AP Calculus AB Quiz: Connecting A Function And Its Derivatives

What this quiz covers

This quiz focuses on Connecting A Function And Its Derivatives, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus AB.

How to use this quiz

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.

Question 1

1. Increasing on $(-2,1)$, then decreasing on $(1,4)$; concave down on $(-2,4)$. (correct answer)
2. Increasing on $(-2,1)$, then decreasing on $(1,4)$; concave up on $(-2,4)$.
3. Decreasing on $(-2,1)$, then increasing on $(1,4)$; concave down on $(-2,4)$.
4. Increasing on $(-2,4)$; concave down on $(-2,4)$.
5. Decreasing on $(-2,4)$; concave down on $(-2,4)$.

Explanation: This question requires connecting the signs of f′ and f″ to the behavior of f. Given f′(x)>0 for x<1, f′(1)=0, and f′(x)<0 for x>1, the function f increases on (-2,1) then decreases on (1,4). Since f″(x)<0 throughout, f is concave down on (-2,4). A tempting distractor might be choice B, which correctly identifies the increasing/decreasing behavior but incorrectly claims f is concave up when f″<0 indicates concave down. The key strategy is to match the sign of f′ to monotonicity (positive = increasing, negative = decreasing) and the sign of f″ to concavity (negative = concave down).

Question 2

1. $f$ has a local maximum at $x=2$. (correct answer)
2. $f$ has a local minimum at $x=2$.
3. $f$ has an inflection point at $x=2$.
4. $f$ is increasing at $x=2$.
5. $f$ is decreasing at $x=2$.

Explanation: This question applies the second derivative test to interpret the nature of a critical point. Given f′(x)=0 at x=2 (critical point) and f″(2)<0, the second derivative test confirms that f has a local maximum at x=2 since the second derivative is negative at the critical point. Choice B might be tempting as it incorrectly applies the second derivative test, but f″<0 at a critical point always indicates a local maximum. The reliable strategy for these problems is to remember the second derivative test: if f′(c)=0 and f″(c)<0, then f has a local maximum at x=c.

Question 3

1. Local maximum at $x=-1$, local minimum at $x=2$, inflection at $x=0$. (correct answer)
2. Local minimum at $x=-1$, local maximum at $x=2$, inflection at $x=0$.
3. Local maximum at $x=-1$, local maximum at $x=2$, no inflection point.
4. Local minimum at $x=-1$, local minimum at $x=2$, inflection at $x=-1$.
5. No extrema, inflection at $x=2$.

Explanation: This problem requires interpreting the signs of f′ and f″ to identify critical points and inflection points. The given information shows f′ changes from positive to negative at x=-1 (local maximum), then from negative to positive at x=2 (local minimum). Since f″>0 on (-3,0) and f″<0 on (0,3), there's an inflection point at x=0 where concavity changes from up to down. Choice B might seem appealing but reverses the types of extrema at x=-1 and x=2. The reliable strategy is to identify where f′ changes sign for extrema, determine the type using the first derivative test, and find inflection points where f″ changes sign.

Question 4

1. $f'(x)<0$ on $(0,3)$ and increases on $(0,2)$ then decreases on $(2,3)$. (correct answer)
2. $f'(x)>0$ on $(0,3)$ and increases on $(0,2)$ then decreases on $(2,3)$.
3. $f'(x)<0$ on $(0,3)$ and decreases on $(0,2)$ then increases on $(2,3)$.
4. $f'(x)>0$ on $(0,3)$ and decreases on $(0,2)$ then increases on $(2,3)$.
5. $f'(x)$ changes sign from negative to positive at $x=2$.

Explanation: This problem involves analyzing how the behavior of f determines the properties of f′. Since f is decreasing on (0,3), we know f′(x)<0 throughout that interval. The given information that f″(x)>0 for x<2 and f″(x)<0 for x>2 tells us that f′ increases on (0,2) then decreases on (2,3). Choice E might seem tempting as it focuses on sign changes, but f′ doesn't change sign since f is decreasing throughout. To solve these problems systematically, remember that if f is monotonic on an interval, f′ maintains its sign, and the behavior of f′ itself is determined by f″.

Question 5

1. $f'$ decreases on $(1,3)$ then increases on $(3,5)$ and stays positive. (correct answer)
2. $f'$ increases on $(1,3)$ then decreases on $(3,5)$ and stays positive.
3. $f'$ is negative on $(1,3)$ and positive on $(3,5)$.
4. $f'$ is constant on $(1,5)$.
5. $f'$ has a local maximum at $x=3$.

