ฝึกแคลคูลัส AP AB

Substitute x = 3 into the linear expression: 2(3) + 1 = 6 + 1 = 7. Because the expression is continuous, direct substitution gives the limit without any factoring.

Direct substitution gives x^2 - 4 = 0^2 - 4 = -4. Since x^2 - 4 is a polynomial, it is continuous everywhere, so the limit equals the function value at x = 0.

Factor the numerator: (x - 2)(x + 2)/(x - 2) = x + 2, so the limit is 2 + 2 = 4.

This is a standard limit: lim x->0 sin(x)/x = 1.

The numerator approaches 0 faster than x, so the limit is 0.

Divide numerator and denominator by x: the limit is 5/2.

As x grows without bound, the numerator stays at 1 while the denominator increases, so the fraction shrinks toward 0. The horizontal asymptote is therefore y = 0.

Factor to (x - 1)(x + 1)/(x + 1) = x - 1, so the limit is -1 - 1 = -2.

Rewrite as (3/2) x (2x/sin(2x)); since sin(2x)/(2x) -> 1, the limit is 3/2.

Factor x^2 - 9 = (x - 3)(x + 3), cancel x - 3, and get 1/(3 + 3) = 1/6.

tan(x)/x = sin(x)/x x 1/cos(x); both factors approach 1, so the limit is 1.

Since sqrt(x) is continuous at x = 4, direct substitution gives sqrt(4) = 2. Options 4, 16, and 8 confuse the input, the square, and other values with the actual output.

The leading terms have equal degree, so the asymptote is y = 3/1 = 3.

Because the values change sign, IVT guarantees at least one c in (1, 3) with f(c) = 0.

Continuity requires all three: f(c) exists, the limit exists, and the two are equal.

Apply the power rule: bring down the exponent 3 and reduce it by one, giving 3x^(3-1) = 3x^2. The derivative can be verified because integrating 3x^2 returns x^3 plus a constant.

Use the power rule with the constant multiple: 5 x 4 x^(4-1) = 20x^3. The exponent decreases by one, and the coefficient becomes 20, matching option a.

The derivative measures the rate of change, and a constant function never changes no matter what x is. Therefore d/dx(7) = 0, while the other options mistake the constant value for a variable term.

Rewrite 1/x as x^-1; the derivative is -x^-2 = -1/x^2.

sqrt(x) = x^(1/2), so the derivative is (1/2)x^(-1/2) = 1/(2 sqrt(x)).

The exponential function e^x is unique because its derivative equals itself: d/dx e^x = e^x. Options x e^x and e^(x - 1) would require different rules that do not apply here.

For x > 0, the derivative of the natural logarithm ln(x) is 1/x. This follows from differentiating e^(ln x) = x with the chain rule, which gives 1/x as the result.

The derivative of sin(x) is cos(x), one of the standard trigonometric differentiation rules. Options sin(x) and -sin(x) confuse the function with its second derivative or with cosine.

The derivative of cos(x) is -sin(x), which follows from the standard differentiation formulas. Option sin(x) misses the negative sign, and cos(x) confuses the function with its derivative.

Apply the product rule: derivative of x^2 times sin(x) plus x^2 times derivative of sin(x), giving 2x sin(x) + x^2 cos(x). Option b changes the sign, and the others omit one product-rule term.

Quotient rule: ((x + 1) - x)/(x + 1)^2 = 1/(x + 1)^2.

Chain rule: 5(3x + 1)^4 x 3 = 15(3x + 1)^4.

Use the chain rule: differentiate the outer exponential and multiply by the derivative of the inner function 2x. This gives e^(2x) x 2 = 2 e^(2x), matching option a.

Use the chain rule: differentiate ln(2x) by taking 1/(2x) and multiplying by the derivative of 2x, which is 2. The product simplifies to 1/x, so option a is correct.

The derivative is 2x, so at x = 3 the slope is 2(3) = 6.

The slope is 4 and the point is (2, 4), so y - 4 = 4(x - 2), which simplifies to y = 4x - 4.

If f prime is positive on an interval, the slope of the tangent line is positive there, so function values rise as x increases. Thus f is increasing, not decreasing or constant.

A negative second derivative means the graph is concave down.

Critical numbers occur where the derivative is 0 or does not exist.

A sign change from positive to negative indicates a local maximum.

f prime = 3x^2 - 3 = 0 when x^2 = 1, so x = 1 and x = -1.

Implicit differentiation gives 2x + 2y dy/dx = 0, so dy/dx = -x/y = -3/4.

A = pi r^2, so dA/dt = 2 pi r dr/dt = 2 pi x 5 x 2 = 20 pi.

This is the derivative of e^x at 0, which is 1.

Product rule: 1 x ln(x) + x x 1/x = ln(x) + 1.

