Concavity of Functions Over Their Domains - AP Calculus AB

1. If $f''(x)>0$ for $x<1$ and $f''(x)<0$ for $x>1$, where is $f$ concave up and concave down?
   Answer: Concave up on $(-\infty,1)$; concave down on $(1,\infty)$ This question tests your understanding of how the second derivative determines concavity. When f''(x) > 0, the function is concave up (shaped like a cup), and when f''(x) < 0, the function is concave down (shaped like a cap). Since f''(x) > 0 for x < 1, the function is concave up on (-∞, 1), and since f''(x) < 0 for x > 1, the function is concave down on (1, ∞). Choice E incorrectly relates concavity to the first derivative's sign rather than the second derivative. Remember: positive second derivative means concave up, negative second derivative means concave down.
2. On $[1,7]$, $f'(x)$ increases on $(1,4)$, is constant on $(4,5)$, and decreases on $(5,7)$; where is $f$ concave up?
   Answer: Concave up on $(1,4)$; concave down on $(5,7)$ This question analyzes concavity based on the behavior of f'(x) over different intervals. When f'(x) is increasing, f''(x) > 0 and f is concave up; when f'(x) is decreasing, f''(x) < 0 and f is concave down. Since f'(x) increases on (1, 4), the function is concave up on (1, 4), and since f'(x) is constant on (4, 5), f''(x) = 0 there (neither concave up nor down). Choice A incorrectly suggests f is concave down on (4, 7), but we only know f' decreases on (5, 7). The key: constant f' means zero concavity.
3. If $f'(x)$ decreases for $x<0$ and increases for $x>0$, where is $f$ concave up and concave down?​​
   Answer: Concave down on $(-\infty,0)$; concave up on $(0,\infty)$ This question tests concavity through the monotonicity of the first derivative. When f'(x) is decreasing, f''(x) < 0, so f is concave down; when f'(x) is increasing, f''(x) > 0, so f is concave up. Since f'(x) decreases for x < 0, the function is concave down on (-∞, 0), and since f'(x) increases for x > 0, the function is concave up on (0, ∞). Choice A reverses these intervals, confusing the relationship between f' behavior and concavity. The mnemonic: decreasing slopes mean downward curvature.
4. Given $f''(x)<0$ on $(-2,3)$ and $f''(x)>0$ on $(3,6)$, where is $f$ concave down?
   Answer: Concave down on $(-2,3)$; concave up on $(3,6)$ This problem asks you to identify where f is concave down based on given information about f''(x). A function is concave down wherever its second derivative is negative. We're told that f''(x) < 0 on (-2, 3), which means f is concave down on the interval (-2, 3). On the interval (3, 6), we have f''(x) > 0, so f is concave up there, not concave down. Choice A reverses the concavity intervals, incorrectly stating that f is concave down on (3, 6). The key strategy: concave down occurs exactly where f''(x) < 0.
5. For a differentiable $f$, $f'(x)$ is increasing on $(-3,1)$ and decreasing on $(1,4)$; where is $f$ concave up?
   Answer: $(-3,1)$ only This question tests understanding of concavity through the behavior of the first derivative. A function f is concave up where f''(x) > 0, which occurs when f'(x) is increasing. Since f'(x) is increasing on (-3,1) and decreasing on (1,4), we have f''(x) > 0 on (-3,1) and f''(x) < 0 on (1,4). Therefore, f is concave up only on the interval (-3,1). A common error is thinking f is concave up on both intervals or on their union, but decreasing f'(x) means negative f''(x) and thus concave down. Remember: when f' increases, f'' is positive and f is concave up.
