AP Calculus AB

AP Calculus AB Practice Test: Practice Test 4

Practice Test 4 for AP Calculus AB: real questions and explanations from the Varsity Tutors practice-test pool.

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Question 1 of 25

A function $v$ is continuous on $[-4,4]$ and satisfies $v(-4)=2$ and $v(4)=2$ . Which statement must be true?

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Question 1

A function $v$ is continuous on $[-4,4]$ and satisfies $v(-4)=2$ and $v(4)=2$ . Which statement must be true?

$v$ has an absolute maximum at an interior point of $(-4,4)$ .
$v$ has an absolute minimum at an interior point of $(-4,4)$ .
$v$ has at least one absolute maximum and at least one absolute minimum on $[-4,4]$ . (correct answer)
$v$ must have a critical point in $(-4,4)$ .
$v$ cannot have any local extrema because the endpoint values are equal.

Explanation: This question applies the Extreme Value Theorem to a function with equal endpoint values. Since v is continuous on the closed interval [-4,4], the EVT guarantees that v must attain both an absolute maximum and an absolute minimum somewhere on [-4,4]. The fact that v(-4)=v(4)=2 doesn't prevent extrema from existing; v could rise above 2 or dip below 2 in the interior. Choice D incorrectly assumes equal endpoints require an interior critical point, but v could be constant at 2 throughout. The EVT guarantee: continuous function + closed interval = both absolute extrema exist, regardless of endpoint values or behavior.

Question 2

If $h'(1)=0$ and $h$ is concave down on $(0,2)$ , what is the classification of $x=1$ ?

Local minimum at $x=1$ because concave down implies decreasing.
Point of inflection at $x=1$ because concavity is given.
Local maximum at $x=1$ because concave down with a critical point. (correct answer)
Neither; concavity alone cannot classify a critical point.
Absolute maximum at $x=1$ because $h'(1)=0$ .

Explanation: This problem applies the Second Derivative Test using concavity information. We know h'(1) = 0, so x = 1 is a critical point. Since h is concave down on (0,2), we know h''(x) < 0 for all x in this interval, including h''(1) < 0. By the Second Derivative Test, when f'(c) = 0 and f''(c) < 0, the function has a local maximum at x = c. The concave down shape creates an inverted U, placing a peak at the critical point. Choice A incorrectly claims a minimum despite the downward concavity. Remember: concave down at a critical point always means local maximum, regardless of how the concavity is described.

Question 3

A slope field is for $\frac{dy}{dx}=x-y^2$ . At the point $(0,2)$ , the solution curve is locally

increasing because slope is positive.
decreasing because slope is negative. (correct answer)
constant because slope is zero.
undefined because slope is infinite.
increasing because slope is zero.

Explanation: This question uses slope field reasoning to determine local solution behavior at specific points. For dy/dx = x - y² at point (0,2), we evaluate: dy/dx = 0 - 2² = -4 < 0. A negative slope indicates the solution curve is locally decreasing at this point. The slope value of -4 means the solution decreases at a relatively steep rate near (0,2). Choice A fails because the slope is negative, not positive, so the solution decreases rather than increases. When analyzing local behavior in slope fields, substitute the coordinates of the specific point into the differential equation to find the slope sign and magnitude.

Question 4

The temperature of a chemical solution is $T(t)$ °C after $t$ minutes. What is the meaning of $T'(7)$ ?

The temperature at $t=7$ , in °C
The change in temperature from $t=0$ to $t=7$ , in °C
The rate of temperature change at $t=7$ , in °C per minute (correct answer)
The time when the temperature first reaches 7°C, in minutes
The rate of time change with respect to temperature at $t=7$ , in minutes per °C

Explanation: This question tests derivative interpretation for a temperature function. Since T(t) represents temperature in °C at time t minutes, T'(7) represents the instantaneous rate of change of temperature with respect to time at t = 7 minutes. This tells us how fast the temperature is changing at that exact moment, measured in °C per minute. A common mistake is thinking T'(7) represents the temperature at t = 7 (that would be T(7)) or the total change in temperature over 7 minutes. Another error is confusing units - the derivative of temperature with respect to time has units of °C per minute, not just °C. When interpreting derivatives, always ask: what is changing (temperature) and with respect to what (time)?

