Determining Intervals on Increasing, Decreasing Functions - AP Calculus BC

1. For $f'(x)=\dfrac{(x-1)(x+4)}{(x+4)^2}$, on which interval(s) is $f$ decreasing?
   Answer: $(-4,1)$ 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 - 1)/(x + 4) < 0 (simplified, x ≠ -4). This holds on (-4, 1), where numerator and denominator opposite signs. Sign chart confirms. A tempting distractor is choice A (-∞, -4), 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.
2. If $f'(x)=\dfrac{(x-3)(x+1)}{x-2}$, on which interval(s) is $f$ decreasing?
   Answer: $(-\infty,-1)\cup(2,3)$ 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.
3. If $q'(x)=\sin x$ for $0<x<2\pi$, on which interval is $q$ increasing?
   Answer: $(0,\pi)$ This problem tests understanding of trigonometric derivatives and increasing intervals. Given q'(x) = sin x for 0 < x < 2π, we need to find where sin x > 0 for q to be increasing. On the interval (0, 2π), sin x is positive when 0 < x < π and negative when π < x < 2π. Therefore, q is increasing only on (0, π). Choice C might attract students who think the function increases on both intervals where the derivative is defined, forgetting that sin x changes sign at x = π. Choice E is similar but incorrectly lists the intervals as separate rather than a union. The strategy for trigonometric derivatives: sketch the derivative function and identify where it's above the x-axis.
4. If $f'(x)=3(x-1)^2(x+2)$, on which interval(s) is $f$ increasing?
   Answer: $(-2,\infty)$ 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) = 3(x-1)²(x+2), the roots are at x=1 (double) and x=-2. Since (x-1)² is always non-negative, f' ≥0 everywhere except where (x+2)<0 and the square is positive, but actually sign depends on (x+2), zero at x=1 and x=-2. f' >0 when x>-2 (since square positive, x+2>0), and f'<0 when x<-2 (x+2<0). At x=1, f'=0 but doesn't change sign, so increasing through there. Thus increasing on (-2, ∞). A tempting distractor like E, (-∞,-2)∪(1,∞), fails because it includes decreasing interval and splits unnecessarily at x=1 where sign doesn't change. Always create a sign chart for the derivative to systematically determine intervals of increase and decrease.
5. For $f'(x)=\dfrac{x-5}{(x+1)^2}$, on which interval(s) is $f$ decreasing?
   Answer: $(-\infty,-1)\cup(-1,5)$ 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.
6. If $f(x)=2x^3+3x^2-12x$, on which interval(s) is $f$ increasing?
   Answer: $(-\infty,-2)\cup(1,\infty)$ 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)=2x³ +3x² -12x, f'(x)=6x² +6x -12=6(x² +x -2)=6(x+2)(x-1), roots -2,1. f'>0 when (x+2)(x-1)>0, i.e., x<-2 or x>1. Thus increasing on (-∞,-2)∪(1,∞). A tempting distractor like B, (-2,1), fails because f'<0 there, decreasing. Always create a sign chart for the derivative to systematically determine intervals of increase and decrease.
7. For $f'(x)=-(x-2)^2(x+3)$, on which interval(s) is $f$ increasing?
   Answer: $(-\infty,-3)$ 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.
8. If $f'(x)=-(x-2)^2(x+3)$, on which interval(s) is $f$ increasing?
   Answer: $(-\infty,-3)$ This problem involves a derivative with a repeated factor. The derivative f'(x) = -(x - 2)²(x + 3) has zeros at x = -3 and x = 2. Note the negative sign in front and that (x - 2)² is always non-negative. Testing intervals: for x = -4, f'(-4) = -(36)(-1) = 36 > 0; for x = 0, f'(0) = -(4)(3) = -12 < 0; for x = 3, f'(3) = -(1)(6) = -6 < 0. Therefore, f is increasing only on (-∞, -3). Students often miss that x = 2 is not a sign change point because (x - 2)² doesn't change sign there—it just touches zero. The key insight: squared factors affect the magnitude but not the sign of the derivative except at their zero.
9. If $f'(x)=\dfrac{x-4}{(x+1)^2}$, on which interval is $f$ increasing?
   Answer: $(4,\infty)$ 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-4)/$(x+1)^2$ indicates that the function f(x) is increasing where f'(x) > 0 and decreasing where f'(x) < 0, with a vertical asymptote at x = -1. The critical point is x = 4, and the denominator is always positive except at x = -1, so the sign follows the numerator, dividing intervals into (-∞, -1), (-1, 4), and (4, ∞). Testing shows f'(x) < 0 in (-∞, -1) and (-1, 4), while f'(x) > 0 in (4, ∞), meaning f is increasing there. A tempting distractor is choice E, (-1, 4) and (4, ∞), but (-1, 4) is where f is decreasing since f'(x) < 0. 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.
10. Given $f'(x)=e^x-3$, on which interval(s) is $f$ decreasing?
    Answer: $(-\infty,\ln 3)$ 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)=$e^x$ - 3, >0 when $e^x$ >3, x>ln3, increasing; <0 when x<ln3, decreasing there. Yes. A tempting distractor like B, (ln3,∞), fails because that's increasing, not decreasing. Always create a sign chart for the derivative to systematically determine intervals of increase and decrease.
