AP Calculus AB Semester 1 Exam Review 2024

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Questions and Answers

Estimate the limit, if it exists: lim f(x) where f(x) is represented by the given x->3 graph:

-1

The limit does not exist. (correct)

For what value of k is the function f(x) = {2x² + 5x - 3, x≠-3 k, x=-3 continuous at x = -3?

7/6

5/6 (correct)

-7/6

Find the derivative of the function y = 12/x³

-12/x⁴ (correct)

12/x²

12/x⁴

-12/x²

Find the derivative of the function y = 4x/(3x+2)

12/(3x + 2)² Signup and view all the answers

Find the derivative of y = e^x/cos(x)

e^x (cos x + sin x)/ cos²x Signup and view all the answers

Find the derivative of the function y = 3e^x sec x

3e^x sec x(1 + tan x) Signup and view all the answers

Find the derivative of the function y = √(x² - 2x)

(x - 1)/2√(x² - 2x) Signup and view all the answers

Find dy/ dx if 3xy = 4x + y²

(4 - 3y)/(2y - 3x) Signup and view all the answers

Find dy/dx for y = 4 sin²(3x)

24 sin(3x) cos(3x) Signup and view all the answers

If f(2) = -3, f’(2) = 3/4, and g(x) = f⁻¹(x), what is the equation of the tangent line to g(x) at x = -3?

y - 2 = 4/3(x + 3) Signup and view all the answers

If y = tan⁻¹(x² + 3x), then dy/dx =

(2x + 3)/(1 + (x² + 3x)²) Signup and view all the answers

If g(x) = f⁻¹(x), f(2) = 1, f(3) = 4, and f’(3) = 1/5, then which of the following must be true?

g’(4) = 5 Signup and view all the answers

Find the derivative of the following function: f(x) = sin²(x²)

2x sin(x²) cos(x²) Signup and view all the answers

Find the derivative of the following function: f(x) = 2 tan⁻¹(√x)

1/(2√x(1 + x)) Signup and view all the answers

Lim x-> 0 sin 4x / x² + 8x

0 Signup and view all the answers

When the height of a cylinder is 12 cm and the radius is 4 cm, the circumference of the cylinder is increasing at a rate of π/4 cm/min, and the height of the cylinder is increasing four times faster than the radius. How fast is the volume of the cylinder changing?

20π cm³/min Signup and view all the answers

Which of the following is false?

The speed of the particle at t = 7 is 2 Signup and view all the answers

On which of the following intervals is the particle speeding up?

(3, 6) Signup and view all the answers

The function f(x) is twice differentiable with f(2) = 6, f’(2) = 3 and f” (2) = 15. Using the tangent line to the graph of f(x) at x = 2, approximate f(1.9)

5.7 Signup and view all the answers

For the function f(x) = 12x⁵ - 5x⁴, how many of the inflection points of the function are also extrema?

None Signup and view all the answers

For time 0 ≤ t ≤ 10, a particle moves along the x-axis with position given by x(t) = t³ - 7t² + 8t + 5. During what time intervals is the speed of the particle increasing?

2/3 < t ≤ 7/3 and 4 < t < 10 Signup and view all the answers

Consider the curve defined by 3x² - xy + 2y = -36. Show that dy/dx = (y - 6x)/(2 - x)

Differentiating both sides of the equation 3x² - xy + 2y = -36 with respect to x, treating y as a function of x gives us 6x - y - xy’ + 2y’ = 0. Rearranging and factoring out y’ gives us y’(2 - x) = (y - 6x), and then dividing both sides by (2-x) we obtain dy/dx = (y - 6x)/(2 - x). Signup and view all the answers

Consider the curve defined by 3x² - xy + 2y = -36. Show that there are no points where the line tangent to the curve is vertical.

