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. 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.

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. 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.

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.

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.

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.

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.

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.