Calculus and Vectors Challenge

Perform the integration by substitution: ∫(2x)(x^2 + 1)dx

• (1/3)(x^2 + 1)^3 + C

• (x^2 + 1) + C

• (1/4)(x^2 + 1)^4 + C

• (1/2)(x^2 + 1)^2 + C

Find the critical points of the function f(x) = x^3 - 6x^2 + 9x.

• x = 4

• x = 0

• x = 2

• x = 1, x = 3

Determine the local minima of the function f(x) = x^4 - 4x^3 + 4x.

• Local minima occur at x ≈ 0.2679 and x = 3.

• Local minima occur at x = 1 and x = 2.

• Local minima occur at x = -1 and x = 4.

• Local minima occur at x = 0 and x = 5.

Determine the local maxima of the function f(x) = -x^2 + 4x - 3.

• (2, 1)

• (1, 0)

• (3, 0)

• (0, -3)

Calculate the vector components of the vector A = 5i + 3j.

• A_x = 3, A_y = 5

• A_x = 0, A_y = 8

• A_x = 5, A_y = 0

• A_x = 5, A_y = 3

Add the vectors A = 2i + 4j and B = 3i - 2j.

• 4i + 6j

• 2i + 2j

• 5i - 2j

• 5i + 2j

Subtract the vector B = 1i + 2j from A = 4i - 3j.

• 4i + 2j

• 5i - 1j

• 2i + 5j

• 3i - 5j

Perform the integration by substitution: ∫(3x^2)(x^3 + 1)^4 dx

• (1/5)(x^3 + 1)^5 + C

• (1/3)(x^3 + 1)^3 + C

• (1/6)(x^3 + 1)^6 + C

• (1/4)(x^3 + 1)^4 + C

Find the critical points of the function f(x) = 2x^2 - 8x + 5.

• x = 2

• x = 3

• x = 0

• x = 1

Determine the local minima of the function f(x) = x^2 - 4x + 4.

• The local minima is at (2, 0).

• The local minima is at (0, 4).

• The local minima is at (1, 1).

• The local minima is at (4, 0).