If for , on which interval(s) is increasing?
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AP Calculus BC Quiz
Practice Determining Intervals On Increasing Decreasing Functions 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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If f′(x)=x1−x for x>0, on which interval(s) is f 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 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.
If f′(x)=x1−x for x>0, on which interval(s) is f increasing?
Explanation: Determining the intervals on which a function is increasing or decreasing is a key skill in calculus that relies on analyzing the sign of the first derivative. The function f is increasing where f'(x) = (1 - x)/√x > 0 for x > 0. Since √x > 0, this holds when 1 - x > 0 or x < 1, so on (0, 1). At x = 1, f' = 0, and for x > 1, f' < 0. A tempting distractor is choice B (1, ∞), but that's where f' < 0, so f decreases there. In general, to analyze the monotonicity of a function, find the critical points where the derivative is zero or undefined, then test the sign of the derivative in each interval determined by those points.
Let f′(x)=(x+5)(x−1)2. On which interval(s) is f decreasing?
Explanation: Determining the intervals on which a function is increasing or decreasing is a key skill in AP Calculus BC that relies on analyzing the first derivative. Given f'(x)=(x+5)(x-1)², roots x=-5, x=1 (double). (x-1)² ≥0, zero at 1; so sign determined by (x+5), but at double root sign doesn't change. f'>0 when x>-5 (except zero at 1), f'<0 when x<-5. Thus decreasing on (-∞,-5). Wait, no: when x<-5, x+5<0, (x-1)²>0, product <0, decreasing; x>-5, >0, increasing. Yes. A tempting distractor like B, (-5,1), fails because f'>0 there, increasing. Always create a sign chart for the derivative to systematically determine intervals of increase and decrease.
Given f′(x)=−(x−3)2(x+2), on which interval is f decreasing?
Explanation: This question tests the skill of determining intervals where a function is decreasing by analyzing the sign of its first derivative. The derivative f'(x) = -(x-3)^2(x+2) indicates that the function f(x) is decreasing where f'(x) < 0 and increasing where f'(x) > 0. The critical points are x = -2 and x = 3, dividing the real line into intervals: (-∞, -2), (-2, 3), and (3, ∞), with f'(x) = 0 at x = 3 but not changing sign there. Sign analysis shows f'(x) > 0 in (-∞, -2) and f'(x) < 0 in (-2, ∞), including through x = 3 where it touches zero but remains decreasing overall. A tempting distractor is choice E, (-∞, -2) and (3, ∞), but (-∞, -2) is where f is increasing, not decreasing. A transferable sign-analysis strategy is to identify critical points, test intervals with representative points, and determine where the derivative is positive for increasing behavior or negative for decreasing behavior.
Given f′(x)=x−2 with domain x≥0, on which interval(s) is f decreasing?
Explanation: This problem tests finding decreasing intervals with a radical derivative. The derivative f'(x) = √x - 2 equals zero when √x = 2, which gives x = 4. For f to be decreasing, we need f'(x) < 0, which means √x < 2. Since √x is an increasing function and the domain is x ≥ 0, we have √x < 2 precisely when 0 ≤ x < 4. Therefore, f is decreasing on [0, 4). A common mistake is writing (0, 4) and excluding x = 0, but since f'(0) = -2 < 0, the function is indeed decreasing at x = 0. When dealing with derivatives involving radicals, solve the inequality algebraically and remember to check endpoint behavior.
Let f′(x)=x(x−6)(x+2). On which interval(s) is f increasing?
Explanation: Determining the intervals on which a function is increasing or decreasing is a key skill in AP Calculus BC that relies on analyzing the first derivative. Given f'(x)=x(x-6)(x+2), roots at 0,6,-2. Ordered: -2,0,6. Sign changes at each, cubic with positive leading coefficient. f'>0 in (-2,0) and (6,∞), f'<0 in (-∞,-2) and (0,6). Yes. A tempting distractor like D, (-2,6), fails because it combines increasing and decreasing intervals. Always create a sign chart for the derivative to systematically determine intervals of increase and decrease.
Given f′(x)=x2(x−1)(x−3), on which interval is f increasing?