A vertical tangent line occurs when the derivative dy/dx is undefined. The derivative dy/dx = (y - 6x)/(2 - x) is undefined when the denominator (2 - x) is equal to 0. Simplifying, we get x = 2. To see if there are any points on the curve for x = 2, we can substitute x = 2 into the original equation and solve for y: 3(2)² - 2y + 2y = -36. This simplifies to 12 = -36, which is not a valid equation. Therefore, there are no points on the curve where x = 2, meaning that there are no points where the line tangent to the curve is vertical. Signup and view all the answers

Consider the curve defined by 3x² - xy + 2y = -36. Find the two points where the curve has a horizontal tangent line.

A line is horizontal when its slope is zero, in other words when dy/dx = 0. Using the derivative we calculated earlier: dy/dx = (y - 6x)/(2 - x), we can find the points where this derivative is equal to zero. A fraction is equal to zero when the numerator is equal to zero. So we want to find the points where y - 6x = 0. We’ll rearrange this to get y = 6x and then plug this y value into our original equation 3x² - xy + 2y = -36: 3x² - x(6x) + 2(6x) = -36. Simplifying this equation we get 3x² - 6x² + 12x + 36 = 0, and then combine like terms to get -3x² + 12x + 36 = 0. Divide both sides by -3, and we obtain x² - 4x - 12 = 0. Factor this equation into (x - 6)(x +2) = 0, so we see that x = 6 or x = -2. We can plug these values back into the equation y = 6x to obtain the corresponding y values. When x = 6, we have y = 6(6) = 36 and when x = -2, we have y = 6(-2) = -12. Therefore, there are two points where the curve has a horizontal tangent line: (6, 36) and (-2, -12) . Signup and view all the answers

Consider the curve defined by 3x² - xy + 2y = -36. Find the value of d²y/dx² at each of the points found in Part C. Determine if the curve has a relative minimum, relative maximum, or neither at each of the points. Explain your reasoning.

We know that dy/dx = (y - 6x)/(2 - x). We can differentiate once more to find: d²y/dx² = [ (2-x)(dy/dx - 6) - (y - 6x)(-1) ]/(2 - x)² Simplifying this equation we get: [d²y/dx² = 2dy/dx - 2xy/dx + 12 + y - 6x ]/(2 - x)² Substituting the equations for dy/dx and y in terms of x we get: [ (2 - x) * ((y - 6x)/(2 - x)) - (2x(y - 6x))/(2 - x) + 12 + (6x) - 6x ]/(2 - x)² Simplifying this equation we get: (2y - 12x - 2xy + 12)/(2 - x)² Substituting the equation for y in terms of x we get: [2(6x) - 12x - 2x(6x) + 12]/(2-x)² Simplifying: (12)/(2 - x)² We can plug in the values we calculated earlier to get the second derivative at both points: At x = 6 we obtain d²y/dx² = (12)/(2-6)² = 12/-16 = -3/4 At x = -2 we obtain d²y/dx² = (12)/(2 - (-2))² = 12/16 = 3/4 Using the second derivative test, we know that if the second derivative is positive at a critical point, then the function has a local minimum. And, if the second derivative is negative at a critical point, then the function has a local maximum. So, at the point (6, 36), the curve has a local maximum since the second derivative is negative. At the point (-2, -12), the curve has a local minimum since the second derivative is positive. Signup and view all the answers

Consider the curve defined by 3x² - xy + 2y = -36. Write an equation of the line tangent to the curve when x = 0.

When x = 0, we can plug this into the original equation to find the corresponding y value. We obtain: 3(0)² - (0)y + 2y = -36 This simplifies to 2y = -36, or y = -18. So the point on the curve where x = 0 is (0, -18). Now, we need to find the slope of the tangent line at this point. We know that dy/dx = (y - 6x)/(2 - x). Plugging in x = 0 and y = -18, we obtain: dy/dx = (-18 - 6(0))/(2 - 0) = -18/2 = -9. So, the slope of the tangent line at (0, -18) is -9. Finally, we can use point-slope form to find the equation of the tangent line: y - (-18) = -9(x - 0) Simplifying, we get: y + 18 = -9x Subtracting 18 from both sides we get: y = -9x - 18 Signup and view all the answers

Study Notes