Let . On which interval(s) is increasing?
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AP Calculus AB Quiz
Practice Determining Intervals On Increasing Decreasing Functions in AP Calculus AB with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.
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Let h′(x)=(x+3)2(x−2). On which interval(s) is h increasing?
This quiz focuses on Determining Intervals On Increasing Decreasing Functions, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus AB.
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.
Let h′(x)=(x+3)2(x−2). On which interval(s) is h increasing?
Explanation: This question asks where function h is increasing, which occurs when h'(x) = (x+3)²(x-2) > 0. The derivative equals zero at x = -3 (with multiplicity 2) and x = 2, but only x = 2 changes the sign of h'(x) since (x+3)² is always non-negative. For x < 2, the factor (x-2) is negative while (x+3)² ≥ 0, making h'(x) ≤ 0 (strictly negative except at x = -3). For x > 2, the factor (x-2) is positive and (x+3)² > 0, making h'(x) > 0. Students might incorrectly select option B by treating x = -3 as a sign-changing point, but even-powered factors don't change sign. The crucial insight is recognizing that squared factors only create horizontal tangents without sign changes, so only odd-powered factors determine where the derivative changes from positive to negative.
For f with f′(x)=x2+1x(x−6), on which interval(s) is f increasing?
Explanation: This question tests finding where f is increasing by analyzing when f'(x) = x(x-6)/(x²+1) > 0. The zeros of f'(x) occur at x = 0 and x = 6, and since x²+1 is always positive, we only need to consider the sign of x(x-6). Testing intervals: for x < 0, x is negative and (x-6) is negative, so the product is positive and f'(x) > 0; for 0 < x < 6, x is positive and (x-6) is negative, so f'(x) < 0; for x > 6, both factors are positive, so f'(x) > 0. A common mistake is thinking the denominator x²+1 could be zero somewhere, but x²+1 ≥ 1 for all real x. The strategy for rational functions is to focus on sign changes in the numerator when the denominator is always positive.
For v with v′(x)=(x+1)21−1, on which interval(s) is v increasing?
Explanation: This problem tests finding where function v is increasing by analyzing where v'(x) = 1/(x+1)² - 1 > 0. We need 1/(x+1)² > 1, which means (x+1)² < 1, equivalent to |x+1| < 1 or -1 < x+1 < 1, giving -2 < x < 0. We must exclude x = -1 where v'(x) is undefined. Testing intervals: for x < -2, (x+1)² > 1 so v'(x) < 0; for -2 < x < -1, (x+1)² < 1 so v'(x) > 0; for -1 < x < 0, (x+1)² < 1 so v'(x) > 0; for x > 0, (x+1)² > 1 so v'(x) < 0. Students might incorrectly choose option A thinking the function increases on the outer intervals, but careful algebraic manipulation shows it increases only near x = -1. The strategy for derivatives involving differences with fractions is to set up and solve the appropriate inequality algebraically.
For f with f′(x)=(x−1)(x+4)(x−3), on which interval(s) is f decreasing?
Explanation: To find where f is decreasing, we need f'(x) = (x-1)(x+4)(x-3) < 0. The zeros are at x = -4, x = 1, and x = 3, dividing the real line into four intervals. Testing each: for x < -4, all three factors are negative, so f'(x) < 0; for -4 < x < 1, (x+4) > 0 while the other two are negative, so f'(x) > 0; for 1 < x < 3, (x-1) > 0 and (x+4) > 0 while (x-3) < 0, so f'(x) < 0; for x > 3, all factors are positive, so f'(x) > 0. Students sometimes count factors incorrectly or lose track of signs with three factors. The systematic approach is to create a sign chart showing each factor's sign in each interval, then multiply to get f'(x)'s sign.
If f′(x)=(x−4)2(x+3)2(x−1) for x=4, on which interval(s) is f increasing?
Explanation: This problem tests your ability to determine the intervals where a function is increasing or decreasing by analyzing the sign of its first derivative. A function f is increasing on intervals where f'(x) > 0 and decreasing where f'(x) < 0. For f'(x) = (x+3)^2 (x-1) / (x-4)^2, both numerator and denominator have squared terms, but the sign follows (x-1) since other factors are non-negative except at points. Critical points are x=-3, x=1, and x=4, with positivity when x > 1 (excluding x=4). Thus, f is increasing on (1,4) ∪ (4,∞). A tempting distractor is choice E (-∞,4) ∪ (4,∞), but it includes x < 1 where f'(x) < 0, ignoring the sign change at x=1. Always create a sign chart marking roots and undefined points, then test a point in each interval to determine the derivative's sign systematically.