Explanation: This question tests the skill of determining intervals where a function is increasing by analyzing the sign of its first derivative. The derivative f'(x) = (x-1)(x-3)/x^2 indicates that the function f(x) is increasing where f'(x) > 0 and decreasing where f'(x) < 0, with a vertical asymptote at x = 0. The critical points are x = 1 and x = 3, and the denominator is positive for x ≠ 0, so the sign follows the numerator, dividing intervals into (-∞, 0), (0, 1), (1, 3), and (3, ∞). Testing shows f'(x) > 0 in (-∞, 0), (0, 1), and (3, ∞), while negative in (1, 3), meaning f is increasing there. A tempting distractor is choice A, (0, 1) and (3, ∞), but it omits (-∞, 0) where f'(x) > 0 as well. A transferable sign-analysis strategy is to identify critical points, test intervals with representative points, and determine where the derivative is positive for increasing behavior or negative for decreasing behavior.
For f′(x)=(x+1)2x−5, on which interval(s) is f decreasing?
Explanation: Determining the intervals on which a function is increasing or decreasing is a key skill in AP Calculus BC that relies on analyzing the first derivative. For f'(x) = (x-5)/(x+1)², critical points are x=5 (numerator zero) and x=-1 (denominator zero, undefined). The sign of f' depends on (x-5) over positive (x+1)². f'<0 when x-5<0, i.e., x<5, but excluding x=-1 where undefined. So decreasing on (-∞,-1) and (-1,5). A tempting distractor like B, (-1,5), fails because it omits the left interval where f' is also negative. Always create a sign chart for the derivative to systematically determine intervals of increase and decrease.
For f(x)=x3−6x2+9x+2, on which intervals is f increasing?
Explanation: This question tests your ability to determine where a function is increasing by analyzing the sign of its derivative. To find where f(x) = x³ - 6x² + 9x + 2 is increasing, we first compute f'(x) = 3x² - 12x + 9 = 3(x² - 4x + 3) = 3(x - 1)(x - 3). The derivative equals zero when x = 1 and x = 3, dividing the number line into three intervals: (-∞, 1), (1, 3), and (3, ∞). Testing values in each interval: for x = 0, f'(0) = 3(-1)(-3) = 9 > 0; for x = 2, f'(2) = 3(1)(-1) = -3 < 0; for x = 4, f'(4) = 3(3)(1) = 9 > 0. Choice B might tempt students who confuse increasing with decreasing intervals. The key strategy is to factor the derivative completely, find critical points, then test the sign in each resulting interval.
Given f′(x)=(x−3)2+1(x−3)(x+1), on which interval(s) is f increasing?
Explanation: Determining the intervals on which a function is increasing or decreasing is a key skill in calculus that relies on analyzing the sign of the first derivative. The function f is increasing where f'(x) = (x - 3)(x + 1)/((x - 3)² + 1) > 0. Denominator always positive, numerator positive on (-∞, -1) ∪ (3, ∞). Thus, f increases there. A tempting distractor is choice B (-1, 3), but there f' < 0, so f decreases. In general, to analyze the monotonicity of a function, find the critical points where the derivative is zero or undefined, then test the sign of the derivative in each interval determined by those points.
Given f(x)=xx2−1 with x=0, on which interval(s) is f increasing?
Explanation: Determining the intervals on which a function is increasing or decreasing is a key skill in calculus that relies on analyzing the sign of the first derivative. For f(x) = (x² - 1)/x (x ≠ 0), f'(x) = 1 + 1/x² > 0 always where defined. This is positive on (-∞, 0) ∪ (0, ∞). No regions where negative. A tempting distractor is choice E (-1, 1), but f' > 0 there too, though it's not the full set. In general, to analyze the monotonicity of a function, find the critical points where the derivative is zero or undefined, then test the sign of the derivative in each interval determined by those points.
If f′(x)=(x−2)(x−2)(x+1), on which interval(s) is f decreasing?
