derivative Test determines local extrema by examining sign changes in f'(x). For function v, v'(x) > 0 on (-6, -1) and v'(x) < 0 on (-1, 2), showing the derivative changes from positive to negative at x = -1. This positive-to-negative transition indicates the function increases before x = -1 and decreases after x = -1, establishing a local maximum at x = -1. Choice B incorrectly suggests a minimum at x = -1, but the sign pattern (+ to -) specifically characterizes maxima, not minima. When f' changes from positive to negative at a critical point, that point represents the function's local peak.
For $v$ , $v'(x)<0$ on $( -7,-3)$ , $v'(x)>0$ on $(-3,2)$ , and $v'(x)>0$ on $(2,6)$ ; what occurs at $x=-3$ ?
Answer: A local minimum at $x=-3$ The First Derivative Test determines local extrema by examining sign changes in the derivative. For function v, v'(x) < 0 on (-7, -3), v'(x) > 0 on (-3, 2), and v'(x) > 0 on (2, 6), showing the derivative transitions from negative to positive at x = -3. This negative-to-positive sign change means the function decreases before x = -3 and increases afterward, confirming a local minimum at x = -3. Choice A incorrectly suggests a maximum at x = -3, but the sign change pattern (- to +) definitively indicates a minimum, not a maximum. When applying this test, remember that a sign change from negative to positive always produces a local minimum.
On $(0,2)$ , $f'(x)<0$ ; on $(2,6)$ , $f'(x)>0$ ; and $f'(2)=0$ . Where is a local minimum?
Answer: A local minimum at $x=2$ The First Derivative Test identifies local extrema through derivative sign changes around critical points. Given f'(x) < 0 on (0, 2), f'(2) = 0, and f'(x) > 0 on (2, 6), the derivative changes from negative to positive at x = 2. This negative-to-positive transition indicates the function decreases before x = 2 and increases afterward, confirming a local minimum at x = 2. Choice A incorrectly identifies a maximum at x = 2, but the sign pattern (- to +) specifically characterizes minima, not maxima. When applying this test, always match the sign change pattern: negative-to-positive signals a minimum, positive-to-negative signals a maximum.
For differentiable $f$ on $(-2,4)$ , $f'(x)<0$ on $(-2,1)$ and $f'(x)>0$ on $(1,4)$ . Where is a local minimum?
Answer: A local minimum at $x=1$ This problem requires applying the First Derivative Test to identify local extrema. When f'(x) < 0 on (-2,1), the function f is decreasing on this interval. When f'(x) > 0 on (1,4), the function f is increasing on this interval. At x = 1, the derivative changes from negative to positive, which means f changes from decreasing to increasing—this is the signature of a local minimum. Some students might incorrectly choose option A (local maximum at x = 1) by confusing the sign change pattern. Remember: when f' changes from negative to positive at a critical point, you have a local minimum; when f' changes from positive to negative, you have a local maximum.
For $g$ , $g'(x)>0$ on $( -6,-3)$ and $g'(x)<0$ on $(-3,0)$ ; where is a local maximum?
Answer: A local maximum at $x=-3$ The First Derivative Test determines local extrema by examining sign changes in f'(x). For function g, g'(x) > 0 on (-6, -3) and g'(x) < 0 on (-3, 0), showing the derivative changes from positive to negative at x = -3. This positive-to-negative transition indicates the function increases before x = -3 and decreases after x = -3, establishing a local maximum at x = -3. Choice A incorrectly identifies a minimum at x = -3, but the sign pattern (+ to -) specifically characterizes maxima, not minima. When f' changes from positive to negative at a critical point, that point represents the function's local peak.
For $u$ , $u'(x)>0$ on $( -1,2)$ , $u'(2)=0$ , and $u'(x)<0$ on $(2,8)$ ; where is a local maximum?
Answer: A local maximum at $x=2$ The First Derivative Test determines local extrema by examining sign changes in f'(x). For function u, u'(x) > 0 on (-1, 2), u'(2) = 0, and u'(x) < 0 on (2, 8), showing the derivative transitions from positive to negative at x = 2. This positive-to-negative sign change indicates the function increases before x = 2 and decreases after x = 2, establishing a local maximum at x = 2. Choice A incorrectly identifies a minimum at x = 2, but the sign change pattern (+ to -) definitively characterizes maxima, not minima. When f' changes from positive to negative at a critical point, that point represents the function's local peak.
A differentiable $g$ has $g'(x)<0$ on $( -2,2)$ , $g'(2)=0$ , and $g'(x)>0$ on $(2,9)$ ; where is a local minimum?