6. Suppose $f''(x)<0$ on $( -5,-1)$ and $(2,6)$, and $f''(x)>0$ on $(-1,2)$; where is $f$ concave up?
   Answer: Concave up on $(-1,2)$; concave down on $(-5,-1)$ and $(2,6)$ This question tests the skill of analyzing the concavity of functions over their domains. The second derivative f'' directly tells us about concavity: positive for concave up and negative for concave down. Given f'' < 0 on (-5,-1) and (2,6), f is concave down there. On (-1,2), f'' > 0, so f is concave up. Therefore, concave up occurs on (-1,2). A tempting distractor is choice A, which swaps the concavity directions. A transferable concavity-sign strategy is to always associate positive second derivative with concave up and negative with concave down across any domain.
7. Given $f''(x)>0$ on $( -10,-4)$ and $f''(x)<0$ on $(-4,3)$ and $(3,8)$, where is $f$ concave up?
   Answer: Concave up on $(-10,-4)$; concave down on $(-4,3)$ and $(3,8)$ This question tests the skill of analyzing the concavity of functions over their domains. Concavity is governed by the sign of f'': positive for concave up and negative for concave down. Here, f'' > 0 on (-10,-4), so concave up there. On (-4,3) and (3,8), f'' < 0, so concave down. The question asks for where f is concave up, which is (-10,-4). A tempting distractor is choice B, which incorrectly assigns concave up to the negative f'' intervals. A transferable concavity-sign strategy is to directly map positive f'' to concave up and negative to concave down in all cases.
8. Suppose $f''(x)<0$ on $(-5,-1)\cup(2,5)$ and $f''(x)>0$ on $(-1,2)$; where is $f$ concave up?
   Answer: Concave up on $(-1,2)$; concave down on $(-5,-1)\cup(2,5)$ This problem directly provides information about the sign of f''(x) on different intervals. When f''(x) > 0, the function is concave up, and when f''(x) < 0, the function is concave down. Since f''(x) > 0 on (-1, 2), the function is concave up on (-1, 2), and since f''(x) < 0 on (-5, -1) ∪ (2, 5), the function is concave down on those intervals. Choice A incorrectly swaps the concavity on these intervals, misinterpreting the relationship between f'' and concavity. Remember: the sign of the second derivative directly determines concavity direction.
9. Given $f''(x)=0$ at $x=-1,2$; $f''(x)>0$ on $(-\infty,-1)$ and $(2,\infty)$, $f''(x)<0$ on $(-1,2)$, where is $f$ concave down?​​
   Answer: Concave down on $(-1,2)$; concave up on $(-\infty,-1)\cup(2,\infty)$ This problem provides the sign of f''(x) on different intervals separated by points where f''(x) = 0. The function is concave up where f''(x) > 0 and concave down where f''(x) < 0. Since f''(x) < 0 on (-1, 2), the function is concave down specifically on the interval (-1, 2). Choice A incorrectly identifies where f is concave down, listing the intervals where f''(x) > 0 instead. Always match: negative second derivative corresponds to concave down intervals.
10. Given $f''(x)>0$ for $0<x<3$ and $f''(x)>0$ for $3<x<8$, where is $f$ concave down?
    Answer: Concave down on no interval; concave up on $(0,3)$ and $(3,8)$ This question tests the skill of analyzing the concavity of functions over their domains. The sign of the second derivative f'' determines concavity: positive indicates concave up. Here, f'' > 0 on both (0,3) and (3,8), so f is concave up everywhere in (0,8). There are no intervals where f'' < 0, meaning no concave down regions. The question asks specifically for where f is concave down, which is nowhere. A tempting distractor is choice E, which incorrectly assumes concave down throughout despite positive f''. A transferable concavity-sign strategy is to remember positive f'' always signals concave up, and negative signals concave down.
11. If $f''(x)>0$ on $(-4,-2)\cup(1,6)$ and $f''(x)<0$ on $(-2,1)$, where is $f$ concave down?
    Answer: Concave down on $(-2,1)$; concave up on $(-4,-2)\cup(1,6)$ This question asks where f is concave down given information about where f''(x) is positive and negative. A function is concave down wherever its second derivative is negative. We're told that f''(x) < 0 on (-2, 1), which means f is concave down on the interval (-2, 1). On the intervals (-4, -2) and (1, 6), we have f''(x) > 0, so f is concave up on these intervals, not concave down. Choice A incorrectly identifies the concave up intervals as concave down. Remember: concave down occurs exactly where f''(x) < 0.
