Find the cubic function f(x)=ax^3+bx^2+cx+d that has a local max value of 3 at -2 and a local min values of 0 at 1.
I have no idea how to solve this. Could you please give step by step answers and explanations for the steps. Thanks!
The local extrema correspond to zeroes of the derivative. That derivative is a wauadratic function. So, you should write down a quadratic function that has its zeroes at x = -2 and x = 1. Then integrate that function. This function is fixed up to an overall factor.
Then you integrate this function and fix the overall factor and the integration constant by demanding that the correct values are assumed at the local extrema.
so
f(1)=a+b+c+d=0
f(-2)=-8a+4b-2c+d=3
f'(1)=3a+2b+c=0
f'(-2)=12a-4b+c=0
Now do I solve the system of equations for a,b,c,d? But is it possible to solve for this system of equations with 4 unknowns? Can I only substitute derivative into derivative and original into original or does it not matter?
you can solve them anyway you want. I would use matrix algebra.
I would suggest you start with this:
f'(x) = C(x+2)(x-1)
Then what is the general form of f(x)?
Could you explain how to solve? How would you do matrix algebra here?
f'(x) = C(x+2)(x-1) =
C(x^2 + x - 2)
f(x) = C(1/3 x^3 + 1/2 x^2 - 2 x) + p
We need to satisfy the equations:
f(-2) = 3
f(1) = 0
So, we have:
10/3 C + p = 3
-7/6 C + p = 0
The solution is:
C = 2/3 and p = 7/9
To find the cubic function that satisfies the given conditions, you can use a process called differentiation. This involves finding the derivative of the cubic function, setting it equal to zero, and then solving for the values of the coefficients a, b, c, and d.
Step 1: The first step is to differentiate the function f(x) with respect to x.
f(x) = ax^3 + bx^2 + cx + d
To differentiate, you need to use the power rule for each term of the polynomial:
f'(x) = 3ax^2 + 2bx + c
Step 2: Next, you need to find the critical points of the function by setting the derivative equal to zero, since the local extremas occur at these points.
3ax^2 + 2bx + c = 0
Step 3: Now, you have two conditions: a local maximum at -2, with a value of 3, and a local minimum at 1, with a value of 0.
Substituting these values into the function and its derivative, you get:
f(-2) = -8a + 4b - 2c + d = 3
f'(1) = 3a + 2b + c = 0
Step 4: You now have two equations with four unknowns. To solve this system, one approach is to use substitution.
From f'(1) = 3a + 2b + c = 0, you can express c in terms of a and b:
c = -3a - 2b
Substitute this value into f(-2) = -8a + 4b - 2c + d = 3:
-8a + 4b - 2(-3a - 2b) + d = 3
Simplify and rearrange the equation:
-2a + 8b - d = 3
Step 5: At this point, you can choose any arbitrary value for a or b to solve for the other variables. Let's assume a = 1:
-2(1) + 8b - d = 3
-2 + 8b - d = 3
8b - d = 5
Now, let's assume b = 1:
8(1) - d = 5
8 - d = 5
d = 3
Step 6: With d = 3, you can now solve for b:
8b - 3 = 5
8b = 8
b = 1
Step 7: Finally, substitute the values obtained for b and d into the equation from Step 4 to solve for a:
-2a + 8(1) - 3 = 3
-2a + 8 - 3 = 3
-2a + 5 = 3
-2a = -2
a = 1
Therefore, the cubic function that satisfies the given conditions is:
f(x) = x^3 + x^2 - 3x + 3