Explanation: Determining the intervals on which a function is increasing or decreasing is a key skill in calculus that relies on analyzing the sign of the first derivative. The function f is decreasing where f'(x) = (x - 2)²(x + 1) < 0. Since (x - 2)² ≥ 0, the sign follows (x + 1), negative for x < -1. Thus, f decreases on (-∞, -1). A tempting distractor is choice C (2, ∞), but there f' > 0, so f increases. In general, to analyze the monotonicity of a function, find the critical points where the derivative is zero or undefined, then test the sign of the derivative in each interval determined by those points.
For p(x)=x4−4x2, on which intervals is p decreasing?
Explanation: To find where p(x) = x⁴ - 4x² is decreasing, we need to analyze where p'(x) < 0. Computing p'(x) = 4x³ - 8x = 4x(x² - 2) = 4x(x - √2)(x + √2), we find critical points at x = 0, x = √2, and x = -√2. Testing signs in each interval: for x < -√2 (try x = -2): p'(-2) = 4(-2)(2)(2 - √2) < 0; for -√2 < x < 0 (try x = -1): p'(-1) = 4(-1)(-1)(-1 - √2) > 0; for 0 < x < √2 (try x = 1): p'(1) = 4(1)(-1)(1 - √2) < 0; for x > √2 (try x = 2): p'(2) = 4(2)(2)(2 - √2) > 0. Choice C might tempt students who only consider where x² - 2 is positive. The key insight: factor completely and carefully track the sign of each factor across all intervals.
Given f′(x)=∣x−3∣−2, on which interval(s) is f decreasing?
Explanation: To find where f decreases given f'(x) = |x - 3| - 2, we need f'(x) < 0, which means |x - 3| < 2. The absolute value inequality |x - 3| < 2 is equivalent to -2 < x - 3 < 2, which gives 1 < x < 5. We can verify: for x < 1, |x - 3| > 2 so f'(x) > 0; for 1 < x < 5, |x - 3| < 2 so f'(x) < 0; for x > 5, |x - 3| > 2 so f'(x) > 0. Therefore, f decreases on (1, 5). Students often struggle with absolute value derivatives, forgetting to solve the inequality |x - 3| < 2 rather than just finding where the derivative equals zero. The strategy for absolute value problems: rewrite as an inequality and solve algebraically.
If f′(x)=x−2(x−3)(x+1), on which interval(s) is f decreasing?
Explanation: This problem involves analyzing a rational derivative to find decreasing intervals. The derivative f'(x) = (x - 3)(x + 1)/(x - 2) has zeros at x = -1 and x = 3, and is undefined at x = 2. Testing intervals: for x = -2, f'(-2) = (-5)(-1)/(-4) = -5/4 < 0; for x = 0, f'(0) = (-3)(1)/(-2) = 3/2 > 0; for x = 2.5, f'(2.5) = (-0.5)(3.5)/(0.5) = -3.5 < 0; for x = 4, f'(4) = (1)(5)/(2) = 5/2 > 0. Therefore, f is decreasing on (-∞, -1) ∪ (2, 3). A tempting error is to think the function cannot decrease through the discontinuity at x = 2, but we analyze each continuous piece separately. For rational derivatives, create a complete sign chart including zeros and discontinuities, then determine the sign on each interval.
Given f′(x)=x2+1(x−2)(x+4), on which interval(s) is f increasing?
Explanation: Given f'(x) = (x - 2)(x + 4)/(x² + 1), we need f'(x) > 0 for f to be increasing. The denominator x² + 1 is always positive, so the sign depends only on the numerator (x - 2)(x + 4). This equals zero at x = 2 and x = -4, creating three intervals. For x < -4: both factors are negative, so f'(x) > 0. For -4 < x < 2: (x + 4) > 0 and (x - 2) < 0, so f'(x) < 0. For x > 2: both factors are positive, so f'(x) > 0. Therefore, f increases on (-∞, -4) ∪ (2, ∞). Students might incorrectly think the denominator creates additional critical points, but x² + 1 never equals zero. When the denominator is always positive, focus only on the numerator's sign.
If f′(x)=(x−2)(x+1) for all real x, on which interval(s) is f increasing?