12. Given $f''(x)=(x+2)(x-1)$ for all $x$, on which intervals is $f$ concave up and concave down?
    Answer: Concave up on $(-\infty,-2)\cup(1,\infty)$; concave down on $(-2,1)$ This problem tests your ability to determine concavity by analyzing the sign of the second derivative. Since f''(x) = (x+2)(x-1), we need to find where f''(x) > 0 (concave up) and where f''(x) < 0 (concave down). The second derivative equals zero at x = -2 and x = 1, dividing the number line into three intervals: (-∞,-2), (-2,1), and (1,∞). Testing a point in each interval: for x < -2, both factors are negative so f''(x) > 0; for -2 < x < 1, the first factor is positive and second is negative so f''(x) < 0; for x > 1, both factors are positive so f''(x) > 0. A common error is confusing the sign analysis or mixing up which sign corresponds to which concavity. Remember: positive second derivative means concave up (like a smile), negative second derivative means concave down (like a frown).
13. If $f''(x)<0$ for $x<0$ and $f''(x)>0$ for $0<x<8$, where is $f$ concave up?
    Answer: Concave up on $(0,8)$; concave down on $(-\infty,0)$ This problem provides direct information about the sign of f''(x) and asks where f is concave up. A function is concave up wherever its second derivative is positive. We're told that f''(x) < 0 for x < 0, which means f is concave down on (-∞, 0), and f''(x) > 0 for 0 < x < 8, which means f is concave up on (0, 8). Therefore, f is concave up on the interval (0, 8). Choice E incorrectly relates concavity to the monotonic behavior of f'(x) in a reversed manner. The fundamental rule: f is concave up where f''(x) > 0.
14. If $f'(x)$ is increasing on $( -\infty,-4)$ and $(2,\infty)$ but decreasing on $(-4,2)$, where is $f$ concave down?​​
    Answer: Concave down on $(-4,2)$; concave up on $(-\infty,-4)\cup(2,\infty)$ This problem determines concavity from the monotonicity of f'(x). When f'(x) is increasing, f''(x) > 0 and f is concave up; when f'(x) is decreasing, f''(x) < 0 and f is concave down. Since f'(x) decreases on (-4, 2), the function is concave down on (-4, 2), and since f'(x) increases on (-∞, -4) and (2, ∞), the function is concave up on those intervals. Choice A incorrectly identifies the concave down intervals as where f' is increasing. Remember: decreasing first derivative means negative second derivative, which means concave down.
15. Given $f'(x)$ is increasing on $(-2,0)$ and decreasing on $(0,5)$, where is $f$ concave up and concave down?
    Answer: Concave up on $(-2,0)$; concave down on $(0,5)$ This question tests the skill of analyzing the concavity of functions over their domains. Concavity can be determined from the behavior of the first derivative f', since f'' is the derivative of f'. If f' is increasing on an interval, then f'' > 0 there, so f is concave up. If f' is decreasing, then f'' < 0, making f concave down. Thus, f is concave up on (-2,0) and concave down on (0,5). A tempting distractor is choice A, which reverses the intervals, confusing increasing with decreasing behavior. A transferable concavity-sign strategy is to note that where the first derivative increases, concavity is up, and where it decreases, concavity is down.
16. For $-3<x<3$, $f''(x)=x(x-1)$; on which intervals is $f$ concave up and concave down?
    Answer: Concave up on $(-3,0)\cup(1,3)$; concave down on $(0,1)$ This question provides an explicit formula for f''(x) = x(x-1) on the interval (-3, 3). To find concavity, we need to determine where this expression is positive or negative. Setting f''(x) = 0 gives x = 0 or x = 1, dividing the interval into three parts. Testing signs: for x < 0, both factors are negative so f''(x) > 0; for 0 < x < 1, x > 0 and (x-1) < 0 so f''(x) < 0; for 1 < x < 3, both factors are positive so f''(x) > 0. Therefore, f is concave up on (-3, 0) ∪ (1, 3) and concave down on (0, 1). Choice B reverses these intervals. The strategy: factor, find zeros, then test signs between zeros.
