A cubic polynomial function f is defined by

f(x)= 4x^3+ab^2+bx+k

where a, b, k, are constants. The function f has a local minimum at x=-2

A. Fine the vales of a and b

B. If you integrate f(x) dx =32 from o to 1, what is the value of K?

I assume you typed the question wrong and meant f(x)= 4x^3 +ax^2 +bx+ k

The derivative f'(x) must be zero at x = -2, so
12x^2 + 2ax + b = 0 at x=2
48 -4a +b = 0
You need more information to be able to calculate both a and b.

To find the values of a and b, we need to make use of the fact that the function f has a local minimum at x = -2.

1. Local Minimum Condition: For a local minimum to occur at x = -2, the derivative of the function f must be zero at that point, and the second derivative must be positive.

a. Finding the Derivative:
Take the derivative of f(x) with respect to x by applying the power rule of differentiation:

f'(x) = 12x^2 + 2abx + b

b. Evaluating at x = -2:
Set x = -2 in the derivative equation and set it equal to zero:

0 = 12(-2)^2 + 2ab(-2) + b

Simplify and solve for b:

0 = 48 - 4ab + b

2. Second Derivative Test: To ensure that the local minimum occurs at x = -2, we need to check whether the second derivative is positive at x = -2.

a. Finding the Second Derivative:
Take the derivative of f'(x) with respect to x:

f''(x) = 24x + 2ab

b. Evaluating at x = -2:
Set x = -2 in the second derivative equation:

f''(-2) = 24(-2) + 2ab

3. Solving for a and b:
Since the local minimum occurs at x = -2, both the first and second derivatives evaluated at x = -2 must be zero.

a. First Derivative:
Substitute b = 48 - 4ab into the first derivative equation:

0 = 12(-2)^2 + 2ab(-2) + (48 - 4ab)

Simplify and solve for a:

0 = 48 - 4ab - 4ab + 48 - 4ab
0 = 96 - 12ab

Divide both sides by -12:

12ab - 96 = 0
12ab = 96
ab = 8

b. Second Derivative:
Substitute b = 48 - 4ab into the second derivative equation:

f''(-2) = 24(-2) + 2ab
0 = -48 + 2ab

Substitute ab = 8:

0 = -48 + 2(8)
0 = -48 + 16
0 = -32

Since the second derivative is not positive, it violates the condition for a local minimum at x = -2. This means that there is an error in the given information or calculations.

Moving on to the second question:

To find the value of k, we need to integrate f(x) with respect to x from 0 to 1, and set it equal to 32.