If f′(x)=(x+3)2(x−2)(x−7), on which interval(s) is f increasing?
Explanation: This problem requires finding where f is increasing, meaning where f'(x) = (x-2)(x-7)/(x+3)² > 0. The numerator zeros are x = 2 and x = 7, and there's a vertical asymptote at x = -3. Since (x+3)² is always positive (where defined), the sign of f'(x) depends only on (x-2)(x-7). Testing intervals: for x < -3, both factors in the numerator are negative, so f'(x) > 0; for -3 < x < 2, still both negative, so f'(x) > 0; for 2 < x < 7, (x-2) > 0 and (x-7) < 0, so f'(x) < 0; for x > 7, both positive, so f'(x) > 0. A common mistake is thinking the squared denominator might change the analysis, but squared terms never change sign. The key is recognizing that f increases on (-∞,-3) ∪ (-3,2) ∪ (7,∞), which simplifies to (-∞,2) ∪ (7,∞) excluding x = -3.
Given f′(x)=(x+5)(x−1)(x−2)2 for x=−5,1, where is f decreasing?
Explanation: This problem tests your ability to determine the intervals where a function is increasing or decreasing by analyzing the sign of its first derivative. A function f is decreasing on intervals where f'(x) < 0 and increasing where f'(x) > 0. For f'(x) = (x-2)^2 / ((x+5)(x-1)), the numerator is non-negative, so the sign follows the reciprocal of the denominator's sign, negative when denominator is negative. Critical points are x=-5, x=1, and x=2, with negativity in (-5,1). Thus, f is decreasing on (-5,1). A tempting distractor is choice A (-∞,-5) ∪ (1,∞), but that's where f'(x) > 0, mistaken if ignoring the squared numerator's effect. Always create a sign chart marking roots and undefined points, then test a point in each interval to determine the derivative's sign systematically.
For x=0, f′(x)=x2(x−3)(x+2). On which interval(s) is f increasing?
Explanation: This problem tests your ability to determine the intervals where a function is increasing or decreasing by analyzing the sign of its first derivative. A function f is increasing on intervals where f'(x) > 0 and decreasing where f'(x) < 0. For f'(x) = (x-3)(x+2)/x^2, the denominator is positive except at x=0 (undefined), so the sign matches the numerator (x-3)(x+2). Critical points are x=-2, x=0, and x=3, with positivity in (-∞,-2) and (3,∞). Thus, f is increasing on (-∞,-2) ∪ (3,∞). A tempting distractor is choice B (-2,0) ∪ (0,3), but that's where f'(x) < 0, possibly from overlooking the denominator's consistent positivity. Always create a sign chart marking roots and undefined points, then test a point in each interval to determine the derivative's sign systematically.
If f′(x)=(x−1)(x−2)(x+3) for x=1, where is f increasing?
Explanation: This question tests the skill of determining intervals on which a function is increasing or decreasing using its first derivative. To find where f is increasing, identify where f'(x) > 0, as a positive derivative indicates the function's slope is upward. The critical points are x = -3, x = 2 (zeros), and x = 1 (undefined), dividing the real line into intervals (-∞, -3), (-3, 1), (1, 2), and (2, ∞). Testing signs shows f'(x) > 0 in (-3, 1) and (2, ∞), while f'(x) < 0 in (-∞, -3) and (1, 2). A tempting distractor like (-∞, -3)∪(1, 2) fails because the derivative is negative there, incorrectly suggesting increase instead of decrease. Always create a sign chart for the derivative by identifying roots and testing intervals to determine where the function is increasing or decreasing.
Suppose f′(x)=ex(x−1)(x+2). On which interval is f decreasing?