17. If $f''(x)=\dfrac{(x-1)(x+3)}{x^2+1}$ for all $x$, where is $f$ concave up and concave down?
    Answer: Concave up on $(-\infty,-3)\cup(1,\infty)$; concave down on $(-3,1)$ This problem involves analyzing a rational second derivative to determine concavity. Given f''(x) = (x-1)(x+3)/(x²+1), we need to find where f''(x) > 0 (concave up) and f''(x) < 0 (concave down). The denominator x²+1 is always positive, so the sign of f''(x) depends only on the numerator (x-1)(x+3). Setting the numerator equal to zero gives x = 1 and x = -3, creating three intervals. For x < -3, both factors are negative so f''(x) > 0; for -3 < x < 1, the first factor is negative and second is positive so f''(x) < 0; for x > 1, both factors are positive so f''(x) > 0. A tempting error is to worry about where the denominator equals zero, but x²+1 is never zero for real x. The strategy for rational functions is to focus on sign changes in the numerator when the denominator is always positive.
18. For $f(x)=\int_0^x g(t)\,dt$, where $g$ is decreasing on $(-2,3)$ and increasing on $(3,6)$, where is $f$ concave down?
    Answer: On $(-2,3)$ only This question tests understanding of concavity for a function defined as an integral. Since f(x) = ∫₀ˣ g(t)dt, we have f'(x) = g(x) by the Fundamental Theorem of Calculus, and thus f''(x) = g'(x). The function f is concave down where f''(x) < 0, which means where g'(x) < 0, occurring when g is decreasing. Since g is decreasing on (-2,3) and increasing on (3,6), we have f concave down on (-2,3) only. Students might confuse this with where g itself is negative, but concavity of f depends on the monotonicity of g. Key insight: for f(x) = ∫ₐˣ g(t)dt, f is concave up/down where g is increasing/decreasing.
19. Suppose $f'(x)$ is increasing on $(-5,-1)$ and decreasing on $(-1,4)$; where is $f$ concave up?
    Answer: Concave up on $(-5,-1)$; concave down on $(-1,4)$ This question requires understanding the relationship between the behavior of f'(x) and the concavity of f. When f'(x) is increasing, this means f''(x) > 0, so f is concave up; when f'(x) is decreasing, this means f''(x) < 0, so f is concave down. Since f'(x) is increasing on (-5, -1), the function f is concave up on (-5, -1). Since f'(x) is decreasing on (-1, 4), the function f is concave down on (-1, 4). Choice D incorrectly relates concavity to whether f'(x) is positive or negative rather than whether f'(x) is increasing or decreasing. Remember: f' increasing means f'' > 0 (concave up), f' decreasing means f'' < 0 (concave down).
20. If $f''(x)=-(x-5)^2$ for all $x$, on which intervals is $f$ concave up and concave down?
    Answer: Concave up on no interval; concave down on $(-\infty,\infty)$ This problem involves analyzing concavity when the second derivative is always non-positive. Given f''(x) = -(x-5)², we need to determine where f''(x) > 0 (concave up) and where f''(x) < 0 (concave down). Since (x-5)² ≥ 0 for all x, we have -(x-5)² ≤ 0 for all x. The expression equals zero only when x = 5, and is strictly negative for all x ≠ 5. Therefore, f is concave down on (-∞,5) and on (5,∞), which together form (-∞,∞) except at the single point x = 5. Students might think f is concave up somewhere because they see a squared term, forgetting about the negative sign. The key insight is that when f''(x) ≤ 0 everywhere and equals zero at only isolated points, the function is concave down on its entire domain.

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