Question 5

A rumor spreads so that the rate $\dfrac{dP}{dt}$ is proportional to $P(1000-P)$ ; which differential equation models this?

$\dfrac{dP}{dt}=kP(1000-P)$ (correct answer)
$\dfrac{dP}{dt}=k(1000-P)$
$\dfrac{dP}{dt}=k\dfrac{P}{1000-P}$
$\dfrac{dP}{dt}=kP(1000+P)$
$\dfrac{dt}{dP}=kP(1000-P)$

Explanation: This problem requires modeling logistic growth with a differential equation. The rumor spread rate dP/dt is given as proportional to P(1000 - P), which directly translates to dP/dt = kP(1000 - P). This is the logistic model where P represents people who know the rumor and (1000 - P) represents those who don't. The product P(1000 - P) captures how spread rate depends on both groups: more knowers mean more spreaders, but fewer unknowers mean fewer targets. Choice B incorrectly omits the P factor, missing the dependence on current spreaders. When modeling bounded growth, look for rates proportional to both the current amount and the remaining capacity.

Question 6

A function $f$ is continuous on $[0,2]$ and differentiable on $(0,2)$ with $f(0)=1$ and $f(2)=1$ . Does MVT guarantee some $c$ with $f'(c)=0$ ?

Yes, because $f(0)=f(2)$ , so the derivative must be zero everywhere.
No, because MVT only applies when $f(0)\ne f(2)$ .
Yes, because $f$ is continuous on $[0,2]$ , differentiable on $(0,2)$ , and $\frac{f(2)-f(0)}{2-0}=0$ . (correct answer)
Yes, because $0$ is between $f(0)$ and $f(2)$ .
No, because $f'(c)=0$ requires $f$ to have a maximum or minimum.

Explanation: This is a special case of MVT known as Rolle's Theorem, which applies when f(a) = f(b). Here, f is continuous on [0,2] and differentiable on (0,2), with f(0) = f(2) = 1. The average rate of change is (f(2)-f(0))/(2-0) = (1-1)/2 = 0/2 = 0. Since MVT guarantees a point where the derivative equals this average rate, there must exist at least one c in (0,2) where f'(c) = 0. This doesn't mean the derivative is zero everywhere (choice A's error), just at least at one point. A common misconception is that f'(c) = 0 requires a maximum or minimum, but it could also occur at an inflection point. When endpoints have equal function values, MVT simplifies to guaranteeing a horizontal tangent somewhere.

Question 7

A function is $K(x)=\sin\!\left(\dfrac{1}{1+\cos x}\right)$ . What is $K'(x)$ ?

$\cos\!\left(\dfrac{1}{1+\cos x}\right)$
$\dfrac{\sin x}{(1+\cos x)^2}\cos\!\left(\dfrac{1}{1+\cos x}\right)$ (correct answer)
$-\dfrac{\sin x}{(1+\cos x)^2}\cos\!\left(\dfrac{1}{1+\cos x}\right)$
$-\dfrac{\sin x}{1+\cos x}\cos\!\left(\dfrac{1}{1+\cos x}\right)$
$\cos\!\left(\dfrac{1}{1+\cos x}\right)\cdot \dfrac{1}{1+\cos x}$

Explanation: To differentiate $K(x) = \sin\left(\dfrac{1}{1 + \cos x}\right)$ , chain rule. Outer $\sin(n)$ , derivative $\cos(n)$ , $n = (1 + \cos x)^{-1}$ . $n' = - (1 + \cos x)^{-2} \cdot (-\sin x) = \dfrac{\sin x}{(1 + \cos x)^2}$ . So $K'(x) = \cos(n) \cdot \dfrac{\sin x}{(1 + \cos x)^2}$ . A common omission is sign error in $n'$ . Pattern: Trig of reciprocal trig needs careful power chain.

Question 8

A continuous function $b$ has $b(6)=1$ and $b(9)=9$ . Which is guaranteed for some $c\in(6,9)$ ?

$b(c)=0$
$b(c)=10$
$b(c)=5$ (correct answer)
$b(c)=-5$
No value is guaranteed because $b$ could be discontinuous.

