On , is decreasing for , increasing for , then decreasing for ; where is concave up?
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AP Calculus BC Quiz
Practice Concavity Of Functions Over Their Domains in AP Calculus BC with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.
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On (−6,6), f′(x) is decreasing for x<−2, increasing for −2<x<3, then decreasing for x>3; where is f concave up?
This quiz focuses on Concavity Of Functions Over Their Domains, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus BC.
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.
On (−6,6), f′(x) is decreasing for x<−2, increasing for −2<x<3, then decreasing for x>3; where is f concave up?
Explanation: This problem requires analyzing the concavity of a function over its domain based on the monotonicity of the first derivative. The function f is concave up where f'(x) is increasing and concave down where f'(x) is decreasing. Given that f'(x) is decreasing on (-6,-2) and (3,6), and increasing on (-2,3), this means concave down on (-6,-2) ∪ (3,6) and concave up on (-2,3). These behaviors directly inform the concavity without needing the second derivative explicitly. A tempting distractor like choice A swaps the concavity intervals, failing because it incorrectly matches increasing f' with concave down instead of up. In general, map regions where f' increases (concave up) and decreases (concave down) using given monotonicity to analyze concavity effectively.
Given f′′(x)<0 for x<1 and f′′(x)>0 for x>1, where is f concave down?
Explanation: This question tests the skill of analyzing the concavity of functions over their domains using the second derivative test. The second derivative f'' determines concavity: when f''(x) < 0, the function is concave down because the slope of the tangent lines is decreasing. In this case, f''(x) < 0 for x < 1, directly indicating concave down on (-∞,1), while f''(x) > 0 for x > 1 means concave up there. The point x=1 is likely an inflection point where concavity changes. A tempting distractor like choice A, which picks (1,∞) for concave down, fails because it confuses the signs and mistakenly assigns concave down to where f'' > 0. Always remember the transferable strategy: test the sign of the second derivative or check if f' is increasing to determine where the function is concave up.
Suppose f′′(x)=0 at x=1 and x=4, with f′′(x)>0 on (1,4) and f′′(x)<0 elsewhere; where is f concave down?
Explanation: This question tests the skill of analyzing the concavity of functions over their domains. The second derivative's sign dictates concavity: positive for up, where the function holds water, and negative for down. Zeros at x=1 and x=4 mark potential inflection points, but the sign changes determine the intervals. With f''(x) > 0 on (1,4) (concave up) and < 0 elsewhere (concave down), concave down is on (-∞,1) ∪ (4,∞). A tempting distractor like choice B omits parts of the down intervals, failing to account for the full 'elsewhere'. Always link the sign of f''(x) or the monotonicity of f'(x) to concavity: positive for up, negative for down.
For −5<x<5, f′′(x)=−(x+1)(x−4); on which intervals is f concave up and concave down?
Explanation: This problem requires analyzing the concavity of a function over its domain using the second derivative. The second derivative f''(x) = -(x+1)(x-4) has roots at x=-1 and x=4, marking potential inflection points. The sign of f''(x) is positive between -1 and 4 (since the parabola opens downward) and negative outside, indicating concave up on (-1,4) and concave down on (-5,-1) ∪ (4,5) within the domain. This sign analysis confirms the intervals accurately. A tempting distractor like choice A reverses the concavity, failing because it misidentifies the regions where f'' is positive versus negative. In general, construct a sign chart for f''(x) by identifying roots and testing points to determine concave up (positive) and concave down (negative) intervals.
For −4<x<4, f′′(x)=3x(x−2); on which intervals is f concave up and concave down?
Explanation: This problem requires analyzing the concavity of a function over its domain using the second derivative. The second derivative f''(x) = 3x(x-2) has roots at x=0 and x=2, which are potential inflection points. To determine concavity, examine the sign of f''(x): it is positive when x < 0 or x > 2, indicating concave up, and negative when 0 < x < 2, indicating concave down. Within the interval (-4,4), this means concave up on (-4,0) ∪ (2,4) and concave down on (0,2). A tempting distractor like choice B swaps the intervals, failing because it incorrectly identifies where f'' is positive versus negative. In general, create a sign chart for f''(x) by testing intervals between roots to reliably determine where the function is concave up (positive) or down (negative).
On (−3,5), f′(x) increases on (−3,1), is constant on (1,3), and decreases on (3,5); where is f concave down?
Explanation: This problem requires analyzing the concavity of a function over its domain based on the behavior of the first derivative. The function is concave up where f'(x) is increasing and concave down where it is decreasing, with constant f' implying f'' = 0, often included in non-strict concavity. Given f' increases on (-3,1), constant on (1,3), and decreases on (3,5), this suggests concave up on (-3,1) ∪ (1,3) and concave down on (3,5). The constant interval is grouped with up in this context. A tempting distractor like choice A reverses parts, failing by misplacing the constant interval's concavity. In general, classify intervals where f' increases or is constant (concave up) versus decreases (concave down) for comprehensive analysis.
If f′′(x)=4−2x for all x, on which interval is f concave down?