Explanation: This question tests the skill of determining intervals on which a function is increasing or decreasing using its first derivative. To find where f is decreasing, identify where f'(x) < 0, as a negative derivative indicates the function's slope is downward. The critical points are x = -2 and x = 1 (zeros, with e^x always positive), dividing the real line into intervals (-∞, -2), (-2, 1), and (1, ∞). Testing signs shows f'(x) < 0 in (-2, 1), while f'(x) > 0 in (-∞, -2) and (1, ∞). A tempting distractor like (-∞, -2)∪(1, ∞) fails because the derivative is positive there, incorrectly suggesting decrease instead of increase. Always create a sign chart for the derivative by identifying roots and testing intervals to determine where the function is increasing or decreasing.
For f with f′(x)=(x+2)(x−1), on which interval is f increasing?
Explanation: This question tests the skill of determining intervals on which a function is increasing or decreasing using its first derivative. To find where f is increasing, identify where f'(x) > 0, as a positive derivative indicates the function's slope is upward. The critical points are x = -2 and x = 1, dividing the real line into intervals (-∞, -2), (-2, 1), and (1, ∞). Testing signs shows f'(x) > 0 in (-∞, -2) and (1, ∞), while f'(x) < 0 in (-2, 1). A tempting distractor like (-∞, 1) fails because it includes (-2, 1) where the derivative is negative, incorrectly suggesting increase there. Always create a sign chart for the derivative by identifying roots and testing intervals to determine where the function is increasing or decreasing.
Given f′(x)=(x+4)(x−2)2, on which interval(s) is f decreasing?
Explanation: This problem tests your ability to determine the intervals where a function is increasing or decreasing by analyzing the sign of its first derivative. A function f is decreasing on intervals where f'(x) < 0 and increasing where f'(x) > 0. For f'(x) = (x+4)(x-2)^2, the factor (x-2)^2 is always non-negative, so the sign is determined by (x+4), making f'(x) < 0 only when x < -4. Critical points are x=-4 and x=2, and sign testing confirms negativity solely left of x=-4. Thus, f is decreasing on (-∞, -4). A tempting distractor is choice B (-4,2), but f'(x) >= 0 there due to the squared term keeping it non-negative. Always create a sign chart marking roots and undefined points, then test a point in each interval to determine the derivative's sign systematically.
For x=−3,0,2, f′(x)=x(x+3)(x−2). On which intervals is f decreasing?
Explanation: This problem tests your ability to determine the intervals where a function is increasing or decreasing by analyzing the sign of its first derivative. A function f is decreasing on intervals where f'(x) < 0 and increasing where f'(x) > 0. For f'(x) = (x+3)(x-2)/x, undefined at x=0, the critical points are x=-3, x=0, and x=2. Sign testing in (-∞,-3), (-3,0), (0,2), and (2,∞) shows negativity in (-∞,-3) and (0,2). Thus, f is decreasing on (-∞,-3) ∪ (0,2). A tempting distractor is choice D (-3,2), but it ignores the sign change at x=0 where the function is undefined. Always create a sign chart marking roots and undefined points, then test a point in each interval to determine the derivative's sign systematically.
For x=−1,3, f′(x)=(x+1)2x−3. On which interval(s) is f increasing?
Explanation: This problem tests your ability to determine the intervals where a function is increasing or decreasing by analyzing the sign of its first derivative. A function f is increasing on intervals where f'(x) > 0 and decreasing where f'(x) < 0. For f'(x) = (x-3)/(x+1)^2, the denominator is always positive except at x=-1 where undefined, so the sign matches (x-3). Critical points are x=-1 and x=3, and testing shows f'(x) > 0 only for x > 3. Thus, f is increasing on (3, ∞). A tempting distractor is choice E (-∞,-1) ∪ (3,∞), but f'(x) < 0 left of x=-1, perhaps confused by sign change at the asymptote. Always create a sign chart marking roots and undefined points, then test a point in each interval to determine the derivative's sign systematically.
If g′(x)=−x(x−4), on which interval(s) is g decreasing?
Explanation: This problem requires finding where function g is decreasing by analyzing where g'(x) < 0. We need to determine where -x(x-4) < 0, which can be rewritten as x(x-4) > 0 for clarity. The derivative equals zero at x = 0 and x = 4, creating three test intervals: (-∞,0), (0,4), and (4,∞). Testing each interval: for x < 0, we have negative times negative equals positive, so x(x-4) > 0 and thus g'(x) < 0; for 0 < x < 4, we have positive times negative equals negative, so x(x-4) < 0 and thus g'(x) > 0; for x > 4, we have positive times positive equals positive, so x(x-4) > 0 and thus g'(x) < 0. A common error would be choosing option A, thinking the function decreases between the zeros, but the sign analysis reveals it actually increases there. Remember to always check the sign of the derivative in each interval by substituting a test point.
