Test de práctica AP Calculus AB

Marca todas las que correspondan.

• A. lim x->0 of 6/x
• B. lim x->3 of (x^2 - 9)/(x - 3)
• C. lim x->0 of sin(6x)/x
• D. lim x->6 of (x - 6)/x
• E. lim x->infinity of (6x^2 + 1)/(x^2 + 5)

Respuesta: B. lim x->3 of (x^2 - 9)/(x - 3), C. lim x->0 of sin(6x)/x, E. lim x->infinity of (6x^2 + 1)/(x^2 + 5)

For the second, factor x^2 - 9 into (x - 3)(x + 3), cancel the common factor and substitute: x + 3 becomes 6. For the third, sin(kx)/x approaches k as x approaches 0, and here k is 6. For the last, divide top and bottom by x^2: the limit is 6/1 = 6 because the leading coefficients decide the end behavior. The first does not exist, since 6/x grows without bound from the right and falls without bound from the left. The fourth is 0/6 = 0 at x = 6.

• A. 2
• B. 3
• C. 4
• D. 5

Respuesta: B. 3

Continuity at x = 2 needs the left limit, the right limit and f(2) to agree. From the left, x^2 + 1 approaches 4 + 1 = 5. At and from the right, kx - 1 gives 2k - 1. Setting 2k - 1 = 5 gives k = 3. Always check all three pieces: a function can have equal one-sided limits and still be undefined or different at the point itself.

• A. e
• B. e + 1
• C. 2e + 1
• D. 2e

Respuesta: D. 2e

Use the product rule: the derivative of x times e^x is (1)(e^x) + (x)(e^x) = e^x (1 + x). At x = 1 that is e * 2 = 2e. The common slip is to differentiate each factor and multiply the results, which gives e^x and loses the x e^x term.

• A. 0
• B. 8
• C. 12
• D. The limit does not exist

Respuesta: C. 12

This is the limit definition of the derivative: it equals f'(2) for f(x) = x^3, and f'(x) = 3x^2 gives 3(4) = 12. You can also expand directly: (2 + h)^3 = 8 + 12h + 6h^2 + h^3, so the numerator is 12h + 6h^2 + h^3. Dividing by h leaves 12 + 6h + h^2, which approaches 12. Direct substitution gives 0/0, which means the form is indeterminate, not that the limit is 0.

• A. 12(3x + 1)^3
• B. 4(3x + 1)^3
• C. 12(3x + 1)^4
• D. 3(3x + 1)^4

Respuesta: A. 12(3x + 1)^3

Apply the chain rule. The outside function is something to the 4th power, so its derivative is 4 times the inside raised to the 3rd power. Then multiply by the derivative of the inside, which is 3. That gives 4 * 3 * (3x + 1)^3 = 12(3x + 1)^3. The answer 4(3x + 1)^3 is what you get if you forget the factor from the inside function.

• A. 3/4
• B. -3/4
• C. 4/3
• D. -4/3

Respuesta: B. -3/4

Differentiate both sides with respect to x and remember that y depends on x: 2x + 2y(dy/dx) = 0. Solving gives dy/dx = -x/y. At (3, 4) that is -3/4. It also makes geometric sense: the radius to (3, 4) has slope 4/3, and a tangent to a circle is perpendicular to the radius, so its slope is the negative reciprocal, -3/4.

Marca todas las que correspondan.

• A. The acceleration is positive at t = 1
• B. The particle is at rest at t = 1 and at t = 3
• C. The particle is speeding up at t = 2
• D. The particle is moving to the left at t = 2
• E. The particle is slowing down at t = 0.5

Respuesta: B. The particle is at rest at t = 1 and at t = 3, D. The particle is moving to the left at t = 2, E. The particle is slowing down at t = 0.5

Velocity is s'(t) = 3t^2 - 12t + 9 = 3(t - 1)(t - 3), which is 0 at t = 1 and t = 3. Acceleration is s''(t) = 6t - 12. At t = 2, velocity is 3(1)(-1) = -3, so the particle moves left, and acceleration is 0, so it is neither speeding up nor slowing down at that instant. At t = 1, acceleration is -6, which is negative. At t = 0.5, velocity is positive (3.75) and acceleration is negative (-9), so the signs differ and the particle is slowing down. A particle speeds up only when velocity and acceleration have the same sign.