Explanation: This question tests the skill of analyzing the concavity of functions over their domains using the second derivative test. Concave down occurs where f''(x) < 0, meaning the first derivative is decreasing and the graph bends downward. Solving 4 - 2x < 0 gives x > 2, so f is concave down on (2,∞), while for x < 2, f'' > 0 indicates concave up. The point x=2 is where concavity changes, an inflection point. A tempting distractor like choice A, which selects (-∞,2) for concave down, fails by reversing the inequality and confusing where f'' is negative. Always remember the transferable strategy: test the sign of the second derivative or check if f' is increasing to determine where the function is concave up.
For a differentiable f, f′(x) is increasing on (−4,−1), decreasing on (−1,2), increasing on (2,5); where is f concave up?
Explanation: This question tests the skill of analyzing the concavity of functions over their domains using the behavior of the first derivative. Concavity is determined by whether the first derivative f' is increasing or decreasing, as an increasing f' indicates the function's slope is rising, meaning the graph bends upward. Specifically, f is concave up where f' is increasing, which occurs on intervals where the second derivative would be positive if it exists. Here, f' is increasing on (-4,-1) and (2,5), so f is concave up there, while it is decreasing on (-1,2), indicating concave down. A tempting distractor like choice C, which suggests concave up on the entire (-4,5), fails because it ignores the interval (-1,2) where f' is decreasing and thus f is concave down. Always remember the transferable strategy: test the sign of the second derivative or check if f' is increasing to determine where the function is concave up.
A differentiable f has f′(x) increasing on (−∞,0) and constant on (0,3); where is f concave up?
Explanation: This question tests the skill of analyzing the concavity of functions over their domains using the behavior of the first derivative. A function is concave up where its first derivative f' is increasing, as this shows the slopes are rising and the graph bends upward. Here, f' is increasing on (-∞,0), so concave up there, but constant on (0,3), meaning no change in slope and thus neither concave up nor down, as f is linear on that interval. The transition at x=0 may indicate an inflection or change in concavity behavior. A tempting distractor like choice D, which includes (-∞,0) ∪ (0,3) for concave up, fails because a constant f' implies f'' = 0, not positive, so it is not concave up. Always remember the transferable strategy: test the sign of the second derivative or check if f' is increasing to determine where the function is concave up.
If f′′(x)=x(x−1)(x+2), on which intervals is f concave down?
Explanation: This question tests the skill of analyzing the concavity of functions over their domains using the second derivative test. Concave down is where f''(x) < 0, indicating decreasing tangent slopes and a downward bend in the graph. For f''(x) = x(x-1)(x+2), a cubic with roots at -2,0,1 and positive leading coefficient, sign analysis shows negative on (-∞,-2) and (0,1), so concave down there. It is positive on (-2,0) and (1,∞), meaning concave up in those intervals. A tempting distractor like choice C, which picks (-∞,-2) ∪ (1,∞) for concave down, fails by misidentifying the signs in the intervals, particularly confusing positive regions for negative. Always remember the transferable strategy: test the sign of the second derivative or check if f' is increasing to determine where the function is concave up.
Given f′′(x)=sinx for 0<x<2π, on which intervals is f concave up and concave down?
Explanation: This question tests concavity analysis when f''(x) = sin x on the interval (0, 2π). To determine concavity, we need to find where sin x is positive (concave up) or negative (concave down). On the interval (0, 2π), sin x > 0 when 0 < x < π, and sin x < 0 when π < x < 2π. Therefore, f is concave up on (0, π) and concave down on (π, 2π). Students might confuse this with the intervals where sin x is increasing or decreasing, or with the intervals for cos x. The key strategy: for trigonometric second derivatives, carefully identify where the function is positive versus negative over the given domain.
Given f′′(x)=(x−2)2x for x=2, where is f concave up and concave down?
Explanation: This question requires analyzing f''(x) = x/(x-2)² where x ≠ 2 to determine concavity. The denominator (x-2)² is always positive for x ≠ 2, so the sign of f'' depends only on the numerator x. When x < 0, f''(x) < 0 (concave down); when x > 0 and x ≠ 2, f''(x) > 0 (concave up). Note that x = 2 is not in the domain, creating a vertical asymptote that doesn't affect concavity on either side. Students might incorrectly think the function changes concavity at x = 2, but it doesn't - the function remains concave up on both (0,2) and (2,∞). The strategy: when analyzing rational functions, check sign changes only at zeros of the numerator and consider domain restrictions separately.
If f′(x) is increasing on (−4,0) and decreasing on (0,3), where is f concave up and concave down?
Explanation: This problem requires understanding the relationship between the behavior of f'(x) and the concavity of f(x). When f'(x) is increasing, this means f''(x) > 0, so f is concave up; when f'(x) is decreasing, f''(x) < 0, so f is concave down. Since f'(x) is increasing on (-4,0), f is concave up on (-4,0). Since f'(x) is decreasing on (0,3), f is concave down on (0,3). Students often confuse the relationship and think increasing f' means f is increasing, not concave up. The key insight: the monotonicity of the first derivative determines the sign of the second derivative, which determines concavity.
If f′(x)<0 on (−3,1) and f′(x) is increasing there, where is f concave up or down?
