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.

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.

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.

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.

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.

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.

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.

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.

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.

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.