Explanation: The Intermediate Value Theorem (IVT) ensures continuous b with b(6) = 1 and b(9) = 9 takes every value between 1 and 9. Since 1 < 5 < 9, there exists c in [6, 9] with b(c) = 5, and as 5 ≠ 1, 5 ≠ 9, c is in (6, 9). This is a standard intermediate value application. A common error is assuming discontinuity allows skipping values, but the question states continuity. Selecting 0 or 10, outside the range, is incorrect. Transferable IVT checklist: 1. Confirm the function is continuous on [a, b]. 2. Verify k is between f(a) and f(b). 3. If yes, there exists c in [a, b] with f(c) = k; if k is strictly between and ≠ f(a), ≠ f(b), then c is in (a, b).

Question 9

A chemical amount changes at rate $A'(t)=4t^9+11$ . Which expression is a general antiderivative of $A'(t)$ ?

$\frac{2}{5}t^{10}+11t+C$ (correct answer)
$\frac{4}{10}t^{10}+11t+C$
$\frac{2}{5}t^{10}+11t$
$\frac{4}{9}t^{10}+11t+C$
$4t^{10}+11t+C$

Explanation: This problem requires finding a general antiderivative of A'(t) = 4t⁹ + 11, demonstrating basic antiderivative reasoning. To find the antiderivative, we apply the power rule for integration: ∫tⁿdt = t^(n+1)/(n+1) + C. For 4t⁹, we get 4t¹⁰/10 = (2/5)t¹⁰; for the constant 11, we get 11t. Combining these terms with the constant of integration gives (2/5)t¹⁰ + 11t + C. Choice B ((4/10)t¹⁰ + 11t + C) shows the unsimplified fraction 4/10 instead of the reduced form 2/5, though both are mathematically equivalent. The key integration strategy is to apply the power rule carefully and simplify fractions when possible for cleaner expressions.

Question 10

In the context of motion, calculus allows us to define instantaneous velocity. This is a profound shift from algebra, which can only be used to compute what type of velocity?

Terminal velocity, where forces are balanced.
Initial velocity, at the beginning of motion.
Average velocity, over a non-zero interval of time. (correct answer)
Constant velocity, where the rate does not change.

Explanation: Algebraic tools are sufficient for calculating rates of change over a discrete, non-zero interval, which defines average velocity. The concept of an instantaneous velocity, the velocity at a single moment in time, requires the tool of limits, which is the foundation of calculus.

Question 11

A function is $E(x)=\csc(x^2-x+4)$ . What is $E'(x)$ ?

$-(2x-1)\csc(x^2-x+4)\cot(x^2-x+4)$ (correct answer)
$(2x-1)\csc(x^2-x+4)\cot(x^2-x+4)$
$-\csc(x^2-x+4)\cot(x^2-x+4)$
$-(2x-1)\sec(x^2-x+4)\tan(x^2-x+4)$
$-(2x-1)\csc^2(x^2-x+4)$

Explanation: This problem tests the skill of differentiating reciprocal trigonometric functions like cotangent, secant, and cosecant. For E(x) = csc(x² - x + 4), the derivative is -csc(u) cot(u) u' where u = x² - x + 4. Then u' = 2x - 1. So E'(x) = -csc(x² - x + 4) cot(x² - x + 4) * (2x - 1). Choice E substitutes csc² incorrectly, tempting if mixing with cot's derivative, but csc uses csc cot with a negative. Consistently recall that csc and cot derivatives both have negatives, but their forms differ.

Question 12

For $f(x)=|x-4|+|x-4|^2$ , does $f'(4)$ exist, and why?

Yes; adding $|x-4|^2$ removes the corner, so $f'(4)$ exists.
No; the $|x-4|$ term creates a corner at $4$ , so $f'(4)$ does not exist. (correct answer)
Yes; $f$ is continuous at $4$ , so $f'(4)$ exists.
No; $f$ has a jump discontinuity at $4$ .
Yes; corners imply the derivative equals $1$ .

Explanation: This question assesses the connection between differentiability and continuity in AP Calculus AB, evaluating combined absolute value functions. For f(x) = |x-4| + |x-4|^2, it is continuous at x=4 with value 0. However, the left-hand derivative is -1, while the right-hand is 1, due to the |x-4| term. This mismatch creates a corner, so f'(4) does not exist. A tempting distractor might say yes because of continuity, but the slopes must match. Remember this differentiability checklist: verify continuity at the point, then check that left- and right-hand derivatives exist and are equal.