Explanation: This problem connects the signs and behavior of f″ to properties of f′. Given f′(x)>0 on (1,5) and f″(x)=0 only at x=3 where f″ changes from negative to positive, derivative f′ has a local minimum at x=3. Since f′>0 throughout, it decreases on (1,3) then increases on (3,5) while staying positive. Choice B might be tempting as it correctly identifies f′ staying positive but incorrectly describes f′ increasing then decreasing, which is opposite to the f″ sign pattern. The systematic approach is to remember that f″<0 means f′ decreasing and f″>0 means f′ increasing.

Question 6

1. $f$ has a local maximum at $x=2$.
2. $f$ has a point of inflection at $x=2$.
3. $f$ is decreasing on $(-1,2)$ and increasing on $(2,5)$.
4. $f'$ is increasing on $(-1,5)$, and $f$ has a local minimum at $x=2$. (correct answer)
5. $f'$ is decreasing on $(-1,5)$, and $f$ has a local minimum at $x=2$.

Explanation: This problem tests multi-representation reasoning by connecting second derivative information to function behavior. Since f''(x) > 0 on (-1,5), the function f is concave up throughout this interval, which means f' is increasing on (-1,5). Given f'(2) = 0 and f'' > 0 at x = 2, the second derivative test confirms f has a local minimum at x = 2. Choice E incorrectly claims f' is decreasing, but positive f'' means f' must be increasing. When f'' > 0 everywhere and f'(c) = 0, the critical point at x = c must be a local minimum by the second derivative test.

Question 7

1. Decreasing and concave down. (correct answer)
2. Decreasing and concave up.
3. Increasing and concave down.
4. Increasing and concave up.
5. Constant and concave down.

Explanation: This question requires matching the signs of f′ and f″ to describe the behavior of f. Given f′(x)<0 and f″(x)<0 on (-6,0), function f is both decreasing (since f′<0) and concave down (since f″<0). Choice B might seem tempting as it correctly identifies decreasing behavior but incorrectly claims concave up when f″<0 indicates concave down. To approach these problems systematically, remember that f′<0 means decreasing and f″<0 means concave down, describing a function that slopes downward and curves downward like a frown.

Question 8

1. Decreasing and concave down on $(2,8)$.
2. Increasing and concave up on $(2,8)$.
3. Decreasing and concave up on $(2,8)$. (correct answer)
4. Increasing and concave down on $(2,8)$.
5. Decreasing with an inflection point in $(2,8)$.

Explanation: This problem requires multi-representation reasoning to match derivative conditions to graph characteristics. Since f'(x) < 0 for all x in (2,8), the function f is decreasing throughout this interval. Since f''(x) > 0 for all x in (2,8), the function f is concave up throughout this interval. Therefore, f is decreasing and concave up on (2,8). Choice A incorrectly states f is concave down, which would require f'' < 0. When matching derivative signs to graphs, remember that f' < 0 means decreasing and f'' > 0 means concave up, creating a graph that falls at a decreasing rate.

Question 9

1. Decreasing and concave up. (correct answer)
2. Decreasing and concave down.
3. Increasing and concave up.
4. Increasing and concave down.
5. Constant and concave up.

Explanation: This problem requires matching the signs of f′ and f″ to the behavior of f. Given f′(x)<0 on (0,6), the function f is decreasing on this interval. Since f″(x)>0 on (0,6), the function f is concave up throughout. Choice B might seem tempting as it correctly identifies decreasing behavior but incorrectly claims concave down when f″>0 indicates concave up. To solve these problems reliably, remember that f′<0 means decreasing and f″>0 means concave up, giving a decreasing function that curves upward like a downward-sloping smile.

Question 10

1. Local minimum at $x=0$; concave up on $(-3,3)$. (correct answer)
2. Local maximum at $x=0$; concave up on $(-3,3)$.
3. Local minimum at $x=0$; concave down on $(-3,3)$.
4. Local maximum at $x=0$; concave down on $(-3,3)$.
5. No extremum at $x=0$; concave up on $(-3,3)$.

Explanation: This question involves interpreting a graph of $f'$ to determine extrema and concavity of f. Since $f'$ is negative on $(-3,0)$ and positive on $(0,3)$, function f has a local minimum at $x=0$. The fact that $f'$ is increasing on $(-3,3)$ means $f'' > 0$, so f is concave up throughout. Choice B might seem tempting as it correctly identifies concavity but incorrectly claims a local maximum when $f'$ changes from negative to positive, which indicates a local minimum. The reliable strategy is to use the first derivative test: $f'$ changing from negative to positive creates a local minimum, and $f'$ increasing means $f'' > 0$ (concave up).

