Find a cubic function f(x)=ax^3+cx^2+d that has a local maximum value of 9 at -4 and a local minimum value of 6 at 0.
Find a, c, and d.
f'(-4) = 9
f'(0) = 6
f'(x) = 3ax^2 + 2cx
f'(-4) = 48a - 8c = 9
f'(0) = 0 + 0 = 6
Oh, oh, something fishy here.
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To find the cubic function that satisfies the given conditions, we need to use the information about the local maximum and minimum points.
First, let's start by finding the critical points of the function. Critical points occur where the derivative of the function is equal to zero or undefined. Since we are dealing with a cubic function, we need to find its derivative to find the critical points.
The derivative of f(x) is given by f'(x) = 3ax^2 + 2cx + d.
Now, let's solve for the critical points by setting the derivative equal to zero and solving for x:
3ax^2 + 2cx + d = 0.
Since we know that the function has a local maximum at x = -4 and a local minimum at x = 0, we can set up two equations using these points.
For the local maximum at x = -4:
f'(-4) = 0. Substitute x = -4 into the derivative equation:
3a(-4)^2 + 2c(-4) + d = 0.
Simplifying the equation gives:
48a - 8c + d = 0. ----(1)
For the local minimum at x = 0:
f'(0) = 0. Substitute x = 0 into the derivative equation:
3a(0)^2 + 2c(0) + d = 0.
Simplifying the equation gives:
d = 0. ----(2)
Now, we have two equations (1) and (2) that involve the variables a, c, and d. Substituting the value of d from equation (2) into equation (1), we get:
48a - 8c = 0.
Next, we need to use the fact that the function has a local maximum value of 9 at x = -4 and a local minimum value of 6 at x = 0.
Substituting these values into the cubic function, we get:
f(-4) = 9:
a(-4)^3 + c(-4)^2 + d = 9.
-64a + 16c + d = 9. ----(3)
f(0) = 6:
a(0)^3 + c(0)^2 + d = 6.
d = 6. ----(4)
Using equation (4), we can substitute the value of d into equation (3) to eliminate d:
-64a + 16c + 6 = 9.
-64a + 16c = 3. ----(5)
We now have two equations, (5) and (2) (d = 6), with two unknowns a and c. Solving these equations simultaneously will give us the values of a and c:
-64a + 16c = 3. ----(5)
16c = 3.
Simplifying equation (5) gives:
-64a + 16(3) = 3.
-64a + 48 = 3.
-64a = -45.
a = -45/-64.
a = 45/64.
Substituting the value of a into equation (5) to solve for c:
-64(45/64) + 16c = 3.
-45 + 16c = 3.
16c = 3 + 45.
16c = 48.
c = 48/16.
c = 3.
Therefore, the values of a, c, and d are:
a = 45/64,
c = 3, and
d = 6.