• A. 10 pi cm^2 per second
• B. 25 pi cm^2 per second
• C. 100 pi cm^2 per second
• D. 20 pi cm^2 per second

Respuesta: D. 20 pi cm^2 per second

Write the relationship, A = pi r^2, then differentiate with respect to time: dA/dt = 2 pi r (dr/dt). Only now substitute the numbers: 2 pi (5)(2) = 20 pi. Substituting r = 5 before differentiating is the classic related rates mistake, because it turns a changing quantity into a constant. The answer 10 pi comes from leaving out dr/dt, and 25 pi is the area itself at that instant.

• A. 1
• B. 1.5
• C. 2
• D. 2.5

Respuesta: C. 2

The average rate of change is (f(3) - f(1))/(3 - 1) = (9 - 1)/2 = 4. Now solve f'(c) = 2c = 4, which gives c = 2. The theorem needs f to be continuous on [1, 3] and differentiable on (1, 3), and a polynomial meets both. It promises that such a c exists. It does not say how to find it, so you still have to solve for it.

Marca todas las que correspondan.

• A. f is increasing on the interval (-1, 1)
• B. f has a local maximum at x = -1
• C. f is concave up on the interval (-infinity, 0)
• D. f has a point of inflection at x = 0
• E. f has a local minimum at x = -1

Respuesta: B. f has a local maximum at x = -1, D. f has a point of inflection at x = 0

The first derivative is f'(x) = 3x^2 - 3 = 3(x - 1)(x + 1). It is positive for x < -1, negative between -1 and 1, and positive for x > 1. So f rises, then falls, then rises: x = -1 is a local maximum and x = 1 is a local minimum, and f is decreasing on (-1, 1). The second derivative is f''(x) = 6x. It is negative for x < 0, so f is concave down there, and positive for x > 0. Because f'' changes sign at x = 0, that point is an inflection point.

On the closed interval [0, 4], f(x) = x^2 - 4x + 1 has an absolute minimum value of _____ and an absolute maximum value of _____.

Opciones para el hueco 1: 1, -3, -4, -2

Opciones para el hueco 2: -3, 0, 1, 5

Respuesta: -3, 1

Use the candidates test. Set f'(x) = 2x - 4 = 0 to get the critical point x = 2, where f(2) = 4 - 8 + 1 = -3. Then check the endpoints: f(0) = 1 and f(4) = 16 - 16 + 1 = 1. The smallest value is -3 and the largest is 1. On a closed interval the absolute extremes happen at a critical point or an endpoint, so skipping the endpoints is the usual way to lose this question.

• A. 5
• B. 4
• C. 12
• D. 25

Respuesta: A. 5

This is the first part of the Fundamental Theorem of Calculus: when the upper limit is x, the derivative of the accumulated integral is the integrand evaluated at x. So g'(x) = sqrt(x^2 + 9), and g'(4) = sqrt(16 + 9) = sqrt(25) = 5. You never need to find an antiderivative, and trying to would be hard here anyway.

• A. 4
• B. 5
• C. 9
• D. 14

Respuesta: B. 5

Each subinterval has width 1, and a left sum uses the left endpoint of each: x = 0, 1 and 2. The sum is 1 * (0 + 1 + 4) = 5. A right sum would use x = 1, 2 and 3 and give 1 + 4 + 9 = 14. The exact integral is 9, which lies between the two. Because x^2 is increasing on [0, 3], the left sum underestimates and the right sum overestimates.

The definite integral of 3x^2 dx from x = 0 to x = 2 equals _____, so the average value of 3x^2 on that interval equals _____.

Opciones para el hueco 1: 4, 6, 8, 12

Opciones para el hueco 2: 2, 4, 6, 8

Respuesta: 8, 4

An antiderivative of 3x^2 is x^3, so the integral is 2^3 - 0^3 = 8. The average value of a function on [a, b] is the integral divided by the length of the interval: 8/(2 - 0) = 4. Notice that the average value is not the average of the endpoint values. At the endpoints 3x^2 is 0 and 12, and their mean would be 6.

The solution to the differential equation dy/dx = 2y with y(0) = 3 is y = _____ e^(_____x).

Opciones para el hueco 1: 1, 2, 3, 6

Opciones para el hueco 2: 1, 2, 3, 6

Respuesta: 3, 2

This is exponential growth. Separate the variables: dy/y = 2 dx, so ln|y| = 2x + C, which gives y = A e^(2x). The initial condition y(0) = 3 means A e^0 = A = 3, so y = 3e^(2x). The rate constant 2 sits in the exponent and the starting value 3 sits in front. A quick check: differentiating 3e^(2x) gives 6e^(2x), which is 2 times y.