Question 11

1. $f$ has a local minimum at $x=3$ and is concave up everywhere. (correct answer)
2. $f$ has a local maximum at $x=3$ and is concave up everywhere.
3. $f$ has a local minimum at $x=3$ and is concave down everywhere.
4. $f$ has a local maximum at $x=3$ and is concave down everywhere.
5. $f$ is decreasing everywhere and has no concavity.

Explanation: This question requires interpreting a graph of f′ to determine properties of f. Since the graph of f′ is an increasing line crossing the x-axis at x=3, we have f′<0 for x<3 and f′>0 for x>3, indicating f has a local minimum at x=3. Since f′ is increasing everywhere, f″>0 throughout, so f is concave up everywhere. Choice B might be appealing as it correctly identifies concavity but incorrectly claims a local maximum when f′ changes from negative to positive, which creates a local minimum. The systematic approach is to identify where f′ changes sign for extrema and note that f′ increasing means f″>0 (concave up).

Question 12

1. $f$ is increasing, and $f$ changes concavity at $x=2$. (correct answer)
2. $f$ is decreasing, and $f$ changes concavity at $x=2$.
3. $f$ has a local maximum at $x=2$.
4. $f$ has a local minimum at $x=2$.
5. $f$ is constant near $x=2$.

Explanation: This problem involves interpreting a graph of f′ with a local maximum to determine properties of f. Since the graph of f′ is above the x-axis, we have f′>0, so f is increasing. The fact that f′ has a local maximum at x=2 means f″ changes from positive to negative at x=2, creating an inflection point where f changes from concave up to concave down. Choice C might be tempting as it suggests a local maximum for f at x=2, but since f′>0 on both sides of x=2, f continues increasing—there's just an inflection point. The key strategy is to distinguish between extrema of f′ (which create inflection points for f) and extrema of f itself.

Question 13

1. $f$ is decreasing and concave up. (correct answer)
2. $f$ is decreasing and concave down.
3. $f$ is increasing and concave up.
4. $f$ is increasing and concave down.
5. $f$ has a local maximum at every point.

Explanation: This question involves interpreting a graph of f′ to determine properties of f. Since the graph of f′ is below the x-axis on (-1,2), we have f′(x)<0, so f is decreasing on that interval. The fact that f′ is increasing on (-1,2) means f″>0, so f is concave up. Choice B might be tempting as it correctly identifies the decreasing behavior but incorrectly states concave down when f′ increasing indicates f″>0 (concave up). The systematic approach is to read f′<0 as decreasing behavior for f, and f′ increasing as f″>0, which means f is concave up.

Question 14

1. $f'$ is positive and increasing
2. $f'$ is positive and decreasing
3. $f'$ is negative and increasing
4. $f'$ is negative and decreasing (correct answer)
5. $f'$ changes sign at least once

Explanation: This question tests multi-representation reasoning by connecting function and second derivative behavior to first derivative properties. Since f is decreasing on (-1,4), we know f'(x) < 0 throughout this interval (the derivative must be negative). Since f''(x) < 0 on (-1,4), the first derivative f' is decreasing on this interval. Combining these facts: f' is both negative and decreasing on (-1,4). Choice C incorrectly claims f' is increasing, which would require f'' > 0. When both f and f'' are negative on an interval, f' must be negative (from f decreasing) and decreasing (from f'' < 0).

Question 15

1. Increasing and concave up for $x<1$; decreasing and concave up for $x>1$ (local maximum at $x=1$).
2. Increasing and concave down for $x<1$; decreasing and concave down for $x>1$ (local maximum at $x=1$). (correct answer)
3. Decreasing and concave down for $x<1$; increasing and concave down for $x>1$ (local minimum at $x=1$).
4. Increasing and concave down for $x<1$; increasing and concave down for $x>1$ (no extremum at $x=1$).
5. Increasing and concave up for $x<1$; decreasing and concave down for $x>1$ (local maximum at $x=1$).

Explanation: This question tests multi-representation reasoning by connecting derivative information to the graph of f. Since f'(x) > 0 for x < 1, the function f is increasing on that interval. Since f'(x) < 0 for x > 1, the function f is decreasing on that interval. The condition f''(x) < 0 throughout means f is concave down everywhere. With f'(1) = 0 and f' changing from positive to negative at x = 1, there's a local maximum at x = 1. Choice E incorrectly claims f is concave up for x < 1, failing to match the given f''(x) < 0. When matching derivative conditions to graphs, systematically check the sign of f' for increasing/decreasing behavior and the sign of f'' for concavity.