Answer: A local minimum at $x=2$ The First Derivative Test determines local extrema by examining sign changes in the derivative. For function g, g'(x) < 0 on (-2, 2), g'(2) = 0, and g'(x) > 0 on (2, 9), showing the derivative transitions from negative to positive at x = 2. This negative-to-positive sign change means the function decreases before x = 2 and increases afterward, confirming a local minimum at x = 2. Choice B incorrectly suggests a maximum at x = 2, but the sign change pattern (- to +) definitively indicates a minimum, not a maximum. When applying this test, remember that a sign change from negative to positive always produces a local minimum.
For $s$ , $s'(x)>0$ on $(-2,0)$ and $s'(x)<0$ on $(0,3)$ with $s'(0)=0$ ; what occurs at $x=0$ ?
Answer: A local maximum at $x=0$ The First Derivative Test identifies local extrema through derivative sign changes. Given s'(x) > 0 on (-2, 0), s'(0) = 0, and s'(x) < 0 on (0, 3), the derivative transitions from positive to negative at x = 0. This positive-to-negative sign change indicates the function increases before x = 0 and decreases afterward, establishing a local maximum at x = 0. Choice B incorrectly concludes no extremum exists, but the clear sign change from positive to negative definitively creates a maximum at x = 0. When f' changes from positive to negative at a critical point, that point is always a local maximum.
For $u$ , $u'(x)>0$ on $( -8,-4)$ , $u'(x)<0$ on $(-4,2)$ , and $u'(x)>0$ on $(2,6)$ ; where is a local maximum?
Answer: A local maximum at $x=-4$ The First Derivative Test locates extrema by analyzing derivative sign changes around critical points. For function u, u'(x) > 0 on (-8, -4), u'(x) < 0 on (-4, 2), and u'(x) > 0 on (2, 6), showing sign changes at both x = -4 and x = 2. At x = -4, the derivative changes from positive to negative (indicating a maximum), while at x = 2, it changes from negative to positive (indicating a minimum). Choice A incorrectly identifies a maximum at x = 2, but the sign change at x = 2 is negative-to-positive, which indicates a minimum, while the maximum occurs at x = -4 where the sign change is positive-to-negative. Always match the sign change pattern to the correct extremum type.
For a differentiable function $f$ , $f'(x)<0$ on $(-3,-1)$ and $f'(x)>0$ on $(-1,2)$ ; where is a local minimum?
Answer: A local minimum at $x=-1$ The First Derivative Test determines local extrema by analyzing sign changes in f'(x). Given that f'(x) < 0 on (-3, -1) and f'(x) > 0 on (-1, 2), the derivative changes from negative to positive at x = -1. This sign change from negative to positive indicates the function is decreasing before x = -1 and increasing after x = -1, confirming a local minimum at x = -1. Choice A might seem tempting since x = -1 is a critical point, but the sign change pattern (- to +) specifically indicates a minimum, not a maximum. When f' changes from negative to positive at a critical point, always conclude there's a local minimum.
For $r$ , $r'(x)<0$ on $(-6,-3)$ , $r'(x)>0$ on $(-3,1)$ , and $r'(x)<0$ on $(1,4)$ ; where is a local maximum?
Answer: A local maximum at $x=1$ The First Derivative Test determines extrema by examining sign changes in the derivative. For function r, r'(x) < 0 on (-6, -3), r'(x) > 0 on (-3, 1), and r'(x) < 0 on (1, 4), showing the derivative changes from positive to negative at x = 1. This positive-to-negative transition indicates the function increases before x = 1 and decreases after x = 1, confirming a local maximum at x = 1. Choice D incorrectly identifies a minimum at x = -3, but at x = -3 the sign change is negative-to-positive (indicating a minimum), while at x = 1 it's positive-to-negative (indicating the maximum). Focus on the specific sign change pattern at each critical point to identify the correct extremum type.
A differentiable function $q$ satisfies $q'(x)<0$ on $(1,3)$ and $q'(x)>0$ on $(3,7)$ . What occurs at $x=3$ ?
Answer: A local minimum at $x=3$ Applying the First Derivative Test requires examining how the derivative's sign changes at critical points. The function q has q'(x) < 0 on (1,3), so q is decreasing before x = 3. Then q'(x) > 0 on (3,7), so q is increasing after x = 3. This negative-to-positive sign change in q' means the function transitions from decreasing to increasing at x = 3, which is the defining characteristic of a local minimum. Option B (local maximum) would require the opposite sign change pattern. Remember the mnemonic: "negative to positive" means the function goes from falling to rising, creating a valley (minimum).
A differentiable function $g$ has $g'(x)>0$ on $(-3,0)$ and $g'(x)<0$ on $(0,5)$ . What occurs at $x=0$ ?
Answer: A local maximum at $x=0$ The First Derivative Test helps us classify critical points by examining sign changes in the derivative. Since g'(x) > 0 on (-3,0), the function g is increasing before x = 0. Since g'(x) < 0 on (0,5), the function g is decreasing after x = 0. This means g changes from increasing to decreasing at x = 0, which indicates a local maximum. Students might mistakenly choose option A (local minimum) by reversing the interpretation of the sign change. The key strategy: positive-to-negative derivative change means local maximum, while negative-to-positive means local minimum.