Explanation: This problem combines information about f'(x) being negative with f'(x) being increasing to determine concavity. The key insight is that if f'(x) is increasing, then f''(x) > 0, which means f is concave up. The fact that f'(x) < 0 tells us f is decreasing, but this doesn't affect concavity - concavity depends on whether f'(x) is increasing or decreasing, not on its sign. Since f'(x) is increasing on (-3,1), f is concave up on (-3,1). Students often confuse f being decreasing (f' < 0) with f being concave down (f'' < 0), but these are independent properties. Remember: concavity is about the rate of change of the slope, not the slope itself.
Suppose f′′(x)=x21−1 for x=0; where is f concave up?
Explanation: This question tests the skill of analyzing the concavity of functions over their domains using the second derivative test. The sign of f''(x) dictates concavity: positive values mean concave up, as the function's graph lies above its tangent lines. For f''(x) = 1/x² - 1 > 0 when 1/x² > 1, which holds for |x| < 1 (excluding x=0), so on (-1,0) ∪ (0,1). Outside |x| > 1, f''(x) < 0, indicating concave down. A tempting distractor like choice A, which picks (-1,1) including x=0, fails because f'' is undefined at x=0, so concavity cannot be assessed there. Always remember the transferable strategy: test the sign of the second derivative or check if f' is increasing to determine where the function is concave up.
If f′′(x)=(x+2)(x−3) for all x, on which intervals is f concave up?
Explanation: This question tests the skill of analyzing the concavity of functions over their domains using the second derivative test. To determine concavity, find where f''(x) > 0 for concave up, as a positive second derivative means the first derivative is increasing and the graph bends upward. Here, f''(x) = (x+2)(x-3) is a quadratic that opens upward with roots at x=-2 and x=3, so it is positive outside the roots on (-∞,-2) ∪ (3,∞). It is negative between the roots on (-2,3), indicating concave down there. A tempting distractor like choice A, which selects (-2,3) for concave up, fails because that is where f'' < 0, confusing the sign analysis. Always remember the transferable strategy: test the sign of the second derivative or check if f' is increasing to determine where the function is concave up.
For twice-differentiable f, f′′(x)>0 on (−3,0) and (2,6), and f′′(x)<0 elsewhere; where is f concave up?
Explanation: This question tests the skill of analyzing the concavity of functions over their domains using the given signs of the second derivative. Concave up is directly where f''(x) > 0, as this means the tangent slopes are increasing and the graph curves upward. The problem states f'' > 0 on (-3,0) and (2,6), so concave up there, and f'' < 0 elsewhere, meaning concave down outside those intervals. These intervals are separated, indicating possible inflection points at x=-3,0,2,6. A tempting distractor like choice C, which merges to (-3,6) for concave up, fails because it includes (0,2) where f'' < 0 and f is actually concave down. Always remember the transferable strategy: test the sign of the second derivative or check if f' is increasing to determine where the function is concave up.
For 0<x<8, f′′(x)=x(x−3)(x−6); where is f concave up and concave down?
Explanation: This problem requires analyzing the concavity of a function over its domain using the second derivative. The second derivative f''(x) = (x-3)(x-6)/x has critical points at x=3, x=6, and a discontinuity at x=0, but within (0,8). Sign analysis shows f'' > 0 on (0,3) ∪ (6,8) and f'' < 0 on (3,6), indicating concave up there and concave down on (3,6). Testing points in each subinterval confirms this. A tempting distractor like choice B reverses the concavity, failing because it ignores the positive product in (0,3) and (6,8). In general, build a sign chart considering roots and discontinuities to pinpoint where f'' is positive (concave up) or negative (concave down).
For all real x, f′′(x)=2sinx; on which intervals is f concave up and concave down on [0,2π]?
Explanation: This problem requires analyzing the concavity of a function over its domain using the second derivative. The second derivative f''(x) = 2 sin x is positive where sin x > 0 (0, π) and negative where sin x < 0 (π, 2π) on [0, 2π]. This indicates concave up on (0, π) and concave down on (π, 2π). The periodic nature of sine determines the signs clearly. A tempting distractor like choice B reverses the intervals, failing because sin x is positive in the first half and negative in the second. In general, evaluate the sign of f''(x) over intervals to identify concave up (positive) and concave down (negative) regions reliably.
A differentiable f has f′(x)>0 on (−2,4) and f′(x) decreasing on (−2,1), increasing on (1,4); where is f concave up?
Explanation: This question tests the skill of analyzing the concavity of functions over their domains using the behavior of the first derivative. Concavity is revealed by the monotonicity of f': where f' is increasing, f is concave up because the slopes are getting steeper, corresponding to a positive second derivative. In this scenario, f' is decreasing on (-2,1), indicating concave down, and increasing on (1,4), indicating concave up, while f' > 0 overall means f is increasing but does not directly affect concavity. The point x=1 is where the behavior of f' changes, likely an inflection point. A tempting distractor like choice C, which chooses the entire (-2,4) for concave up, fails because it overlooks the decreasing part of f' on (-2,1) where f is actually concave down. Always remember the transferable strategy: test the sign of the second derivative or check if f' is increasing to determine where the function is concave up.