Question 16

1. $f'$ decreases through $0$ and changes sign from positive to negative. (correct answer)
2. $f'$ increases through $0$ and changes sign from negative to positive.
3. $f'$ increases through $0$ and changes sign from positive to negative.
4. $f'$ is positive on both sides of $0$ and increasing.
5. $f'$ is negative on both sides of $0$ and increasing.

Explanation: This question requires interpreting the concavity of f and the behavior of f′ near a horizontal tangent. Since f is concave down on (-1,1) and has a horizontal tangent at x=0, we have f″<0 and f′(0)=0. In a concave down function, f′ must be decreasing, so if f′(0)=0, then f′ changes from positive (left of 0) to negative (right of 0). Choice B might seem appealing but describes behavior consistent with concave up functions. The key insight is that concave down functions have decreasing derivatives, so horizontal tangents occur where the derivative decreases through zero.

Question 17

1. $f'(x)>0$ on $(1,9)$ and decreases on $(1,4)$ then increases on $(4,9)$. (correct answer)
2. $f'(x)<0$ on $(1,9)$ and decreases on $(1,4)$ then increases on $(4,9)$.
3. $f'(x)>0$ on $(1,9)$ and increases on $(1,4)$ then decreases on $(4,9)$.
4. $f'(x)$ changes sign at $x=4$.
5. $f'(x)=0$ on $(1,4)$ and $f'(x)>0$ on $(4,9)$.

Explanation: This problem involves connecting the monotonicity of f and signs of f″ to properties of f′. Since f is increasing on (1,9), we have f′(x)>0 throughout. The given conditions f″(x)<0 on (1,4) and f″(x)>0 on (4,9) mean f′ decreases on (1,4) then increases on (4,9). Choice C might be tempting as it correctly identifies f′>0 but incorrectly states f′ increases then decreases, which is opposite to what the f″ signs indicate. To solve these systematically, use f increasing to conclude f′>0, then use f″ signs to determine whether f′ itself is increasing or decreasing.

Question 18

1. $f$ has exactly one local maximum on $(0,12)$. (correct answer)
2. $f$ has exactly one local minimum on $(0,12)$.
3. $f$ has exactly one inflection point on $(0,12)$.
4. $f$ is increasing on $(0,12)$.
5. $f$ is concave up on $(0,12)$.

Explanation: This question connects the signs of f″ and the behavior of f′ to determine extrema of f. Given f″(x)<0 on (0,12) (f is concave down) and f′ crosses the x-axis once from positive to negative, f has exactly one critical point that must be a local maximum. In concave down functions, critical points where f′=0 are always local maxima. Choice B might be tempting but is impossible since concave down functions cannot have local minima at interior critical points where f′ changes sign. The systematic approach is to remember that in concave down regions, any sign change in f′ from positive to negative creates a local maximum.

Question 19

1. $f$ is increasing and concave down on $(0,6)$. (correct answer)
2. $f$ is decreasing and concave down on $(0,6)$.
3. $f$ is increasing and concave up on $(0,6)$.
4. $f$ has a local maximum in $(0,6)$.
5. $f$ has an inflection point in $(0,6)$.

Explanation: This problem involves matching the signs of f′ and f″ to describe f's behavior. Given f′(x)>0 and f″(x)<0 on (0,6) with f′(x)=0 nowhere in the interval, function f is increasing and concave down throughout. The fact that f′ never equals zero means there are no critical points in the interval. Choice C might be tempting as it correctly identifies increasing behavior but incorrectly claims concave up when f″<0 indicates concave down. To approach these systematically, use f′>0 for increasing behavior and f″<0 for concave down behavior.

Question 20

1. $f'$ increases on $( -4,-1)$, decreases on $( -1,2)$, then increases on $(2,4)$. (correct answer)
2. $f'$ decreases on $( -4,-1)$, increases on $( -1,2)$, then decreases on $(2,4)$.
3. $f'$ increases on $( -4,2)$ and decreases on $(2,4)$.
4. $f'$ decreases on $( -4,2)$ and increases on $(2,4)$.
5. $f'$ is constant on each subinterval.

Explanation: This problem connects the sign pattern of f″ to the behavior of f′. Given f″(x)>0 for x<-1, f″(x)<0 for -1<x<2, and f″(x)>0 for x>2, the derivative f′ increases on (-4,-1), decreases on (-1,2), then increases on (2,4). This creates a pattern where f′ has a local maximum at x=-1 and a local minimum at x=2. Choice B might seem tempting as it reverses this pattern, but the signs of f″ directly determine whether f′ is increasing or decreasing. To solve these systematically, remember that f″>0 means f′ is increasing and f″<0 means f′ is decreasing.