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.

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.

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.

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.

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.

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.

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.

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.

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.

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.