For f with f′(x)=(x−1)(x+2), on which interval(s) is f increasing?
Explanation: This question tests determining intervals where a function is increasing by analyzing the sign of its derivative. A function f is increasing when f'(x) > 0, so we need to find where (x-1)(x+2) > 0. The derivative equals zero at x = -2 and x = 1, which divide the number line into three intervals: (-∞,-2), (-2,1), and (1,∞). Testing values in each interval: for x < -2, both factors are negative so f'(x) > 0; for -2 < x < 1, the factors have opposite signs so f'(x) < 0; for x > 1, both factors are positive so f'(x) > 0. Students might incorrectly choose option A by thinking the function increases between the critical points, but sign analysis shows it actually decreases there. The key strategy is to create a sign chart by testing one value from each interval created by the zeros of the derivative.
If q′(x)=x+1x−3, on which interval(s) is q increasing?
Explanation: This question requires determining where function q is increasing by finding where q'(x) = (x-3)/(x+1) > 0. The numerator equals zero at x = 3 and the denominator equals zero at x = -1 (vertical asymptote), creating three intervals to test. For x < -1: numerator is negative and denominator is negative, so q'(x) > 0; for -1 < x < 3: numerator is negative and denominator is positive, so q'(x) < 0; for x > 3: numerator is positive and denominator is positive, so q'(x) > 0. A common mistake would be choosing option B, thinking the function increases between the critical values, but careful sign analysis shows it decreases there. When analyzing rational functions, remember that both zeros of the numerator and denominator create boundaries for sign changes, and test each resulting interval systematically.
If f′(x)=−(x+3)2(x−2), on which interval(s) is f decreasing?
Explanation: To determine where f is decreasing, we analyze where f'(x) = -(x+3)²(x-2) < 0. The critical points are x = -3 and x = 2. Since (x+3)² is always non-negative and the expression has a leading negative sign, we need -(x+3)²(x-2) < 0, which means (x+3)²(x-2) > 0. Testing intervals: for x < -3, (x+3)² > 0 and (x-2) < 0, so the product is negative and f'(x) > 0; for -3 < x < 2, (x+3)² ≥ 0 and (x-2) < 0, so f'(x) > 0; for x > 2, both factors are positive, so the product is positive and f'(x) < 0. Students often forget the leading negative sign flips all inequalities. Remember that f decreases where f'(x) < 0, which here is only on (2, ∞).
For f with f′(x)=(x+2)(x−1)2, on which interval(s) is f increasing?
Explanation: This problem tests your ability to determine where a function is increasing by analyzing the sign of its derivative. A function f is increasing on intervals where f'(x) > 0, so we need to find where (x+2)(x-1)² > 0. The critical points occur where f'(x) = 0, which gives us x = -2 and x = 1. Testing intervals: for x < -2, both factors are negative so f'(x) > 0; for -2 < x < 1, (x+2) > 0 and (x-1)² > 0 so f'(x) > 0; for x > 1, both factors are positive so f'(x) > 0. Many students incorrectly choose option A, thinking the function changes behavior at x = 1, but (x-1)² is always non-negative and equals zero only at x = 1. The key insight is that squared factors don't change the sign of the derivative when crossing their zeros, so f is increasing on (-2, ∞).
A function u has derivative signs: u′(x)>0 for x<−1, u′(x)<0 for −1<x<4, u′(x)>0 for x>4. Where is u increasing?
Explanation: This problem directly provides the sign of u'(x) in different intervals, making it straightforward to identify where u is increasing. A function increases where its derivative is positive, so we need the intervals where u'(x) > 0. According to the given information: u'(x) > 0 for x < -1 and u'(x) > 0 for x > 4, while u'(x) < 0 for -1 < x < 4. Therefore, u is increasing on (-∞,-1) ∪ (4,∞). Choice B incorrectly identifies the interval where u is decreasing (where u'(x) < 0) rather than increasing. When given derivative sign information directly, simply match positive derivative regions with increasing intervals and negative derivative regions with decreasing intervals.