Question 13

A graph of $f$ shows a smooth curve except a jump at $x=0$ ; does $f'(0)$ exist?

Yes; smoothness on each side implies $f'(0)$ exists.
No; a jump discontinuity means $f$ is not continuous, so $f'(0)$ does not exist. (correct answer)
Yes; if one-sided derivatives exist, the derivative exists.
No; a jump indicates a corner, so $f'(0)$ does not exist.
Yes; discontinuities imply vertical tangents with defined slope.

Explanation: This question assesses the connection between differentiability and continuity in AP Calculus AB, considering jump discontinuities. A jump at x=0 means the left- and right-hand limits differ, causing discontinuity. Since differentiability requires continuity, f'(0) cannot exist. The smoothness on each side is irrelevant without overall continuity. A tempting distractor might say yes if one-sided derivatives exist, but continuity is prerequisite. Remember this differentiability checklist: verify continuity at the point, then check that left- and right-hand derivatives exist and are equal.

Question 14

The total cost to produce $x$ units of a certain product is given by a differentiable function $C(x)$ .

If the cost to produce 50 units is $C(50) = \$1500$ and the cost to produce 52 units is $C(52) = \$1580$ , which of the following is the best estimate for the marginal cost when 50 units are produced?

$40$ per unit (correct answer)
$80$ per unit
$2$ per unit
$30$ per unit

Explanation: The marginal cost when 50 units are produced is $C'(50)$ . This can be estimated by the average rate of change of the cost function over the interval $[50, 52]$ . The calculation is: $C'(50) \approx \frac{C(52) - C(50)}{52 - 50} = \frac{1580 - 1500}{2} = \frac{80}{2} = 40$ dollars per unit.

Question 15

Given $C(3)=4$ and $C'(3)=-1$ , approximate $C(2.8)$ using local linearity at $3$ .

$3.8$
$4.2$ (correct answer)
$3.98$
$-0.2$
$2.8$

Explanation: In AP Calculus AB, local linearity refers to using the tangent line for approximations near a point, which is the skill applied here. For C(3) = 4 and C'(3) = -1, the approximation is C(x) ≈ 4 -1*(x - 3). For x = 2.8, this yields 4 -1*(-0.2) = 4.2. Conceptually, the tangent line mimics the function's behavior locally, with the derivative as its slope. A frequent error is forgetting the negative sign in the derivative or miscomputing Δx as positive. The general form is f(x) ≈ f(a) + f'(a)(x - a). For any similar problem, calculate the deviation Δx accurately and multiply by the derivative before adding to the function value.

Question 16

Which integral gives the volume when the region under $y=2-x$ from $x=0$ to $x=2$ is revolved about the $x$ -axis?

$\pi\displaystyle\int_{0}^{2}(2-x)\,dx$
$\pi\displaystyle\int_{0}^{2}(2-x)^{2}\,dx$ (correct answer)
$\pi\displaystyle\int_{0}^{2}(x-2)^{2}\,dy$
$\pi\displaystyle\int_{0}^{2}x^{2}\,dx$
$\pi\displaystyle\int_{0}^{2}(2-(2-x))^{2}\,dx$

Explanation: This problem requires the disc method for revolving around the x-axis. When rotating the region under y = 2 - x from x = 0 to x = 2 around the x-axis, we create circular discs perpendicular to the x-axis. Each disc has radius equal to the y-value at that x-position, which is 2 - x, so the disc's area is π(2 - x)². The volume is the integral of these disc areas: π∫₀²(2 - x)²dx. Choice A incorrectly uses π∫₀²(2 - x)dx, which forgets to square the radius function—this is a common error that gives an incorrect result. The disc method for x-axis rotation always requires squaring the function: use π∫[f(x)]²dx, never just π∫f(x)dx.

Question 17

Let $A(x)=\int_{0}^{x} f(t)\,dt$ . If $f(x)<0$ and $f$ is increasing on $(1,4)$ , what describes $A$ on $(1,4)$ ?