A differentiable $v$ has $v'(x)>0$ on $( -1,1)$ , $v'(1)=0$ , and $v'(x)<0$ on $(1,3)$ ; where is a local maximum?
Answer: A local maximum at $x=1$ The First Derivative Test determines local extrema by examining sign changes in the derivative. Since v'(x) > 0 on (-1, 1), v'(1) = 0, and v'(x) < 0 on (1, 3), the derivative changes from positive to negative at x = 1. This positive-to-negative transition indicates the function increases before x = 1 and decreases afterward, establishing a local maximum at x = 1. Choice B incorrectly suggests a minimum at x = 1, but the sign pattern (+ to -) specifically characterizes maxima, not minima. When f' changes from positive to negative at a critical point, that point represents the function's local peak.
Let $h$ be differentiable on $(0,6)$ . If $h'(x)<0$ on $(0,2)$ , $h'(x)>0$ on $(2,5)$ , and $h'(x)<0$ on $(5,6)$ , what happens at $x=5$ ?
Answer: A local maximum at $x=5$ This problem tests the First Derivative Test with multiple critical points. The function h decreases on (0,2) since h'(x) < 0, increases on (2,5) since h'(x) > 0, and decreases again on (5,6) since h'(x) < 0. At x = 5, the derivative changes from positive to negative, indicating that h changes from increasing to decreasing. This positive-to-negative sign change in h' confirms a local maximum at x = 5. Option A (local minimum at x = 5) represents a common error where students confuse the sign change pattern. Always check the derivative's sign change: positive-to-negative yields a maximum, negative-to-positive yields a minimum.
A function $p$ has $p'(x)<0$ on $( -7,-1)$ and $p'(x)>0$ on $(-1,1)$ ; where is a local minimum?
Answer: A local minimum at $x=-1$ The First Derivative Test identifies local extrema through derivative sign changes around critical points. Since p'(x) < 0 on (-7, -1) and p'(x) > 0 on (-1, 1), the derivative transitions from negative to positive at x = -1. This negative-to-positive sign change means the function decreases before x = -1 and increases afterward, confirming a local minimum at x = -1. Choice B incorrectly suggests a maximum at x = -1, but the sign change pattern (- to +) definitively indicates a minimum, not a maximum. When applying this test, always remember that negative-to-positive sign changes produce local minima.
For $w$ , $w'(x)>0$ on $(-7,-2)$ and $w'(x)<0$ on $(-2,-1)$ ; where does a local maximum occur?
Answer: A local maximum at $x=-2$ The First Derivative Test identifies local extrema through derivative sign changes around critical points. Since w'(x) > 0 on (-7, -2) and w'(x) < 0 on (-2, -1), the derivative changes from positive to negative at x = -2. This positive-to-negative transition indicates the function increases before x = -2 and decreases afterward, confirming a local maximum at x = -2. Choice B incorrectly suggests a minimum at x = -2, but the sign pattern (+ to -) specifically indicates a maximum, not a minimum. When applying this test, always remember that a positive-to-negative sign change produces a local maximum.
A function $s$ is differentiable on $( -5,5 )$ . If $s'(x)<0$ on $(-5,-1)$ and $s'(x)<0$ on $(-1,5)$ , what can be concluded?
Answer: No local extremum at $x=-1$ The First Derivative Test requires a sign change in the derivative to establish a local extremum. Here, s'(x) < 0 on both (-5,-1) and (-1,5), meaning s is decreasing on both intervals. Since the derivative maintains the same negative sign on both sides of x = -1, there is no sign change at this point. Without a sign change in s', the First Derivative Test tells us there is no local extremum at x = -1. Students might incorrectly think that the derivative being defined piecewise on intervals automatically creates an extremum at the boundary. The crucial insight: continuity of the derivative's sign (staying negative) through x = -1 means the function continues decreasing, with no local extremum.
A function $q$ has $q'(x)>0$ on $( -10,-8)$ , $q'(x)<0$ on $(-8,-4)$ , and $q'(x)<0$ on $(-4,0)$ ; what occurs at $x=-8$ ?
Answer: A local maximum at $x=-8$ The First Derivative Test locates extrema by analyzing derivative sign changes around critical points. For function q, q'(x) > 0 on (-10, -8), q'(x) < 0 on (-8, -4), and q'(x) < 0 on (-4, 0), showing the derivative changes from positive to negative at x = -8. This positive-to-negative transition indicates the function increases before x = -8 and decreases afterward, establishing a local maximum at x = -8. Choice C incorrectly concludes no extremum exists at x = -8, but the clear sign change from positive to negative definitively creates a maximum there. When f' changes from positive to negative at a critical point, that point is always a local maximum.

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