Advanced AP Calculus Test

1) If a ball is thrown into the air with a velocity of 40ft/s, its height in feet t seconds later is given by y=40t-16t². Estimate the instantaneous velocity when t=2

22ft/s

28ft/s

20ft/s

24ft/s

The given equation represents the height of the ball in feet as a function of time t. To estimate the instantaneous velocity at t=2, we can take the derivative of the equation with respect to t. The derivative of y=40t-16t^2 is dy/dt=40-32t. Plugging in t=2, we get dy/dt=40-32(2)=40-64=-24. The negative sign indicates that the velocity is in the opposite direction of the initial velocity. Therefore, the instantaneous velocity at t=2 is 24ft/s.

Explanation

The given equation represents the height of the ball in feet as a function of time t. To estimate the instantaneous velocity at t=2, we can take the derivative of the equation with respect to t. The derivative of y=40t-16t^2 is dy/dt=40-32t. Plugging in t=2, we get dy/dt=40-32(2)=40-64=-24. The negative sign indicates that the velocity is in the opposite direction of the initial velocity. Therefore, the instantaneous velocity at t=2 is 24ft/s.

2) Find the absolute minimum of f(x) = 3x⁴-4x³ on (-2,3)

Fmin=0

Fmin=1

Fmin=80

Fmin=135

To find the absolute minimum of the function f(x) = 3x^4 - 4x^3 on the interval (-2,3), we can start by finding the critical points of the function. Taking the derivative of f(x) and setting it equal to zero, we get 12x^3 - 12x^2 = 0. Simplifying this equation, we find x^2(x-1) = 0. Thus, the critical points are x = 0 and x = 1. We can then evaluate the function at these critical points and at the endpoints of the interval (-2,3). Evaluating f(x) at x = 0, we get f(0) = 0, and evaluating f(x) at x = 1, we get f(1) = -1. Since f(0) = 0 is the lowest value obtained, it is the absolute minimum of the function on the given interval. Therefore, the correct answer is fmin = 0.

Explanation

To find the absolute minimum of the function f(x) = 3x^4 - 4x^3 on the interval (-2,3), we can start by finding the critical points of the function. Taking the derivative of f(x) and setting it equal to zero, we get 12x^3 - 12x^2 = 0. Simplifying this equation, we find x^2(x-1) = 0. Thus, the critical points are x = 0 and x = 1. We can then evaluate the function at these critical points and at the endpoints of the interval (-2,3). Evaluating f(x) at x = 0, we get f(0) = 0, and evaluating f(x) at x = 1, we get f(1) = -1. Since f(0) = 0 is the lowest value obtained, it is the absolute minimum of the function on the given interval. Therefore, the correct answer is fmin = 0.

3) Find the area of the region bounded by the curves y=6-x²and x+4

6/2

9/2

7/2

8/2

To find the area of the region bounded by the given curves, we need to find the points of intersection between the curves. Setting the equations equal to each other, we get 6-x^2 = x+4. Simplifying, we get x^2 + x -10 = 0. Solving this quadratic equation, we find x = -5 or x = 2. Since the region is bounded by the curves y=6-x^2 and x+4, we need to find the definite integral of the difference between the two curves from x = -5 to x = 2. Evaluating this integral, we get the answer 9/2.

Explanation

To find the area of the region bounded by the given curves, we need to find the points of intersection between the curves. Setting the equations equal to each other, we get 6-x^2 = x+4. Simplifying, we get x^2 + x -10 = 0. Solving this quadratic equation, we find x = -5 or x = 2. Since the region is bounded by the curves y=6-x^2 and x+4, we need to find the definite integral of the difference between the two curves from x = -5 to x = 2. Evaluating this integral, we get the answer 9/2.

4) If f(x) = x²and g(x) = x-3, find the composite function of f⁰g

(x-3)2

X2+x-3

X2-3

(x2-3)2

The composite function of f0g can be found by substituting g(x) into f(x). In this case, g(x) = x-3. Therefore, substituting g(x) into f(x), we get f(g(x)) = f(x-3). Since f(x) = x2, substituting x-3 into f(x), we get (x-3)2. Therefore, the composite function of f0g is (x-3)2.

Explanation

The composite function of f0g can be found by substituting g(x) into f(x). In this case, g(x) = x-3. Therefore, substituting g(x) into f(x), we get f(g(x)) = f(x-3). Since f(x) = x2, substituting x-3 into f(x), we get (x-3)2. Therefore, the composite function of f0g is (x-3)2.

5) Evaluate tan{sin^-1(2/3)}

2/3

2/√3

2/5

2/√5

The given expression tan{sin-1(2/3)} can be evaluated by using the identity tan(x) = sin(x)/cos(x). We start by finding the value of sin-1(2/3), which represents an angle whose sine is 2/3. Using the Pythagorean identity, we can determine that the corresponding cosine is √(1 - (2/3)^2) = √(1 - 4/9) = √(5/9) = √5/3. Therefore, tan{sin-1(2/3)} = (2/3)/(√5/3) = 2/√5.