Decreasing and concave up (correct answer)
Increasing and concave up
Decreasing and concave down
Increasing and concave down
Constant and concave up

Explanation: This problem combines both monotonicity and concavity analysis of accumulation functions. Since $A'(x) = f(x) < 0$ on $(1,4)$ , the function $A$ is decreasing throughout this interval. For concavity, $A''(x) = f'(x)$ , and since $f$ is increasing, we have $f'(x) > 0$ , making $A''(x) > 0$ . This means $A$ is concave up on $(1,4)$ . Choice C might seem appealing since $f < 0$ , but concavity depends on $f'$ , not $f$ itself. To analyze accumulation functions completely, use the integrand for monotonicity and the integrand's derivative for concavity.

Question 18

A company’s cost is $C(x)=7x^5-4x^2+9$ . What is $C'(x)$ ?

$35x^4-8x$ (correct answer)
$35x^5-8x^2$
$7x^4-4x+0$
$35x^4-4x$
$35x^4-8x+9$

Explanation: This problem involves applying the power rule to find the derivative of a cost function. The power rule tells us that the derivative of x^n is nx^(n-1). For C(x) = 7x^5 - 4x^2 + 9, we differentiate each term separately: the derivative of 7x^5 is 7·5x^4 = 35x^4, the derivative of -4x^2 is -4·2x^1 = -8x, and the derivative of the constant 9 is 0. Thus, C'(x) = 35x^4 - 8x + 0 = 35x^4 - 8x. Choice C incorrectly keeps the +0 term and uses wrong coefficients, failing to multiply the original coefficients by the powers. Remember that constants always have a derivative of zero and can be omitted from the final answer.

Question 19

The derivative of $f(x) = \frac{4x^5 - 6x^3}{2x^2}$ for $x \neq 0$ is

$6x^2 - 3$ (correct answer)
$4x - 3$
$\frac{20x^4 - 18x^2}{4x}$
$\frac{2}{3}x^4 - \frac{3}{2}x^2$

Explanation: First, simplify the expression for $f(x)$ by dividing each term in the numerator by the denominator: $f(x) = \frac{4x^5}{2x^2} - \frac{6x^3}{2x^2} = 2x^3 - 3x$ . Now, differentiate the simplified polynomial using the power rule: $f'(x) = 2(3x^2) - 3(1) = 6x^2 - 3$ .

Question 20

For $y=\arcsec(5x)$ , what is $\dfrac{dy}{dx}$ ?

$\dfrac{5}{\sqrt{(5x)^2-1}}$
$\dfrac{5}{|5x|\sqrt{(5x)^2-1}}$ (correct answer)
$-\dfrac{5}{|5x|\sqrt{(5x)^2-1}}$
$\dfrac{1}{|5x|\sqrt{(5x)^2-1}}$
$5\sec^2(5x)$

Explanation: This question tests the skill of differentiating inverse trigonometric functions, specifically the inverse secant. The derivative of arcsec(u) is 1/(|u| √(u² - 1)) times du/dx. u = 5x, du/dx = 5, positive. Secant property. Result: 5 / (|5x| √((5x)² - 1)). Choice C negative, confuses with arccsc. Recognize arcsec positive with absolute.

Question 21

What integral gives the volume if the base is bounded by $y=\tan x$ and $y=0$ on $0\le x\le \frac{\pi}{4}$ , with semicircular cross sections perpendicular to the $x$ -axis?

$\displaystyle \int_{0}^{\pi/4} \frac{\pi}{8}\,\tan x\,dx$
$\displaystyle \int_{0}^{\pi/4} \frac{\pi}{8}\,(\tan x)^2\,dx$ (correct answer)
$\displaystyle \int_{0}^{\pi/4} \frac{\pi}{4}\,(\tan x)^2\,dx$
$\displaystyle \int_{0}^{\pi/4} \frac{1}{2}\,(\tan x)^2\,dx$
$\displaystyle \int_{0}^{\pi/4} \pi\,(\tan x)^2\,dx$

Explanation: This problem involves computing the volume of a solid with semicircular cross-sections perpendicular to the x-axis, a key skill in understanding non-rectangular cross-sections. The diameter of each semicircle is the vertical distance between the curves y=tan x and y=0, which is tan x. The area of a semicircle is (1/2) π r², where r = (tan x)/2. Therefore, the area simplifies to π (tan x)² / 8. The volume is the integral of this area from x=0 to x=π/4. A common mistake is to use π (tan x)² / 4, as in choice C, which would be for full circles with radius (tan x)/2. Remember, for any cross-section shape, determine the base length, apply the area formula correctly, and integrate along the perpendicular axis.