Explanation: Determining the intervals where a function is increasing or decreasing is a key skill in AP Calculus BC, relying on the first derivative test. The function f is increasing where its derivative f'(x) is positive, which occurs when (x-2)(x+1) > 0. By finding the critical points x = -1 and x = 2, we divide the real line into intervals and test the sign of f'(x) in each: positive in (-∞, -1) and (2, ∞), and negative in (-1, 2). Thus, f is increasing on (-∞, -1) ∪ (2, ∞). A tempting distractor like (-1, 2) fails because that's where f'(x) is negative, indicating the function is actually decreasing there. Always create a sign chart using the roots of the derivative to systematically determine where it is positive or negative.
Suppose h′(x)=x2+1(x−4)(x+1). On which interval(s) is h increasing?
Explanation: Determining the intervals where a function is increasing or decreasing is a key skill in AP Calculus BC, relying on the first derivative test. The function h is increasing where its derivative h'(x) = (x-4)(x+1)/(x^2+1) is positive; since the denominator is always positive, the sign matches the numerator. The critical points are x = -1 and x = 4, dividing the line into intervals where the numerator is positive in (-∞, -1) ∪ (4, ∞) and negative in (-1, 4). Thus, h is increasing on (-∞, -1) ∪ (4, ∞). A tempting distractor like (-∞, 4) fails because it includes (-1, 4) where h'(x) is negative, meaning the function decreases there. Always create a sign chart using the roots of the derivative to systematically determine where it is positive or negative.
For f′(x)=−(x−2)2(x+3), on which interval(s) is f increasing?
Explanation: Determining the intervals on which a function is increasing or decreasing is a key skill in AP Calculus BC that relies on analyzing the first derivative. For f'(x)= -(x-2)²(x+3), roots x=2 (double), x=-3. Negative sign, (x-2)² ≥0, (x+3). Product (x-2)²(x+3) has sign of (x+3), zero at 2 and -3. So - of that: when x>-3, positive then negative f'<0; x<-3, negative then negative f'>0? Wait: (x-2)² always pos except zero, times (x+3): pos*(pos)>0 for x>-3, pos*(neg)<0 for x<-3. Then overall - of that: - (>0) = <0 for x>-3; - (<0)= >0 for x<-3. At double root x=2, zero but sign doesn't change (remains neg). So f'>0 on (-∞,-3), f'<0 on (-3,∞). Thus increasing on (-∞,-3). A tempting distractor like C, (2,∞), fails because f'<0 there. Always create a sign chart for the derivative to systematically determine intervals of increase and decrease.
Let f′(x)=(x−2)(x+2)(x−1)(x+1). On which interval(s) is f decreasing?
Explanation: Determining the intervals on which a function is increasing or decreasing is a key skill in calculus that relies on analyzing the sign of the first derivative. The function f is decreasing where f'(x) = (x² - 1)/(x² - 4) < 0. This occurs when numerator and denominator have opposite signs, specifically (-2, -1) ∪ (1, 2). Sign analysis confirms. A tempting distractor is choice A (-∞, -2) ∪ (-1, 1) ∪ (2, ∞), but it includes regions where f' > 0. In general, to analyze the monotonicity of a function, find the critical points where the derivative is zero or undefined, then test the sign of the derivative in each interval determined by those points.
If f′(x)=(x−3)2x+2, on which interval(s) is f increasing?
Explanation: Determining the intervals on which a function is increasing or decreasing is a key skill in AP Calculus BC that relies on analyzing the first derivative. For f'(x)=(x+2)/(x-3)², critical point x=-2 (num zero), x=3 undefined (den zero). Denominator always positive except at 3. Sign of f' same as (x+2). So f'>0 when x>-2, f'<0 when x<-2, but excluding 3. Since at 3 undefined, intervals (-∞,-2), (-2,3), (3,∞). f'<0 on (-∞,-2), >0 on (-2,3)∪(3,∞). Yes, increasing on (-2,3)∪(3,∞). A tempting distractor like A, (-∞,-2), fails because that's decreasing. Always create a sign chart for the derivative to systematically determine intervals of increase and decrease.