The First Derivative Test determines local extrema by examining derivative sign changes. For function g, g'(x) > 0 on (2, 5) and g'(x) < 0 on (5, 9), showing the derivative changes from positive to negative at x = 5. This positive-to-negative transition indicates the function increases before x = 5 and decreases after x = 5, confirming a local maximum at x = 5. Choice A incorrectly suggests a minimum at x = 5, but the sign pattern (+ to -) specifically characterizes maxima, not minima. When f' transitions from positive to negative at a critical point, that point represents the function's local peak.

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.

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.

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.

The First Derivative Test identifies local extrema through derivative sign changes around critical points. For function t, t'(x) > 0 on (-9, -5), t'(x) < 0 on (-5, -2), and t'(x) > 0 on (-2, 0), showing sign changes at both x = -5 and x = -2. At x = -5, 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 = -5, but the question asks specifically for the local minimum, which occurs at x = -2. Always identify the type of extremum that matches the sign change pattern at each critical point.

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.

The First Derivative Test determines local extrema by examining sign changes in the derivative. Given $r'(x) < 0$ on $(-3, 2)$ and $r'(x) > 0$ on $(2, 7)$, 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 $f'$ changes from negative to positive at a critical point, that point is always a local minimum.

The First Derivative Test identifies local extrema through derivative sign changes. Since g'(x) > 0 on (-2, 1) and g'(x) < 0 on (1, 4), the derivative transitions from positive to negative at x = 1. This positive-to-negative sign change means the function increases before x = 1 and decreases after x = 1, establishing a local maximum at x = 1. Choice A incorrectly suggests a minimum at x = 1, but the sign pattern (+ to -) definitively indicates a maximum, not a minimum. To apply this test effectively, remember that a sign change from positive to negative at a critical point always signals a local maximum.

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.

The First Derivative Test requires a sign change in f'(x) to establish local extrema. Here, h'(x) is negative on (-5, -2), zero at x = -2, and negative on (-2, 3), showing no sign change around x = -2. Since h'(x) remains negative on both sides of x = -2, the function continues decreasing through this point without forming an extremum. Choice B might appear correct since h'(-2) = 0, but a zero derivative alone doesn't guarantee an extremum—the derivative must change sign. When f'(x) maintains the same sign on both sides of a critical point, no local extremum occurs there.