find the cubic function,f(x)=ax^3+bx^2+cx^d , has local maximum value of 3, at x=-2 and local minimum value of 0, at x=1.

f' = 3ax^2 + 2bx + c

f'=0 at x = -2
f'=0 at x = 1
sum of roots = -1
product of roots = -2
so, -2b/3a = 1, so 2b = -3a
c/3a = -2, so c = -6a

f(x) = ax^2 - 3a/2 x - 6ax + d

f(-2) = 3 and f(1) = 0, so

4a + 3a + 12a + d = 3
a - 3a/2 - 6a + d = 0

a=2/9 d=7/9, so

f(x) = 2/9 x^3 + 1/3 x^2 - 4/3 x + 7/9

= 2/9 (x^2 + 3x^2 - 12x + 7)

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or, since we know we have a double root at x=1,

f(x) = a(x-k)(x-1)^2
f'(x) = a(x-1)^2 + 2a(x-k)(x-1)
= a(x-1)(x-1 + 2x-2k)
= a(x-1)(3x-2k-1)
we know there's a max at x = -2, so k-7/2, and

f(x) = a(x+7/2)(x-1)^2
Since f(-2) = 3,
a(3/2)(9) = 3
a = 2/9
f(x) = 2/9 (a+7/2)(x-1)^2
= 2/9 x^2 + 1/3 x^2 - 4/3 x + 7/9

see the graph and analysis at

http://www.wolframalpha.com/input/?i=2%2F9+%28x%2B7%2F2%29%28x-1%29^2

To find the cubic function with the given local maximum and minimum values, we can use the information provided about the values of f(x) at certain points. We know that the local maximum value of the function is 3 at x = -2, and the local minimum value is 0 at x = 1.

To find the cubic function, we need to determine the values of a, b, c, and d.

Step 1: Local maximum
At x = -2, the function has a local maximum value of 3. This means that the derivative of the function at x = -2 is equal to zero, since the slope of the function changes from positive to negative at this point.

Let's differentiate the function f(x) with respect to x:
f'(x) = 3ax^2 + 2bx + c

To find the value of a, b, and c, we need to substitute the value of x = -2 into the derivative equation and set it equal to zero.

f'(-2) = 3a(-2)^2 + 2b(-2) + c = 0
12a - 4b + c = 0. Equation (1)

Step 2: Local minimum
At x = 1, the function has a local minimum value of 0. This means that the derivative of the function at x = 1 is equal to zero, since the slope of the function changes from negative to positive at this point.

We differentiate the function f(x) with respect to x again:
f'(x) = 3ax^2 + 2bx + c

Now, substitute the value of x = 1 into the derivative equation and set it equal to zero.

f'(1) = 3a(1)^2 + 2b(1) + c = 0
3a + 2b + c = 0. Equation (2)

Step 3: Solve the equations
We have two equations with three unknowns (a, b, c). This suggests that these equations may not have a unique solution. However, since the equation represents a cubic function, we can introduce another equation by considering the behavior of the function for x → ± ∞.

In this case, we know that a cubic function approaches negative infinity as x → -∞ and approaches positive infinity as x → +∞. This implies that the leading coefficient (a) must be positive.

For simplicity, we can assume d = 3, as it is a common exponent for cubic functions.

Step 4: Substitute d = 3 into Equations (1) and (2):

12a - 4b + c = 0 (equation 1)
3a + 2b + c = 0 (equation 2)

Step 5: Solve the system of equations

Subtract equation 2 from equation 1 to eliminate c:

12a - 4b - (3a + 2b) = 0
9a - 6b = 0
a - (2/3)b = 0

Simplifying, we get:
a = (2/3)b (equation 3)

Step 6: Determine the values of a, b, and c.

Since a cubic function has infinitely many possible solutions, we need to make an arbitrary choice for one of the variables. To simplify the equation, let's assume a constant value for b, say b = 3.

Substituting b = 3 into equation 3, we find:
a = (2/3)(3) = 2

Now we can substitute the values of a and b into equation 2 to find c:

3(2) + 2(3) + c = 0
6 + 6 + c = 0
c = -12

Thus, we have determined the values of a, b, and c as a = 2, b = 3, and c = -12.

Therefore, the cubic function f(x) with a local maximum of 3 at x = -2 and a local minimum of 0 at x = 1 can be represented as:

f(x) = 2x^3 + 3x^2 - 12x