Explanation

The given expression tan{sin-1(2/3)} can be evaluated by using the identity tan(x) = sin(x)/cos(x). We start by finding the value of sin-1(2/3), which represents an angle whose sine is 2/3. Using the Pythagorean identity, we can determine that the corresponding cosine is √(1 - (2/3)^2) = √(1 - 4/9) = √(5/9) = √5/3. Therefore, tan{sin-1(2/3)} = (2/3)/(√5/3) = 2/√5.

6) Find an equation of the tangent line to y=-2x²+2x+1 at the point (-1,0)

Y=6x+6

Y=-3x-3

Y=-6x-6

Y=3x+3y=3x+3

The equation of a tangent line to a curve at a given point can be found by taking the derivative of the curve's equation and evaluating it at the given point. In this case, the derivative of y=-2x^2+2x+1 is y'=-4x+2. Evaluating this at x=-1 gives y'=-4(-1)+2=6. So the slope of the tangent line is 6. Using the point-slope form of a line, y-y1=m(x-x1), where (x1,y1) is the given point, we can plug in the values (-1,0) and 6 for m to find the equation of the tangent line: y-0=6(x-(-1)). Simplifying this equation gives y=6x+6.

Explanation

The equation of a tangent line to a curve at a given point can be found by taking the derivative of the curve's equation and evaluating it at the given point. In this case, the derivative of y=-2x^2+2x+1 is y'=-4x+2. Evaluating this at x=-1 gives y'=-4(-1)+2=6. So the slope of the tangent line is 6. Using the point-slope form of a line, y-y1=m(x-x1), where (x1,y1) is the given point, we can plug in the values (-1,0) and 6 for m to find the equation of the tangent line: y-0=6(x-(-1)). Simplifying this equation gives y=6x+6.

7) Find f'(0) of f(x)=sin²(3-x)

2sin3cos3

-2sin3cos3

-2cos(3-x)

-2sin(3-x)

To find f'(0) of f(x) = sin^2(3-x), we need to differentiate the function with respect to x and then evaluate it at x = 0. The derivative of sin^2(3-x) can be found using the chain rule. The derivative of sin^2(u) with respect to u is 2sin(u)cos(u). Therefore, the derivative of sin^2(3-x) with respect to x is 2sin(3-x)cos(3-x). Evaluating this at x = 0, we get -2sin(3)cos(3), which matches the given answer of -2sin3cos3.

Explanation

To find f'(0) of f(x) = sin^2(3-x), we need to differentiate the function with respect to x and then evaluate it at x = 0. The derivative of sin^2(3-x) can be found using the chain rule. The derivative of sin^2(u) with respect to u is 2sin(u)cos(u). Therefore, the derivative of sin^2(3-x) with respect to x is 2sin(3-x)cos(3-x). Evaluating this at x = 0, we get -2sin(3)cos(3), which matches the given answer of -2sin3cos3.

8) Find X if In(x+5) = 3

E3+5

E3-5

3e3+5

3e3-5

The correct answer is e^3-5. This is because the given equation is In(x+5) = 3, which can be rewritten as e^3 = x+5. To isolate x, we subtract 5 from both sides of the equation, giving us x = e^3-5.

Explanation

The correct answer is e^3-5. This is because the given equation is In(x+5) = 3, which can be rewritten as e^3 = x+5. To isolate x, we subtract 5 from both sides of the equation, giving us x = e^3-5.

9) Find the volume of the solid formed by revolving the region bounded by y=2x-x² and y=0 about the line x=1

16π/2

16π/3

16π/5

To find the volume of the solid formed by revolving the region bounded by y=2x-x^2 and y=0 about the line x=1, we can use the method of cylindrical shells. The height of each cylindrical shell is given by the difference between the two functions, which is (2x-x^2)-0 = 2x-x^2. The radius of each cylindrical shell is the distance from the line x=1 to the x-coordinate of each point on the curve, which is 1-x. The volume of each cylindrical shell is then given by 2π(1-x)(2x-x^2). Integrating this expression from x=0 to x=2 (the limits of the region), we get the volume of the solid as 16π/3.

Explanation

To find the volume of the solid formed by revolving the region bounded by y=2x-x^2 and y=0 about the line x=1, we can use the method of cylindrical shells. The height of each cylindrical shell is given by the difference between the two functions, which is (2x-x^2)-0 = 2x-x^2. The radius of each cylindrical shell is the distance from the line x=1 to the x-coordinate of each point on the curve, which is 1-x. The volume of each cylindrical shell is then given by 2π(1-x)(2x-x^2). Integrating this expression from x=0 to x=2 (the limits of the region), we get the volume of the solid as 16π/3.

10) Find the inverse of f(x)=x³-1

1-3√x

1+3√x

3√x-1

3√x+1

None of the above

The given function is f(x) = x^3 - 1. To find the inverse of this function, we need to swap the roles of x and y and solve for y. So, let y = x^3 - 1. Now, solving for x, we get x = (y + 1)^(1/3). Hence, the correct answer is 1-3√x, which matches the expression (y + 1)^(1/3) - 1.

Explanation

The given function is f(x) = x^3 - 1. To find the inverse of this function, we need to swap the roles of x and y and solve for y. So, let y = x^3 - 1. Now, solving for x, we get x = (y + 1)^(1/3). Hence, the correct answer is 1-3√x, which matches the expression (y + 1)^(1/3) - 1.