11. For $f(x)=x^3-3x^2-9x+1$, on which interval(s) is $f$ increasing?
    Answer: $(-\infty,-1)\cup(3,\infty)$ 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. To find where f(x) = x³ - 3x² - 9x + 1 is increasing, compute f'(x) = 3x² - 6x - 9 and find its roots at x = -1 and x = 3. The derivative changes sign from positive to negative at x = -1 and from negative to positive at x = 3, indicating f is increasing where f' > 0. Test points show f' > 0 on (-∞, -1) and (3, ∞), so f is increasing there. A tempting distractor like choice C, (-1, 3), fails because f' is negative there, meaning f is decreasing. Always create a sign chart for the derivative to systematically determine intervals of increase and decrease.
12. Let $q'(x)=\dfrac{(x+3)^2}{x-2}$ for $x\ne2$. On which interval(s) is $q$ increasing?
    Answer: $(2,\infty)$ 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 q is increasing where its derivative q'(x) = $(x+3)^2$/(x-2) is positive for x ≠ 2; the numerator is always non-negative, so the sign depends on the denominator, positive when x > 2. Testing intervals: negative in (-∞, 2) (except zero at x = -3) and positive in (2, ∞). Thus, q is increasing on (2, ∞). A tempting distractor like (-∞, -3) ∪ (-3, 2) fails because q'(x) is negative there, indicating decreasing behavior, not increasing. Always create a sign chart using the roots of the derivative to systematically determine where it is positive or negative.
13. For $f'(x)=x(x-5)(x+2)$, on which interval is $f$ decreasing?
    Answer: $(-\infty,-2)$ and $(0,5)$ 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(x-5)(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, x = 0, and x = 5, dividing the real line into intervals: (-∞, -2), (-2, 0), (0, 5), and (5, ∞). Sign analysis with test points shows f'(x) < 0 in (-∞, -2) and (0, 5), while positive in the others, meaning f is decreasing in those intervals. A tempting distractor is choice B, (-2, 0) and (5, ∞), but those are where f is increasing since f'(x) > 0 there. 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.
14. Let $f'(x)=\dfrac{(x+3)(x-1)}{x^2}$. On which interval(s) is $f$ increasing?
    Answer: $(-\infty,-3)\cup(1,\infty)$ 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 $f'(x) = \dfrac{(x+3)(x-1)}{x^2}$, critical points are $x=-3$, $x=1$ (numerator zeros), $x=0$ (undefined). Sign analysis: $f' > 0$ when even number of negative factors, which is on $(-\infty,-3) \cup (1, \infty)$, excluding $x=0$. At $x=0$ undefined, but intervals are open. So increasing there. A tempting distractor like D, $(-3,1)$, fails because it includes regions where $f' < 0$, like $(-3,0)$. Always create a sign chart for the derivative to systematically determine intervals of increase and decrease.
15. Given $f'(x)=-(x-3)^2(x+2)$, on which interval is $f$ decreasing?
    Answer: $(-2,\infty)$ 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.
16. For $f'(x)=\dfrac{(x-1)^3}{x+4}$, on which interval(s) is $f$ decreasing?
    Answer: $(-4,1)$ 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 - 1)³/(x + 4) < 0. This occurs when numerator and denominator have opposite signs, specifically in (-4, 1). Sign chart confirms negative in that interval. A tempting distractor is choice A (-∞, -4) ∪ (1, ∞), 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.
17. For $f(x)=\ln(x)+\dfrac{4}{x}$ with $x>0$, on which interval(s) is $f$ increasing?
    Answer: $(4,\infty)$ This problem requires finding where a logarithmic function increases. First, find f'(x) = 1/x - 4/x² = (x - 4)/x². Since x > 0 by the domain restriction, x² is always positive, so the sign of f'(x) depends on (x - 4). When 0 < x < 4, we have x - 4 < 0, making f'(x) < 0. When x > 4, we have x - 4 > 0, making f'(x) > 0. Therefore, f is increasing on (4, ∞). Students might incorrectly think the function increases for all x > 0 because ln(x) is increasing, but the term 4/x decreases fast enough to make f decrease initially. The strategy: combine derivatives algebraically first, then analyze the resulting rational expression's sign.
18. Let $f'(x)=x(x-6)(x+2)$. On which interval(s) is $f$ increasing?
    Answer: $(-2,0)\cup(6,\infty)$ 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.
19. For $f'(x)=2-\sqrt{x}$ with domain $x\ge0$, on which interval is $f$ increasing?
    Answer: $[0,4)$ 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) = 2 - √x for x ≥ 0 indicates that the function f(x) is increasing where f'(x) > 0 and decreasing where f'(x) < 0. The critical point is where √x = 2, so x = 4, dividing the domain into [0, 4) and [4, ∞). Analysis shows f'(x) > 0 in [0, 4) and f'(x) < 0 in (4, ∞), with f'(4) = 0, meaning f is increasing in [0, 4). A tempting distractor is choice A, (0, 4), but this omits x = 0 where f'(0) = 2 > 0. 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.
20. If $f'(x)=\dfrac{x+2}{(x-3)^2}$, on which interval(s) is $f$ increasing?
    Answer: $(-2,3)\cup(3,\infty)$ 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.

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