Determining Intervals on Increasing, Decreasing Functions - AP Calculus AB

1. For $s$ with $s'(x)=\dfrac{x^2-9}{x^2+1}$, on which interval is $s$ increasing?
   Answer: $(-\infty,-3)\cup(3,\infty)$ To find where s is increasing, we need s'(x) = (x²-9)/(x²+1) > 0. The numerator x²-9 = (x-3)(x+3) equals zero at x = ±3, while the denominator x²+1 is always positive (never zero). Testing intervals: for x < -3, both (x-3) < 0 and (x+3) < 0, so the numerator is positive and s'(x) > 0; for -3 < x < 3, (x+3) > 0 but (x-3) < 0, so the numerator is negative and s'(x) < 0; for x > 3, both factors are positive, so s'(x) > 0. Choice A suggests s increases only between -3 and 3, but this is exactly where s'(x) < 0 and s decreases. The strategy for rational derivatives is to factor completely, note that positive denominators don't affect the sign, and focus on where the numerator changes sign.
2. Let $h'(x)=(x+3)^2(x-2)$. On which interval(s) is $h$ increasing?
   Answer: $(2,\infty)$ 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.
3. If $t'(x)=\dfrac{(x-2)^2}{x-6}$, on which interval(s) is $t$ decreasing?
   Answer: $(-\infty,6)$ This problem requires finding where function t is decreasing by determining where t'(x) = (x-2)²/(x-6) < 0. The numerator (x-2)² is always non-negative and equals zero only at x = 2, while the denominator equals zero at x = 6 (vertical asymptote). Since the numerator is always non-negative, t'(x) < 0 only when the denominator is negative, which occurs when x < 6. At x = 2, we have t'(x) = 0 (horizontal tangent), and for x > 6, both numerator and denominator are positive so t'(x) > 0. Students might incorrectly think the function's behavior changes at x = 2, but the squared factor only creates a horizontal tangent without changing the decreasing nature. Remember that when the numerator of a rational derivative has only non-negative factors, the sign depends entirely on the denominator.
4. For $v$ with $v'(x)=\dfrac{1}{(x+1)^2}-1$, on which interval(s) is $v$ increasing?
   Answer: $(-2,0)$ 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.
5. If $f'(x)=(x+1)^3(x-6)$, on which interval(s) is $f$ decreasing?
   Answer: $(-1,6)$ 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) = $(x+1)^3$ (x-6), the critical points are x=-1 (multiplicity 3) and x=6. Sign testing in (-∞,-1), (-1,6), and (6,∞) shows negativity only in (-1,6). Thus, f is decreasing on (-1,6). A tempting distractor is choice A (-∞,-1) ∪ (6,∞), but f'(x) > 0 there, perhaps confused by the odd multiplicity not changing sign 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.
6. If $f'(x)=(x-1)(x-4)(x+2)$, on which interval(s) is $f$ increasing?
   Answer: $(-2,1)\cup(4,\infty)$ 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-1)(x-4)(x+2), the critical points are x=-2, x=1, and x=4. Sign testing in (-∞,-2), (-2,1), (1,4), and (4,∞) shows positivity in (-2,1) and (4,∞). Thus, f is increasing on (-2,1) ∪ (4,∞). A tempting distractor is choice A (-∞,-2) ∪ (1,4), but that's where f'(x) < 0, perhaps from miscounting sign changes in the cubic. Always create a sign chart marking roots and undefined points, then test a point in each interval to determine the derivative's sign systematically.
7. If $f'(x)=\dfrac{(x-2)(x-7)}{(x+3)^2}$, on which interval(s) is $f$ increasing?
   Answer: $(-\infty,2)\cup(7,\infty)$ 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.
8. If $q'(x)=\dfrac{x-3}{x+1}$, on which interval(s) is $q$ increasing?
   Answer: $(-\infty,-1)\cup(3,\infty)$ 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.
9. For $f$ with $f'(x)=(x-1)(x+2)$, on which interval(s) is $f$ increasing?
   Answer: $(-\infty,-2)\cup(1,\infty)$ 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.