Question 22

A model gives $f'(x)=9x^{2}-\frac{5}{2}x^{4}$ . Which is an antiderivative of $f'(x)$ ?

$3x^{3}-\frac{1}{2}x^{5}+C$
$3x^{3}-\frac{5}{8}x^{5}+C$ (correct answer)
$9x^{3}-\frac{5}{8}x^{5}+C$
$3x^{3}-\frac{5}{8}x^{4}+C$
$3x^{3}-\frac{5}{8}x^{5}$

Explanation: This problem tests finding antiderivatives using the power rule for integration on each term separately. For f'(x) = 9x² - 5/2·x⁴, we integrate: the antiderivative of 9x² is 9x³/3 = 3x³, and the antiderivative of -5/2·x⁴ is (-5/2)·x⁵/5 = -5x⁵/10 = -5/8·x⁵. The complete antiderivative is 3x³ - 5/8·x⁵ + C. Choice A (3x³ - 1/2·x⁵ + C) incorrectly computes the second term—it appears to divide 5/2 by 5 incorrectly, getting 1/2 instead of 5/8. When integrating, carefully handle fractions: multiply the coefficient by 1/(n+1) where n is the original exponent, and always include the constant C.

Question 23

The derivative satisfies $k'(x)<0$ on $( -9,-4)$ and $k'(x)>0$ on $(-4,-1)$ . Where does $k$ have a local minimum?

At $x=-9$ , $k$ has a local minimum.
At $x=-4$ , $k$ has a local maximum.
At $x=-4$ , $k$ has a local minimum. (correct answer)
At $x=-1$ , $k$ has a local minimum.
There is no local extremum.

Explanation: This problem involves the First Derivative Test to identify local extrema based on the sign of the derivative. The First Derivative Test indicates that a local minimum occurs where the derivative changes from negative to positive. Here, k'(x) is negative on (-9,-4) and positive on (-4,-1), showing a sign change from negative to positive at x=-4. This means the function decreases before x=-4 and increases after, confirming a local minimum. A tempting distractor is choice B, which claims a local maximum at x=-4, but that would require a positive to negative change, not observed here. Remember, to apply this test effectively, always check for sign changes around critical points where the derivative is zero or undefined.

Question 24

If $r$ is continuous on $[0,5]$ and differentiable on $(0,5)$ with $r(0)=-6$ and $r(5)=-1$ , does MVT guarantee $c$ with $r'(c)=1$ ?

Yes, because $r$ is continuous on $[0,5]$ , so the derivative must equal $1$ somewhere.
No, because the derivative cannot be positive if both endpoint values are negative.
Yes, because $\frac{-1-(-6)}{5-0}=1$ and $r$ is continuous on $[0,5]$ and differentiable on $(0,5)$ . (correct answer)
No, because MVT requires differentiability on $[0,5]$ including endpoints.
Yes, because $r(0)\ne r(5)$ , so Rolle’s Theorem applies.

Explanation: Despite both endpoint values being negative, the Mean Value Theorem still applies because r is continuous on [0,5] and differentiable on (0,5). The average rate of change is (r(5)-r(0))/(5-0) = (-1-(-6))/5 = 5/5 = 1. MVT guarantees there exists at least one c in (0,5) where r'(c) = 1. The derivative being positive means the function is increasing at that point, which makes sense since r(5) > r(0) even though both values are negative. A common misconception is thinking negative function values prevent positive derivatives, but derivatives measure rate of change, not the values themselves. Remember: MVT works regardless of whether function values are positive, negative, or zero.

Question 25

For $s$ with $s'(x)=\dfrac{x^2-9}{x^2+1}$ , on which interval is $s$ increasing?

$(-3,3)$
$(-\infty,-3)\cup(3,\infty)$ (correct answer)
$(-\infty,3)$
$(0,\infty)$
$(-\infty,-3)\cup(-3,3)$

Explanation: 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.

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