Differentiate the outer sine and multiply by the derivative of the inner 2x: cos(2x) x 2 = 2 cos(2x). Option b forgets the chain-rule factor, so option a is correct.

Product rule: 2x e^x + x^2 e^x = e^x(x^2 + 2x).

f prime = 2x - 4 = 0 at x = 2, and f double prime is positive, so it is a minimum.

The first derivative is 4x^3 and the second derivative is 12x^2.

The derivative of tan(x) is sec^2(x), a standard trigonometric differentiation result. Option b shows tan(x) itself, and sec(x) tan(x) is the derivative of sec(x), not tan(x).

Increase the power by 1 and divide by the new power: x^3 + C.

Integrate a constant by multiplying it by x and adding the constant of integration: integral of 2 dx = 2x + C. Option x^2 + C would be the integral of 2x, not of 2.

Increase the power: 4x^3 integrates to x^4 + C.

The exponential function e^x is its own antiderivative, so the integral of e^x dx is e^x + C. Option x e^x + C would require the product rule, and ln(x) is the antiderivative of 1/x.

The integral of 1/x is ln|x| + C, or ln(x) + C for x > 0.

The derivative of sin(x) is cos(x), so reversing the derivative shows that the integral of cos(x) dx is sin(x) + C. Option -sin(x) would be the integral of -cos(x), and tan(x) is different.

Because the derivative of -cos(x) is sin(x), the antiderivative of sin(x) is -cos(x) + C. Option cos(x) has derivative -sin(x), so its sign is wrong.

Recognize the standard arctangent form: the derivative of arctan(x) is 1/(1 + x^2), so its antiderivative is arctan(x) + C. The other options would require integrands such as 1/x or 1/sqrt(1 - x^2).

Let u = 2x + 1, du = 2 dx, so the integral becomes (1/2)(u^4/4) = (1/8)(2x + 1)^4 + C.

Find the antiderivative x^3/3, then evaluate from 0 to 1: (1^3)/3 - (0^3)/3 = 1/3. This definite integral produces a number rather than a family of functions.

Find the antiderivative x^2, then evaluate from 0 to 2: 2^2 - 0^2 = 4. Option 8 would come from evaluating 2x^2 incorrectly, so option a is correct.

[-cos(x)] from 0 to pi = -cos(pi) - (-cos(0)) = 1 + 1 = 2.

The derivative of an accumulation function with integrand t^2 is x^2.

Integrate 4 - x^2 to get 4x - x^3/3, then evaluate from 0 to 2: 8 - 8/3 = 16/3.

The average value is (1/(b - a)) times the definite integral from a to b.

Differentiability at c requires the derivative limit to exist, which forces the function to approach f(c) smoothly. Therefore differentiability implies continuity, although the reverse is not always true.

A constant function keeps the same output for every input, so its rate of change is always 0. Thus its derivative is 0, and the statement is true.

When the derivative is positive on an interval, each small increase in x produces an increase in f, so f is increasing there. The statement correctly describes the first derivative test.

Continuity does not guarantee differentiability; corners and cusps are examples.

Because e^x differentiates to itself, integrating it also returns e^x plus a constant. The statement e^x + C is therefore true and can be checked by differentiating the result.

The second derivative measures how the slope itself changes. When f double prime is positive the graph is concave up, and when it is negative the graph is concave down, so the statement is true.

As x grows large, 1/x approaches 0, not infinity.

At a differentiable local extremum, the tangent line is horizontal, so the derivative is 0.

Any function x^2 + C has derivative 2x; 2 alone has derivative 0.

The derivative of 2x and 2x + 1 is 2; the others are not.

sin(x)/x, x/x, and cos(x) at 0 all approach 1; x/x^2 approaches 0.

The derivative is 2x, the second derivative is 2, it increases for x > 0, and f(-x) = f(x).

Differentiate each proposed antiderivative: x^2 + C gives 2x, 3x + C gives 3, x^4/4 + C gives x^3, and -1/x + C gives 1/x^2. Since every one differentiates back to its integrand, all four are correct.

The derivative 3x^2 - 3 is zero at x = 1 and x = -1.

These are the standard differentiation rules used throughout AP Calculus.

The derivatives of e^x, ln x, sin x, and cos x are e^x, 1/x, cos x, and -sin x.

The standard limits are 1, 0, 0, and 4 respectively.

Match each standard antiderivative: e^x maps to e^x + C, 1/x maps to ln|x| + C, cos(x) maps to sin(x) + C, and sin(x) maps to -cos(x) + C. These four basic forms appear frequently on the AP exam.

Derivatives measure rates, integrals accumulate, limits describe approach, and continuity means no breaks.

Velocity is the first derivative of position, acceleration is the second, displacement is the integral of velocity, and area under a curve is a definite integral.