10. If $g'(x)=-x(x-4)$, on which interval(s) is $g$ decreasing?
    Answer: $(-\infty,0)\cup(4,\infty)$ 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.
11. Given $f'(x)=(x+4)(x-2)^2$, on which interval(s) is $f$ decreasing?
    Answer: $(-\infty,-4)$ 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.
12. If $f'(x)=-(x+3)^2(x-2)$, on which interval(s) is $f$ decreasing?
    Answer: $(2,\infty)$ 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, ∞).
13. If $v'(x)=(x-7)(x-2)(x+1)$, on which interval is $v$ increasing?​​
    Answer: $(-1,2)\cup(7,\infty)$ This problem requires finding where v is increasing given v'(x) = (x-7)(x-2)(x+1). The derivative has three zeros at x = -1, 2, and 7, creating four intervals to test. For x < -1: (-)(-)(-)= (-); for -1 < x < 2: (-)(-)( +) = (+); for 2 < x < 7: (-)(+)(+) = (-); for x > 7: (+)(+)(+) = (+). Therefore, v is increasing on (-1,2) ∪ (7,∞). With three factors, students often lose track of signs or incorrectly assume the function alternates between increasing and decreasing at each zero. The key is to methodically check the sign of each factor in each interval and multiply them together, remembering that an odd number of negative factors yields a negative product.
14. For $x\ne-3,0,2$, $f'(x)=\dfrac{(x+3)(x-2)}{x}$. On which intervals is $f$ decreasing?
    Answer: $(-\infty,-3)\cup(0,2)$ 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.
15. For $f$ with $f'(x)=\dfrac{x(x-6)}{x^2+1}$, on which interval(s) is $f$ increasing?
    Answer: $(-\infty,0)\cup(6,\infty)$ 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.
16. A function has $f'(x)=\dfrac{2-x}{\sqrt{x}}$ for $x>0$. On which interval is $f$ increasing?
    Answer: $(0,2)$ 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 point is x = 2 (zero in numerator), with domain x > 0 and √x > 0, so sign follows (2 - x). Testing shows f'(x) > 0 in (0, 2), while f'(x) < 0 in (2, ∞). A tempting distractor like (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.
17. For $x\ne0$, $f'(x)=\dfrac{(x-3)(x+2)}{x^2}$. On which interval(s) is $f$ increasing?
    Answer: $(-\infty,-2)\cup(3,\infty)$ 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.
18. 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?
    Answer: $(-\infty,-1)\cup(4,\infty)$ 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.
19. Given $f'(x)=\dfrac{(x-4)(x+2)}{(x-1)^2}$, on which interval(s) is $f$ increasing?
    Answer: $(-\infty,-2)\cup(4,\infty)$ This problem asks where f is increasing, which occurs when f'(x) > 0. The derivative f'(x) = (x-4)(x+2)/(x-1)² has zeros at x = 4 and x = -2, with a vertical asymptote at x = 1. Since the denominator (x-1)² is always positive (except at x = 1 where f' is undefined), the sign of f'(x) depends only on the numerator (x-4)(x+2). Testing intervals: for x < -2, both factors in the numerator are negative, so f'(x) > 0; for -2 < x < 1, the numerator is negative; for 1 < x < 4, the numerator is still negative; for x > 4, both factors are positive, so f'(x) > 0. A common error is including x = 1 in the domain, but f' is undefined there. Remember that rational derivatives require checking both where the derivative equals zero and where it's undefined.
20. If $f'(x)=\dfrac{-(x+2)}{(x-3)^2}$ for $x\neq 3$, on which interval is $f$ decreasing?
    Answer: $(-2,3)\cup(3,\infty)$ 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 point is x = 3 (undefined), with sign determined by -(x+2) over positive $(x-3)^2$, negative when x > -2. Testing intervals shows f'(x) < 0 in (-2, 3) and (3, ∞), while f'(x) > 0 in (-∞, -2). A tempting distractor like (-∞, -2)∪(3, ∞) fails because it includes (-∞, -2) where the derivative is positive, incorrectly suggesting decrease there. Always create a sign chart for the derivative by identifying roots and testing intervals to determine where the function is increasing or decreasing.

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